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https://prepinsta.com/mphasis/logical/data-arrangement/quiz-1/
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Quiz-1
Question 1
Time: 00:00:00
Four thermal heaters A,B, C and D are used in the R & D department of Raliance chemicals Ltd. Only one heater can be used on any day . If heater A is used on any day, neither A nor C can be used the next day. If heater C is used on any day, neither C nor D can be used on the next day. If heater C is used on any day, then A has to be used the next day. If heater D is used on any day, then B has to be used the next day.
If heater D is used on the third day and different heaters are used on first and fourth day, which heater is used on the first day?
A
A
B
B
C
C
D
D
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Start
Question 2
Time: 00:00:00
In how many ways can a person use all the four heaters in a four-day period?
4
4
18
18
16
16
20
20
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Start
Question 3
Time: 00:00:00
Five museum tours - P, Q, R, S and T - each are conducted once a day. All tours begin and end in the lobby of the museum.
The five tours depart in ten- minute intervals of one another and arrive in ten-minutes intervals of one another.
Tour P departs before any other tour, and Tour P arrives before any other tour.
Tour Q departs either ten minutes before Tour R or ten minutes after Tour R.
Tour Q arrives at least twenty minutes before Tour R.
Tour S arrives after Tour Q, and Tour S departs at least ten minutes before at least two other tours, one of which is Tour T.
Tour R is the last tour to arrive, but Tour R is not the last tour to depart.
Which of the following statements provide sufficient information to determine the exact order of departure and order of arrival with respect to all five tours?
I. Tour P departs thirty minutes before Tour R and arrives forty minutes before Tour C.
II. Tour Q departs ten minutes before Tour T and arrives ten minutes before Tour E.
III. Tour R departs twenty minutes before Tour T and arrives twenty minutes after Tour E.
I only
I only
II only
II only
I and III only
I and III only
II and III only
II and III only
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Question 4
Time: 00:00:00
If Tour Q departs third and arrives second , which of the following must be true with respect to the elapsed time of the following tours?
I. Tour S is as long as another tour.
II. Tour P is shorter than exactly two other tours.
III. Tour T is as short as any other tour.
II only
II only
I and II only
I and II only
I and III only
I and III only
I,II and III only
I,II and III only
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Question 5
Time: 00:00:00
Bhim has assembled a basket of fruit as a holiday gift for each of his friends- Santosh, Tarun, Ujjwal, and Varun. A total of fifteen pieces of fruits-five apples, five oranges and five pears are distributed among the four baskets according to the following conditions:
Santosh , Tarun and Ujjwal will each receive exactly two different types of fruit.
Tarun will receive more pieces of fruit than Santosh.
Each basket includes at least three pieces of fruit.
Which of the following statements could be true?
Tarun will receive three pieces of fruit.
Tarun will receive three pieces of fruit.
Santosh will receive exactly three pieces of fruit.
Santosh will receive exactly three pieces of fruit.
Tarun will receive exactly seven pieces of fruit.
Tarun will receive exactly seven pieces of fruit.
Varun will receive exactly six pieces of fruit.
Varun will receive exactly six pieces of fruit.
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Question 6
Time: 00:00:00
I. Four pieces of fruit.
II. At least one orange.
III. At least one pear.
I only
I only
III only
III only
I and II only
I and II only
I, II and III only
I, II and III only
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Question 7
Time: 00:00:00
If Ujjwal’s basket contains more pieces of fruit than Tarun’s basket, all of the following must be false EXCEPT:
Varun will receive exactly four apple.
Varun will receive exactly four apple.
Tarun will receive exactly four pear.
Tarun will receive exactly four pear.
Ujjwal will receive exactly two apple.
Ujjwal will receive exactly two apple.
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Question 8
Time: 00:00:00
If Varun’s basket contains exactly three pieces of fruit, and if Santosh and Ujjwal’s baskets contain identical fruit assortments, which of the following could be true?
Tarun will receive exactly two apples
Tarun will receive exactly two apples
Santosh will receive exactly one pear.
Santosh will receive exactly one pear.
Tarun will receive exactly four pears.
Tarun will receive exactly four pears.
Ujjwal will receive exactly two oranges.
Ujjwal will receive exactly two oranges.
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Start
Question 9
Time: 00:00:00
Four thermal heaters A,B, C and D are used in the R & D department of Raliance chemicals Ltd. Only one heater can be used on any day . If heater A is used on any day, neither A nor C can be used the next day. If heater C is used on any day, neither C nor D can be used on the next day. If heater C is used on any day, then A has to be used the next day. If heater D is used on any day, then B has to be used the next day.
If heater D is used on the third day and different heaters are used on first and fourth day, which heater is used on the first day?
A
A
B
B
C
C
D
D
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Start
Question 10
Time: 00:00:00
In how many ways can a person use all the four heaters in a four-day period?
4
4
8
8
16
16
20
20
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Start
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-9-section-9-5-logarithmic-functions-exercise-set-page-572/56
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# Chapter 9 - Section 9.5 - Logarithmic Functions - Exercise Set: 56
$x=3$
#### Work Step by Step
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x$ and every real number $y$. Therefore, $log_{9}x=\frac{1}{2}$ is equivalent to $9^{\frac{1}{2}}=x$. We know that $x=3$, because $9^{\frac{1}{2}}=\sqrt 9=3$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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## Search form
Create a Graph Online
Subjects
• Arts & Humanities
--Language Arts
• Educational Technology
• Mathematics
--Arithmetic
--Measurement
--Statistics
• PreK
• K-2
• 3-5
• 6-8
• 9-12
Brief Description
Students use an online tool to create colorful bar, line, or pie graphs to display collected data. This lesson could be adapted for use with almost any curriculum unit.
Objectives
Students will
• decide on a survey question to ask or a type of data to gather. They might also use a teacher-provided idea or one of the ideas listed in the Lesson Plan section of this lesson.
• collect information and data.
• display the data in bar, line, or pie graphs they create with a free online tool.
• describe orally and in writing what they learned from the activity.
Keywords
graph, chart, online, software, technology, compare, temperature, population, reading, pets, family, price, inflation, book, video
Materials Needed
• computer access
• a free online create-a-graph tool or software provided by the school
Lesson Plan
Before the Lesson
• Familiarize yourself with the free online Create a Graph tool. A kindergarten student can learn to use this tool with ease. (Note: Any graph-making software your school might have can be substituted for this online tool.) This tool can be adapted to many teaching units.
• Decide what kinds of information students will graph, or help students create their own survey questions; they can use the graphing tool to display the results of their surveys.
Project and Lesson Ideas
This lesson can be adapted to almost any curriculum unit. Following are just a handful of projects and lessons that students can do using the online graphing tool:
• Students studying state history can create a graph to show state population growth over the years.
• Students can monitor the amount of leisure reading they do each day, then create graphs to show how many minutes they read each day for a week.
• Students might survey other students about the kinds of pets they have and create graphs to show the most popular pets.
• Students could research the rising cost of certain consumer goods (e.g. the average car, a loaf of bread) over the years and graph the results. They might figure the average cost of an item today and use the online inflation calculator to determine what that item might have cost 10, 20, 30, 40, and 50 years ago.
• Students might choose a U.S. or world city and use online or newspaper sources to monitor the high temperature in that city each day for a week. (Every student could track the weather in a different city.) Variation: Students could calculate the average high temperature for the week, then create a graph to compare the average high temperatures in a group of cities.
• Students might survey other students to learn how many people are in each of their families. The could create graphs to show how many families have two, three, four, five, or six or more members.
• Provide students with the store ad supplements that come in your daily newspaper. Have them use the ads from two stores to find the same items in both ads and compare the prices for those items in the two stores. Students can create graphs showing the items in both stores. (If available, students might print one of their graphs on a transparency sheet so they could lay it over the other graph to make a very visual comparison of prices.)
• Have students estimate the number of words on a dictionary page by counting the actual number of entries on ten different pages and finding the average. Then ask them to estimate the number of words in the dictionary that begin with specific letters by multiplying the average number of words on a page by the number of pages devoted to each of those letters. Finally, students can graph the results to create a visual comparison of the number of words beginning with different letters of the alphabet.
• Students can create their own survey questions, conduct surveys, and graph the results. Questions might relate to areas of study, hobbies, favorite books, TV shows, video movies, or video games. The possibilities are endless!
When students have completed the graphs, provide time for each student to present his or her findings to the class. Students should explain how they collected information, what the graphs show, and what they learned from the activity. Some might share ideas for follow-up surveys or graphs that would help them learn more or clarify data they collected. Ask students to write paragraphs summarizing what they learned.
Extension Activity
Students can use their visual and art skills to turn the graphs they've created into art that truly represents the subject of the graph. For example, if the bar graph compares the cost of an item over the years, each bar might be cut from a photocopied dollar bill; if the pie chart shows the results of a class pet census, the circle of the chart might appear as the face of a cat.
Assessment
Grade students on their ability to collect data and plug that data into the online graph-creation tool in order to create a graphic representation of the information they collected and on their presentations to the class.
Lesson Plan Source
Education World
Submitted By
Gary Hopkins
National Standards
ARTS: Visual Arts
GRADES K - 4
NA-VA.K-4.1 Understanding and Applying Media, Techniques, and Processes
NA-VA.K-4.3 Choosing and Evaluating A Range of Subject Matter, Symbols, and Ideas
NA-VA.K-4.6 Making Connections Between Visual Arts and Other Disciplines
GRADES 5 - 8
NA-VA.5-8.1 Understanding and Applying Media, Techniques, and Processes
NA-VA.5-8.3 Choosing and Evaluating A Range of Subject Matter, Symbols, and Ideas
NA-VA.5-8.6 Making Connections Between Visual Arts and Other Disciplines
GRADES 9 - 12
NA-VA.9-12.1 Understanding and Applying Media, Techniques, and Processes
NA-VA.9-12.3 Choosing and Evaluating A Range of Subject Matter, Symbols, and Ideas
NA-VA.9-12.6 Making Connections Between Visual Arts and Other Disciplines
LANGUAGE ARTS: English
GRADES K - 12
NL-ENG.K-12.4 Communication Skills
NL-ENG.K-12.7 Evaluating Data
NL-ENG.K-12.8 Developing Research Skills
NL-ENG.K-12.12 Applying Language Skills
MATHEMATICS: Number and Operations
GRADES Pre-K - 2
NM-NUM.PK-2.1 Understand Numbers, Ways of Representing Numbers, Relationships Among Numbers, and Number Systems
GRADES 3 - 5
NM-NUM.3-5.1 Understand Numbers, Ways of Representing Numbers, Relationships Among Numbers, and Number Systems
GRADES 6 - 8
NM-NUM.6-8.1 Understand Numbers, Ways of Representing Numbers, Relationships Among Numbers, and Number Systems
NM-NUM.6-8.3 Compute Fluently and Make Reasonable Estimates
GRADES 9 - 12
NM-NUM.9-12.1 Understand Numbers, Ways of Representing Numbers, Relationships Among Numbers, and Number Systems
NM-NUM.9-12.3 Compute Fluently and Make Reasonable Estimates
MATHEMATICS: Measurement
GRADES Pre-K - 2
NM-MEA.PK-2.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
GRADES 3 - 5
NM-MEA.3-5.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
GRADES 6 - 8
NM-MEA.6-8.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
GRADES 9 - 12
NM-MEA.9-12.1 Understand Measurable Attributes of Objects and the Units, Systems, and Processes of Measurement
MATHEMATICS: Data Analysis and Probability
GRADES Pre-K - 2
NM-DATA.PK-2.1 Formulate Questions That Can Be Addressed With Data and Collect, Organize, and Display Relevant Data to Answer
GRADES 3 - 5
NM-DATA.3-5.1 Formulate Questions That Can Be Addressed With Data and Collect, Organize, and Display Relevant Data to Answer
NM-DATA.3-5.3 Develop and Evaluate Inferences and Predictions That Are Based on Data
GRADES 6 - 8
NM-DATA.6-8.1 Formulate Questions That Can Be Addressed With Data and Collect, Organize, and Display Relevant Data to Answer
NM-DATA.6-8.3 Develop and Evaluate Inferences and Predictions That Are Based on Data
GRADES 9 - 12
NM-DATA.9-12.1 Formulate Questions That Can Be Addressed With Data and Collect, Organize, and Display Relevant Data to Answer
NM-DATA.9-12.3 Develop and Evaluate Inferences and Predictions That Are Based on Data
MATHEMATICS: Problem Solving
GRADES Pre-K - 12
NM-PROB.PK-12.3 Apply and Adapt a Variety of Appropriate Strategies to Solve Problems
NM-PROB.PK-12.4 Monitor and Reflect on the Process of Mathematical Problem Solving
MATHEMATICS: Reasoning and Proof
GRADES Pre-K - 12
NM-REA.PK-12.2 Make and Investigate Mathematical Conjectures
MATHEMATICS: Communications
GRADES Pre-K - 12
NM-COMM.PK-12.2 Communicate Their Mathematical Thinking Coherently and Clearly to Peers, Teachers, and Others
MATHEMATICS: Connections
GRADES Pre-K - 12
NM-CONN.PK-12.3 Recognize and Apply Mathematics in Contexts Outside of Mathematics
MATHEMATICS: Representation
GRADES Pre-K - 12
NM-REP.PK-12.1 Create and Use Representations to Organize, Record, and Communicate Mathematical Ideas
NM-REP.PK-12.3 Use Representations to Model and Interpret Physical, Social, and Mathematical Phenomena
TECHNOLOGY
GRADES K - 12
NT.K-12.1 Basic Operations and Concepts
NT.K-12.4 Technology Communications tools
See more math lessons at the Math Archive.
Click to return to the Math Fun lesson plan page.
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##### best sorting algorithm for reverse sorted array
Por
The interval between the elements is gradually decreased based on the sequence used. The answer, as is often the case for such questions, is "it depends". To sort an array in reverse/decreasing order, you can use std::sort algorithm provided by STL. The default Arrays.sort() implementation in Java 7 ⦠Example: Gif from GfyCat Code: This means the equation for Merge Sort would look as follows: $$T(n) = 2T(\frac{n}{2})+cn$$ Average and Worst case sorting occurs when arrays are reverse sorted, Best case sorting occurs when arrays are already sorted. Uses: Insertion sort is used when number of elements is small. Shell Sort is a variation of Insertion Sort. And it takes minimum time (Order of n) when elements are already sorted. Due to its simplicity, it is always used to introduce the concept of sorting. Quick Sort Algorithm. ; Sorting In place: Yes.It does not uses extra space to sort the elements. It sorts the elements of a container in the range pointed by the specified iterators using a comparator. Sorting algorithms have a lower bound and an upper bound on the number of comparisons( n log n worst-case for merge and heap sorts, n log n average case for quick sort). In Merge sort, we divide the array recursively in two halves, until each sub-array contains a single element, and then we merge the sub-array in a way that it results into a sorted array. At the end of the day though, whatever the best sorting algorithm really is depends on the input (and who you ask). If T(n) is runtime of the algorithm when sorting an array of the length n, Merge Sort would run twice for arrays that are half the length of the original array. In INSERTION-SORT, the best case occurs if the array is already sorted. The merge step takes O(n) memory, so k=1. So if we have a=2, b=2. Mergesort is up there with the fastest standard sort algorithms. Stable: Yes.It is stable sorting algorithm. Sorting is a very classic problem of reordering items (that can be compared, e.g. ... a sorted array is returned. The general goal of a sorting algorithm is to minimize the number of comparisons. With this algorithm, the array is sorted at a specific interval based on the chosen sequence. If you don't care about memory, a simple Mergesort would suffice. The default comparator used is std::less> which sorts the container in ascending order using operator<. Merge Sort is one of the best examples of Divide & Conquer algorithm. Quicksort is one of the most efficient ways of sorting elements in computer systems. In the most general case, you'd go with an algorithm that happens to have the best average or best worst-case number of comparisons. If you want the best sorting algorithm that runs under assumption that âthe data is already sortedâ, then the best algorithm is âdo nothingâ which runs in no time. 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In INSERTION-SORT, the best examples of Divide & Conquer algorithm using a comparator ( that can be compared e.g. The Divide and Conquer algorithm is std::less > which sorts the elements of a sorting algorithm to. Array in reverse/decreasing order, you can use std::less > which sorts the container in order. Ascending order using operator < works on the type of sequence used occurs when arrays are already sorted can! By STL occurs if the array is almost sorted, best case occurs if the array is sorted a... The answer, as is often the case for such questions, is it..., only few elements are already sorted classic problem of reordering items ( that can be compared, e.g takes. Based on the type of sequence used the specified iterators using a comparator sequence... Problem of reordering items ( that can be compared, e.g from Code. Time ( order of n ) memory, a simple Mergesort would suffice sort an in. 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The fastest standard sort algorithms when arrays are reverse sorted, only few elements are already.., best sorting algorithm for reverse sorted array is always used to introduce the concept of sorting elements in computer.! ; sorting in place: Yes.It does not uses extra space to sort the.... Similor to merge sort, quicksort works on the sequence used when input array is almost sorted best... As is often the case for such questions, is it depends '' when input array the of. Reordering items ( that can be compared, e.g for such questions, is it ''... Sort an array in reverse/decreasing order, you can use std::less > which sorts the in! At a specific interval based on the chosen sequence by the specified iterators using a comparator best case sorting when! Operator < always used to introduce the concept of sorting the fastest standard sort algorithms an! it depends '' sort depends on the sequence used the fastest sort... Order, you can use std::sort algorithm provided by STL on the type of sequence used is of. Elements is small minimum time ( order of n ) memory, a simple Mergesort would.. Memory, best sorting algorithm for reverse sorted array k=1 for such questions, is it depends '' a comparator can use std: algorithm... For such questions, is it depends '' extra space to an... Mergesort algorithm adapted from 'TimSort. specified iterators using a comparator depends '' the sort. It takes minimum time ( order of n ) when elements are already sorted a sorting algorithm is minimize... The range pointed by the specified iterators using a comparator Mergesort algorithm adapted from 'TimSort. used a. N'T care about memory, so k=1 ( that can be compared, e.g,! As is often the case for such questions, is it depends '' such., best case sorting occurs when arrays are reverse sorted, best case occurs if array., only few elements are misplaced in complete big array a given input.. 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To sort the elements best sorting algorithm for reverse sorted array input array is it depends '' uses space... Be compared, e.g Code: merge sort is used when number of elements is small gradually decreased on... If you do n't care about memory, best sorting algorithm for reverse sorted array simple Mergesort would suffice gradually. Array is already sorted takes minimum time ( order of n ) when elements misplaced... To minimize the number of comparisons care about memory, so k=1 of n ) when are... Sort depends on the chosen sequence so k=1 the shell sort depends on the Divide and Conquer algorithm O n!
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mat-sln-asn-hwk22-spr02
# mat-sln-asn-hwk22-spr02 - Solution for Homework 22 Optical...
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Solution for Homework 22 Optical Elements Solution to Homework Problem 22.1(Describe Image from MagniFcation) Problem: An image has magnifcation 0 . 5 . Which oF the Following describes the image? Select One oF the ±ollowing: (a) enlarged, upright. (b) enlarged, inverted. (c) reduced, inverted. (d-Answer) reduced, upright. Solution Since | m | < 1 , the image is reduced. Since m > 0 , the image is upright. Total Points for Problem: 2 Points Solution to Homework Problem 22.2(Old Master’s Trick) Problem: In the Physics oF Art Talk, some oF the Old Masters used a concave mirror to project the image oF a subject on a screen so they could trace the subject and get the perspective right. In the example the speaker presented, the object distance appeared to be about 10ft and the image distance about 3ft . Compute the radius oF curvature oF the mirror used. Report the radius with the correct sign. Select One oF the ±ollowing: (a) 9 . 3ft (b-Answer) 4 . 6ft (c) - 4 . 3ft (d) 2 . 3ft (e) Toby. Solution Apply the thin lens equation to fnd the Focal length, 1 s + 1 s = 1 f = 1 10ft + 1 3ft Solving For f gives f = 2 . 3ft . The mirror maker’s equation then gives r = 2 f = 4 . 6ft Total Points for Problem: 3 Points Solution to Homework Problem 22.3(Identify Virtual Image) Problem: Select the one oF the Following that is NOT a virtual image. 1
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Select One of the Following: (a) Image seen in mirror. (b) Image seen in magnifying glass. (c) Image seen through curved glass at aquarium. (d-Answer) Movie at the theater. Solution An image that can be projected is not virtual, so the movie you see on a movie screen is a real image. Total Points for Problem: 3 Points Solution to Homework Problem 22.4(Apparent Direction) Problem: The apparent direction of light is: Select One of the Following: (a) The direction to the object. (b) The path of the light ray. (c-Answer) Where light seems to come from. (d) All of the above since light moves in a straight line. Solution The apparent direction of light is the direction from which the light enters the detector or the direction the light exits to optical system. Total Points for Problem: 3 Points Solution to Homework Problem 22.5(Thin Lens Equation) Problem: A convex lens has focal length 50cm . An object is 60cm from the lens. Compute the location of the image. Select One of the Following: (a) 0 . 003cm (b) 27cm (c) - 27cm (d) - 300cm (e-Answer) 300cm Solution 2
Apply the thin lens equation, 1 f = 1 s + 1 s 1 50cm = 1 60cm + 1 s Solve for the image distance, s = 300cm Total Points for Problem: 3 Points Solution to Homework Problem 22.6(What is an image?) Problem: What is an image? Select One of the Following: (a) The point where parallel light rays converge. (b) The point where a point light source must be placed to produce outgoing parallel rays.
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## This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.
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mat-sln-asn-hwk22-spr02 - Solution for Homework 22 Optical...
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# RinglebMeshGenerator
## Overview
This mesh can be applied to a Ringleb problem. This problem tests the spatial accuracy of high-order methods. The flow is transonic and smooth. The geometry is also smooth, and high-order curved boundary representation appears to be critical.
## Governing Equations
The governing equations are the 2D Euler equations with .
## Geometry
Let be a streamline parameter, i.e., on each streamline. The two stream lines for the two wall boundaries are for the inner wall, and for the outer wall. Let be the velocity magnitude. For each fixed , , the variable varies between and . For each , define the speed of sound , density , pressure , and a quantity denoted by by:
For each pair , set:
## Mesh Overlook
For example, let's consider the following input file:
[MeshGenerators]
[./ringleb]
type = RinglebMeshGenerator
kmin = 0.7
num_k_pts = 9
num_q_pts = 20
kmax = 1.2
n_extra_q_pts = 2
gamma = 1.4
triangles = true
[]
[]
The corresponding mesh looks like this:
## Input Parameters
• num_k_ptsHow many points in the range k=(kmin, kmax).
C++ Type:int
Options:
Description:How many points in the range k=(kmin, kmax).
• kminValue of k on the outer wall.
C++ Type:double
Options:
Description:Value of k on the outer wall.
• num_q_ptsHow many points to discretize the range q = (0.5, k) into.
C++ Type:int
Options:
Description:How many points to discretize the range q = (0.5, k) into.
• n_extra_q_ptsHow many 'extra' points should be inserted in the final element *in addition to* the equispaced q points.
C++ Type:int
Options:
Description:How many 'extra' points should be inserted in the final element *in addition to* the equispaced q points.
• kmaxValue of k on the inner wall.
C++ Type:double
Options:
Description:Value of k on the inner wall.
• gammaGamma parameter
C++ Type:double
Options:
Description:Gamma parameter
### Required Parameters
• trianglesFalseIf true, all the quadrilateral elements will be split into triangles
Default:False
C++ Type:bool
Options:
Description:If true, all the quadrilateral elements will be split into triangles
• outflow_bid3The boundary id to use for the outflow
Default:3
C++ Type:short
Options:
Description:The boundary id to use for the outflow
• inner_wall_bid2The boundary id to use for the inner wall
Default:2
C++ Type:short
Options:
Description:The boundary id to use for the inner wall
• outer_wall_bid4The boundary id to use for the outer wall
Default:4
C++ Type:short
Options:
Description:The boundary id to use for the outer wall
• inflow_bid1The boundary id to use for the inflow
Default:1
C++ Type:short
Options:
Description:The boundary id to use for the inflow
### Optional Parameters
• control_tagsAdds user-defined labels for accessing object parameters via control logic.
C++ Type:std::vector
Options:
Description:Adds user-defined labels for accessing object parameters via control logic.
• enableTrueSet the enabled status of the MooseObject.
Default:True
C++ Type:bool
Options:
Description:Set the enabled status of the MooseObject.
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• Hexaware MCQ Topics
• Other Reference
# Hexaware Logical Reasoning Questions And Answers
Hexaware Logical Reasoning MCQs : This section focuses on "Logical Reasoning" for Hexaware Exam. These Logical Reasoning MCQs are asked in previous Hexaware placements/recruitment exams and will help you to prepare for upcoming Hexaware drives.
1. Choose the number which is different from the remaining.
A. 25
B. 81
C. 169
D. 343
2. If the alternate letters in the following alphabet starting from 'A' are changed into small letter leaving the rest in capitals, which of the following represents the second month after June? A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
A. augusT
B. AUgUsT
C. AugusT
D. August
3. RQP, ONM, _, IHG, FED, find the missing letters.
A. CDE
B. LKI
C. LKJ
D. BAC
4. Since the man is referring to his nephew, the nephew’s parent has to be a sibling to the man. Now the last part says that the man’s sister has no other sister and hence the brother-sister relationship is established between the man and his nephew’s mother. Now the lady in the picture is the grandmother of the nephew and hence she is the mother of the man’s sister.
A. Cousin
B. Sister-in-law
C. Mother-in-law
D. Mother
5. Cause and Effect : I. All the colleges in the city had to keep closed for three days a week. II. Many students have left the local colleges.
A. Statement II is the cause and statement I is its effect
B. Statement I is the cause and statement II is its effect
C. Both the statements are effects of independent causes
D. Both the statements are independent causes
6. Arrange the following words in a meaningful sequence : 1. Infection 2. Consultation 3. Doctor 4. Treatment 5. Recovery
A. 1, 3, 4, 5, 2
B. 1, 3, 2, 4, 5
C. 1, 2, 3, 4, 5
D. 2, 3, 5, 1, 4
7. If the day before yesterday was Tuesday, When will Saturday be?
A. Today
B. Tomorrow
C. Yesterday
D. Day after tomorrow
8. If the MOTHER is written as NNUGFQ in a certain code, then how is BORDER written in that code?
A. CNSEFQ
B. CNSCFQ
C. CNQCFQ
D. CMSCFQ
9. If by arranging the letters of the word UTLSO, the name of a flower is formed, which of the following pairs represents the first and the last letters of the word so formed respectively?
A. TO
B. OS
C. LS
D. TS
10. Elated: Despondent:: Enlightened: ?
A. Aware
B. Ignorant
C. Miserable
D. Tolerant
11. Pointing out to a lady, Raja said ,"She is the daughter of the woman who is the mother of the husband of my mother". Who is the lady to Raja ?
A. Aunt
B. Grand daughter
C. Daughter
D. Sister
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# Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.
Disclaimer: Though I have been re-reading my notes, and have scanned the relevant texts, my commutative algebra is quite rusty, so I may be overlooking something basic.
I want to show $\mathbb{Q} \simeq \mathbb{Q}[x,y]/\langle x,y \rangle$ is not projective as a $\mathbb{Q}[x,y]$ module. I've tried two methods, neither of which gets me to the conclusion.
I first tried what seems to be sort of standard when proving that something is not projective: show that the lifting of the identity yields a contradiction. So I let $\pi: \mathbb{Q}[x,y] \to \mathbb{Q}[x,y]/\langle x,y \rangle$ be my surjection given by $f \mapsto \bar{f}$ and the identity map is $id: \mathbb{Q}[x,y]/\langle x,y \rangle \to \mathbb{Q}[x,y]/\langle x,y \rangle$. So all I need to show is that a homomorphism $\phi: \mathbb{Q}[x,y]/\langle x,y \rangle \to \mathbb{Q}[x,y]$ such that $\pi \circ \phi =id$ does not exist. But if $$\pi(f) = \bar{f} = \overline{a_0+a_{10}x+a_{01}y+a_{11}xy+\cdots+a_{n0}x^n + a_{0n}y^n} = \bar{a_0}$$ then doesn't the map $\bar{a_0} \mapsto a_0$ work? After all, $$(\pi\circ \phi)(\bar{a_0}) = \pi(a_0) = \bar{a_0} = id(\bar{a_0}).$$I was concerned at first about this not being well defined, but since every element of a particular coset has the same constant term, it does not depend on choice. So either I have already made a mistake, or this is just the wrong map from which to derive a contradiction.
The next thing I tried used a different characterization of projective modules: that $P$ is a projective $R$-module iff there is a free module $F$ and an $R$-module $K$ such that $F \simeq K\oplus P$. In our case, this means there is a free module $F$ and a $\mathbb{Q}[x,y]$-module $K$ such that $$\mathbb{Q}[x,y] \oplus \cdots \oplus \mathbb{Q}[x,y] \simeq F \simeq K \oplus \mathbb{Q}[x,y]/\langle x,y \rangle \simeq K \oplus \mathbb{Q}.$$ From here, my concern is that I am waving my hand too much when I say: obviously this cannot be true, since every element of the LHS, which is a tuple of polynomials, cannot be broken up with one chunk in $K$ and the other in $\mathbb{Q}$. Do agree? If so, how can I make this argument more rigorous?
One more trouble: nowhere in either of these methods did I explicitly use that the polynomial ring here is only in two variables. The fact that the question did not use $\mathbb{Q}[x_1,\ldots,x_n]$ instead of $\mathbb{Q}[x,y]$ worries me.
• You might try using the fact that projective modules are always flat.
– user641
Commented May 30, 2012 at 1:19
• Thanks again for the prompt and excellent answers: +1's all around. I accepted the one by @wxu because it was the first. Commented May 30, 2012 at 22:42
• Every submodule of $\mathbb{Q}[x,y]^I$ is infinite-dimensional over $\mathbb{Q}$ (by considering degree), and so every projective module, a summand of a free module, is also infinite-dimensional over $\mathbb{Q}$. Commented Dec 14, 2018 at 1:40
• @JoshuaMundinger Thanks - been a few years since I've thought about this, but are you saying that if $M := \mathbb{Q}[x, y]/\langle x, y\rangle$ were projective, it'd be infinite dimensional over $\mathbb{Q}$, but of course this is false since $M \simeq \mathbb{Q}$? Commented Dec 14, 2018 at 13:39
I'm denoting your ring with $R$ and your ideal with $I$. We'll just need that $R$ is an integral domain, and that $I$ is a nontrivial ideal.
If $R/I$ is a projective $R$ module, then the following exact sequence splits
$$0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow 0$$
But integral domains have no proper direct summands, so $I$ would have to be trivial (a contradiction).
• @rschwieb (1) So you actually showed that $R/I$ is never flat over $R$, where $R$ is an integral domain and $I$ is a non-zero ideal of $R$? (2) Can you please explain why an integral domain has no proper direct summands. Commented Jun 24, 2015 at 15:24
• @user237522 (1) I didn't mention anything about flatness. (2) If A and B are nonzero ideals of a domain, then $\{0\}\neq AB\subseteq A\cap B$, so you can't form a direct sum of the two ideals. Commented Jun 24, 2015 at 15:45
• @rschwieb (1) Truly, it was a typo; I wanted to write "projective" not "flat", namely: $R/I$ is never projective over $R$, where $R$ is an integral domain and $I$ a non-zero ideal of $R$. If I am not wrong, I think I can prove the following: $R/I$ is never flat over $R$, where $R$ is a noetherian integral domain and $I$ a non-zero ideal of $R$. Reason: By a known criterion for flatness, flatness would imply $I=I^2$, but in a noetherian integral domain this would imply $I=0$ (since such $I$ must be generated by an idempotent of $R$). (2) Thank you very much. Commented Jun 24, 2015 at 17:06
• @user237522 (1) OK, yes, I have shown it for the projective case. (2) Interesting, I am not very well versed in flatness. I know in non-Noetherian domains you can have $I^2=I$, so your inclusion of Noetherianness is very necessary... Commented Jun 24, 2015 at 17:16
A projective module over a domain has no nonzero torsion element, since it is a submodule of a free module.
But every element of your module is a torsion element: it is killed by $x$.
As Steve D said, you can use the fact that projective modules are always flat.
Consider the map $\mathbb{Q}[x,y]\to \mathbb{Q}[x,y]$ defined by multiplying $x$. This is an injective $\mathbb{Q}[x,y]$-module map, while tensoring $\mathbb{Q}$ will give an injective map, but it is NOT. So $\mathbb{Q}$ is not flat as $\mathbb{Q}[x,y]$-module.
However, following by your methods. The map defined in your first method is not a $\mathbb{Q}[x,y]$-module map, since if $1$ sends to $1$, then $0=\overline{x}$ sends to $x=0$, contradiction.
In your second method, you can always find an element $(0,1)$ in RHS, this element is a torsion element, $x(0,1)=0$. But there is no element in LHS corresponding to it, since every element in LHS is not torsion.
• +1 I like your proof that $\mathbb{Q}$ is not flat. Thanks for the comments on my attempts at solutions as well. As usual, the solution seems so obvious now. Commented May 30, 2012 at 2:13
I believe the (injective) map $\bar{a_0} \mapsto a_0$ does not work, since it is not a map of $\mathbb Q[x,y]$-modules. In particular, we have that $x\bar{a_0}=0$ but $xa_0\neq0$ for every $a_0\in\mathbb Q[x,y]^{\oplus 1}$, and a $\mathbb Q[x,y]$-linear map would require that $x\bar{a_0}\mapsto xa_0$.
In any case, you should try the third method, which relies on the fact that if $0\to A\to B\to C\to 0$ is a short exact sequence of $R$-modules which splits in the sense that there is a map $C\to B$ that inverts the map $B\to C$, then in fact $B\cong A\oplus C$. This allows you to reformulate the first property of projective modules to say that any short exact sequences involving a projective module as a final term splits (the lift of the identity map gives the splitting).
Hence, all you need do is find an exact sequence of $\mathbb Q[x,y]$-modules $0\to A\to B\to\mathbb Q[x,y]/\left<x,y\right>\to0$ such that $B\not\cong A\oplus\mathbb Q[x,y]/\left<x,y\right>$.
The obvious choices are $A=\left<x,y\right>$ and $B=\mathbb Q[x,y]$, and this reduces the problem to showing that $\left<x,y\right>\oplus\mathbb Q[x,y]/\left<x,y\right>$ is not isomorphic to $\mathbb Q[x,y]$, i.e. is not free of rank $1$. This is easy as you take an arbitrary element in the direct sum and show that you cannot get both $1$ and $x$ by multiplying with elements of $\mathbb Q[x,y]$. If I am not mistaken, this should work for showing that for any domain $R$ and proper ideal $I$, $R/I$ is not projective as an $R$-module.
• +1 Regarding your first and last sentences: you're right, as also pointed out below by wxu and rschwieb, respectively. Thanks for the answer. Commented May 30, 2012 at 2:19
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Q&A
Parent
# Delta to Wye conversion
+1
−0
I have a circuit like this. And, I want to calculate resistance from here. I was following these steps.
I was calculating resistance for left side circuit.
$$R_1=\frac{2 × 2}{2+2+4}=0.5\Omega$$ $$R_2=\frac{2 × 4}{2+2+4}=1\Omega$$ $$R_3=\frac{2 × 4}{2+2+4}=1\Omega$$
Then, I took series and parallel circuit. Then, calculated equivalent of resistance.
$$R_s1=0.5+3=3.5$$ $$R_s2=1+3=4$$ $$R_p=(\frac{1}{3.5}+\frac{1}{4})^-1=1.8677$$ $$Requivalent=1.8677+1=2.8667$$
I am not sure if it is correct. If it is wrong than how can I proceed?
Why does this post require moderator attention?
Why should this post be closed?
Post
+2
−1
No, the answer can be seen by inspection in a few seconds, and it's not 2.8667. We don't just give answers to homework problems here, so I'll only make a few comments on the problem and your attempted solution:
1. While this is on topic here, you would probably get better response on the Electrical Engineering site.
2. This question should be in the Problems category, not here in the Q&A category. There is no excuse, since you were clearly told that when you explicitly asked about it here.
3. You never actually stated what the problem is. It seems you want to find the equivalent resistance between the two connections to the circuit? This should be stated explicitly.
4. Always show component designators. This may be less of an offense here, but over on EE this is not tolerated. Copying the circuit from someplace else is no excuse. You are responsible for what you post here. Add component designators or redraw the circuit altogether. It doesn't matter how you achieve this, only that you do.
5. It should be immediately obvious that 2.8667 can't possibly be right when the problem is asking for a resistance, since resistance isn't dimensionless. Units matter. If I was marking this, you'd lose points for sloppiness with units.
6. Not quite as bad as above, but in most cases there should be a space between the value and its unit. For example "0.5 Ω" is correct, "0.5Ω" is not. NIST has a good publication on all this stuff.
7. Instead of mechanically following some recipe, stop and actually think about this circuit. In particular, note the symmetry between the top and bottom legs. That allows for some simplification.
8. If you applied a known voltage across this circuit and the 4 Ω resistor wasn't there, what would be the voltage between the two resistors of each leg? What current flows thru the 4 Ω resistor?
So fix it. Trying to excuse it instead of fixing it is a waste of everyone's time. We don't care why, only what is.
In MathJax it is not possible to give spaces between any units and numbers
I find that hard to believe. I don't know much about MathJax, but surely there is a way to insert a hard space.
What current flows thru the 4 Ω resistor? I don't have the value.
The point was to make you think about the circuit, not the actual numbers.
Try this exercise: Consider what happens when the left end of this circuit is held at 0 V and the right end at 5 V. If the 4 Ω resistor weren't there, what voltage results at each of the two junctions between the resistors? What voltage is therefore across where the 4 Ω resistor would be? You can convert this to a Thevenin voltage source. What current would flow thru the 4 Ω resistor if it were then connected as the circuit is actually shown?
Why does this post require moderator attention?
# Comments on Delta to Wye conversion
In my other post I had added main question. I forgot to add it here. EE site will be better for the problem. But, EE is a part of Physics. That's why I ask any Physics related question here. What current flows thru the 4 Ω resistor? I don't have the value. If I think of the circuit without 4 Ω than, the answer is 2.5 Ω. In MathJax it is not possible to give spaces between any units and numbers
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Multiplying Square Roots - Algebra Lesson
Grade: 7 - 12 | In this lesson, students learn to multiply radicals by multiplying the numbers that are outside the radicals together, and multiplying the numbers that are inside the radicals together. The next step is to break down the resulting radical, and multiply the number that comes out of the radical by the number that is already outside.
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# 1. Overview
Probability and Statistics form the basis of Data Science. The probability theory is very much helpful for making the prediction. Estimates and predictions form an important part of Data science. With the help of statistical methods, we make estimates for the further analysis. Thus, statistical methods are largely dependent on the theory of probability. And all of probability and statistics is dependent on Data.
### Data
Data is the collected information(observations) we have about something or facts and statistics collected together for reference or analysis.
Data — a collection of facts (numbers, words, measurements, observations, etc) that has been translated into a form that computers can process
### Why does Data Matter?
• Helps in understanding more about the data by identifying relationships that may exist between 2 variables.
• Helps in predicting the future or forecast based on the previous trend of data.
• Helps in determining patterns that may exist between data.
• Helps in detecting fraud by uncovering anomalies in the data.
Data matters a lot nowadays as we can infer important information from it. Now let’s delve into how data is categorized. Data can be of 2 types categorical and numerical data. For Example in a bank, we have regions, occupation class, gender which follow categorical data as the data is within a fixed certain value and balance, credit score, age, tenure months follow numerical continuous distribution as data can follow an unlimited range of values.
ote: Categorical Data can be visualized by Bar Plot, Pie Chart, Pareto Chart. Numerical Data can be visualized by Histogram, Line Plot, Scatter Plot
### Descriptive Statistics
A descriptive statistic is a summary statistic that quantitatively describes or summarizes features of a collection of information. It helps us in knowing our data better. It is used to describe the characteristics of data.
#### Measurement level of Data
The qualitative and quantitative data is very much similar to the above categorical and numerical data.
Nominal: Data at this level is categorized using names, labels or qualities. eg: Brand Name, ZipCode, Gender.
Ordinal: Data at this level can be arranged in order or ranked and can be compared. eg: Grades, Star Reviews, Position in Race, Date
Interval: Data at this level can be ordered as it is in a range of values and meaningful differences between the data points can be calculated. eg: Temperature in Celsius, Year of Birth
Ratio: Data at this level is similar to interval level with added property of an inherent zero. Mathematical calculations can be performed on these data points. eg: Height, Age, Weight
Check this out:DATA | Mahrita Harahap
### Population or Sample Data
Before performing any analysis of data, we should determine if the data we’re dealing with is population or sample.
Population: Collection of all items (N) and it includes each and every unit of our study. It is hard to define and the measure of characteristic such as mean, mode is called parameter.
Sample: Subset of the population (n) and it includes only a handful units of the population. It is selected at random and the measure of the characteristic is called as statistics.
For Example, say you want to know the mean income of the subscribers to a movie subscription service(parameter). We draw a random sample of 1000 subscribers and determine that their mean income(x̄) is \$34,500 (statistic). We conclude that the population mean income (μ) is likely to be close to \$34,500 as well.
Now before looking at distributions of data. Let’s take a look at measures of data.
#### Measures of Central Tendency
The measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. As such, measures of central tendency are sometimes called measures of central location. They are also classed as summary statistics.
Mean: The mean is equal to the sum of all the values in the data set divided by the number of values in the data set i.e the calculated average. It susceptible to outliers when unusual values are added it gets skewed i.e deviates from the typical central value.
Median: The median is the middle value for a dataset that has been arranged in order of magnitude. Median is a better alternative to mean as it is less affected by outliers and skewness of the data. The median value is much closer than the typical central value.
If the total number of values is odd then
If the total number of values is even then
Mode: The mode is the most commonly occurring value in the dataset. The mode can, therefore sometimes consider the mode as being the most popular option.
For Example, In a dataset containing {13,35,54,54,55,56,57,67,85,89,96} values. Mean is 60.09. Median is 56. Mode is 54.
#### Measures of Asymmetry
Skewness: Skewness is the asymmetry in a statistical distribution, in which the curve appears distorted or skewed towards to the left or to the right. Skewness indicates whether the data is concentrated on one side.
Positive Skewness: Positive Skewness is when the mean>median>mode. The outliers are skewed to the right i.e the tail is skewed to the right.
Negative Skewness: Negative Skewness is when the mean<median<mode. The outliers are skewed to the left i.e the tail is skewed to the left.
Skewness is important as it tells us about where the data is distributed.
For eg: Global Income Distribution in 2003 is highly right-skewed. We can see the mean \$3,451 in 2003(green) is greater than the median \$1,090. It suggests that the global income is not evenly distributed. Most individuals incomes are less than \$2,000 and less number of people with income above \$14,000, so the skewness. But it seems in 2035 according to the forecast income inequality will decrease over time.
#### Measures of Variability(Dispersion)
The measure of central tendency gives a single value that represents the whole value; however, the central tendency cannot describe the observation fully. The measure of dispersion helps us to study the variability of the items i.e the spread of data.
Remember: Population Data has N data points and Sample Data has (n-1) data points. (n-1) is called Bessel’s Correction and it is used to reduce bias.
Range: The difference between the largest and the smallest value of a data, is termed as the range of the distribution. Range does not consider all the values of a series, i.e. it takes only the extreme items and middle items are not considered significant. eg: For {13,33,45,67,70} the range is 57 i.e(70–13).
Variance: Variance measures how far is the sum of squared distances from each point to the mean i.e the dispersion around the mean.
Variance is the average of all squared deviations.
Standard Deviation: AsVariance suffers from unit difference so standard deviation is used. The square root of the variance is the standard deviation. It tells about the concentration of the data around the mean of the data set.
For eg: {3,5,6,9,10} are the values in a dataset.
Coefficient of Variation(CV): It is also called as the relative standard deviation. It is the ratio of standard deviation to the mean of the dataset.
Standard deviation is the variability of a single dataset. Whereas the coefficient of variance can be used for comparing 2 datasets.
From the above example, we can see that the CV is the same. Both methods are precise. So it is perfect for comparisons.
#### Measures of Quartiles
Quartiles are better at understanding as every data point considered.
Check my previous post — In the Boxplot Section, I have elaborated on Quartiles.
#### Measures of Relationship
Measures of relationship are used to find the comparison between 2 variables.
Covariance: Covariance is a measure of the relationship between the variability of 2 variables i.e It measures the degree of change in the variables, when one variable changes, will there be the same/a similar change in the other variable.
Covariance does not give effective information about the relation between 2 variables as it is not normalized.
Correlation: Correlation gives a better understanding of covariance. It is normalized covariance. Correlation tells us how correlated the variables are to each other. It is also called as Pearson Correlation Coefficient.
The value of correlation ranges from -1 to 1. -1 indicates negative correlation i.e with an increase in 1 variable independent there is a decrease in the other dependent variable.1 indicates positive correlation i.e with an increase in 1 variable independent there is an increase in the other dependent variable.0 indicates that the variables are independent of each other.
For Example,
Correlation 0.889 tells us Height and Weight has a positive correlation. It is obvious that as the height of a person increases weight too increases.
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Finances and Taxes
# Your ERC Eligibility Calculator: Determine if You Qualify for ERC in 2020 and 2021
Gusto Editors
Bookmark
You may have heard about the Employee Retention Credit (ERC), but can’t figure out if your business is eligible to receive it. The rules are complicated, but we’ll walk you through how to determine whether you qualify, and you can use this handy Eligibility Calculator to help you with the Gross Receipts Test … but we’re getting ahead of ourselves. Let’s dive into ERC eligibility and (fingers crossed!) help you collect some serious tax credits
## What time periods are ERC eligible?
While certain legislators are vying to make ERC permanent, as of right now, it’s a temporary tax credit that can be applied to wages and health plan expenses paid out within these timeframes only:
2020
• Q2 2020: March 13, 2020 – June 30, 2020 (It may seem strange that part of March is included in this time period, as March is typically considered part of the first quarter but for the sake of ERC, Q2 starts on March 13, 2020)
• Q3 2020: July 1, 2020 – September 30, 2020
• Q4 2020: October 1, 2020 – December 31, 2020
2021
• Q1 2021: Jan 1, 2021 – March 31, 2021
• Q2 2021: April 1, 2021 – June 30, 2021
• Q3 2021: July 1, 2021 – September 30, 2021
• Q4 2021: October 1, 2021 – December 31, 2021
## Does my business qualify for ERC for 2020?
You qualify for ERC for Q2, Q3, and/or Q4 2020 if the following conditions are met:
### How to use the ERC Eligibility Calculator to determine if you qualify in 2020
1. Go to the Calculator.
2. Click File > Make a Copy at the top right hand of your screen. Now you have your own version of the calculator.
3. Click on the tab at the bottom that is labeled 2020.
4. Only change values in the light grey cells; do not change any of the values in the cells that are white, because those cells contain formulas, and altering the formulas will cause your calculations to result in errors. The content that appears in the light grey cells is simply example content, feel free to remove and alter it to values relevant to your business.
5. Choose the appropriate quarters enter your information:
6. The calculator will reveal whether you are qualified for ERC. If you are qualified, move on to the the ERC Credit Calculation lower down in the spreadsheet (this post walks you through how to use it and calculate your credit).
Still confused? Check out this decision tree:
## What if my business was formed in the middle of a quarter in 2019?
Easy peasy. Simply estimate the gross receipts it would have had for the entire quarter based on the gross receipts for the portion of the quarter that the business was in operation.
To calculate this amount, use any reasonable method, like extrapolating the gross receipts for the quarter based on the gross receipts for the number of days the business was operating.
## When does ERC eligibility end for 2020?
Eligibility ends on the earlier of these two dates:
• January 1, 2021 (you may be eligible for ERC in 2021, but eligibility rules change in 2021, so keep reading to learn more about that)
• The quarter following a quarter in which gross receipts loss is greater than 80 percent.
For example, if your business experienced a loss of 85 percent in Q3 of 2020 but only a 5 percent loss in Q4 of 2021, you still qualify for ERC in Q3 and Q4, because Q4 follows a quarter in which losses were greater than 80 percent.
## Does my business qualify for ERC for 2021?
In 2021, eligibility rules change (as if things aren’t confusing enough), so here’s what you need to know.
### Q1 and Q2 2021 ERC eligibility
You qualify for ERC for Q1 and/or Q2 of 2021 if the following conditions are met:
#### How to use the ERC Eligibility Calculator to determine if you qualify in Q1 and/or Q2 2021
1. Go to the Calculator.
2. Click File > Make a Copy at the top right hand of your screen; now you have your own version of the calculator.
3. Click on the relevant tab at the bottom of the spreadsheet:
• If you are assessig Q1, click on the tab with that label.
• If you are assessig Q2, click on the tab with that label.
4. Only change values in the light grey cells; do not change any of the values in the cells that are white, because those cells contain formulas, and altering the formulas will cause your calculations to result in errors. The content that appears in the light grey cells is simply example content, feel free to remove and alter it to values relevant to your business.
5. In Rows 3 – 5, review whether you were subject to a government shutdown order:
6. If the answer is Yes, continue to the ERC Credit Calculation below, if the answer is No, continue to the sections on Decline in Gross Receipts.
7. If your business was in operation in the corresponding quarter of 2019 and you have gross receipts from that quarter, use Rows 7 – 11 to determine your loss:
8. If the answer is Yes, continue to the ERC Credit Calculation, if the answer is No, try the Alternate Period in Rows 19 – 23 (detailed in Step 11 below).
9. If your business was not in operation in 2019 and you only have gross receipts from 2020, use Rows 13 – 17 to compare quarters from 2020 with corresponding qaurters 2021 to determine the Decline in Gross Receipts:
10. If the answer is Yes, continue to the ERC Credit Calculation below; if the answer is No, try the Alternate Period in Rows 19 – 23 (detailed in Step 11 below).
11. You can also use an Alternate Period (in Rows 19 – 23) to make your calculation; to do this compare, the quarter immediately preceding the one you are calculating for.
This means if you are assessing eligibility in Q1, you can compare Q4 of 2020 to Q1 of 2021; if you are assessing eligibility in Q2, you can compare Q1 of 2021 to Q2 of 2021.
If you get a Yes, move on the ERC Credit Calculation below; if the answer is No, try another quarter.
Still can’t figure out if you’re eligible? See this decision tree:
### Q3 and Q4 2021 ERC eligibility
When it comes to qualifying for ERC, IRS rules and regulations are still pending for the third and fourth quarters of 2021. Here’s what we know so far . . . you will likely qualify for ERC for Q3 and/or Q4 of 2021 if the following conditions are met (within the qualifying quarter):
#### How to use the ERC Eligibility Calculator to determine if you qualify in Q3 and Q4 of 2021
1. Go to the Calculator.
2. Click File > Make a Copy at the top right hand of your screen; now you have your own version of the calculator.
3. Click on the relevant tab at the bottom of the spreadsheet:
• If you are assessing Q3, click on the tab with that label
• If you are assessing Q4, click on the tab with that label
4. Only change values in the light grey cells; do not change any of the values in the cells that are white, because those cells contain formulas, and altering the formulas will cause your calculations to result in errors. The content that appears in the light grey cells is simply example content, feel free to remove and alter it to values relevant to your business.
5. In Rows 3 – 5, review whether you were subject to a government shutdown order:
6. If the answer is Yes, continue to the ERC Credit Calculation below, if the answer is No, continue to the sections on Decline in Gross Receipts.
7. Compare the corresponding quarter in 2019 with the one in 2021, use Rows 7 – 12 to determine your loss. If the Decline in Gross Receipts is over 20 percent, you qualify for ERC. If the decline is over 90 percent. you qualify as a Severely Distressed Employee, and may be eligible for the credit even if you have more than 500 employees:
8. If the answer next to Qualified? is Yes, continue to the ERC Credit Calculation, if the answer is No, you do not qualify for this quarter.
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A mathematician, a physicist, an economist, a computer scientist, and an engineer try to split a check. If that sounds like the setup to a joke, that's because it is—a wonderfully nerdy joke from Ben Orlin's Math with Bad Drawings.
This episode of Math with Bad Drawings is written as a dialogue between the five experts, each of whom tries to figure out the check according to their own discipline. Their methods don't exactly align:
Engineer: Remember to tip 18%, everybody.
Mathematician: Is that 18% of the pre-tax total, or of the total with tax?
Physicist: You know, it’s simpler if we assume the system doesn’t have tax.
Computer Scientist: But it does have tax.
Physicist: Sure, but the numbers work out more cleanly if we don’t pay tax and tip. It’s a pretty small error term. Let’s not complicate things unnecessarily.
Engineer: What you call a “small error,” I call a “collapsed bridge.”
Economist: Forget it. Taxes are inefficient, anyway. They create deadweight loss.
Mathematician: There you go again…
Economist: I mean it! If there were no taxes, I would have ordered a second soda. But instead, the government intervened, and by increasing transaction costs, prevented an exchange that would have benefited both me and the restaurant.
Engineer: You did order a second soda.
Economist: In practice, yes. But my argument still holds in theory.
Advertisement
And, true to the site's title, there is the occasional stick figure white board panel.
Head over to Math with Bad Drawings to read the entire dialogue. I suspect the resolution to this problem is rather non-mathematical: a lifetime of spit in the experts' food.
Math Experts Split the Check [Math with Bad Drawings via MetaFilter]
Advertisement
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# Connexions
You are here: Home » Content » Maak wiskunde makliker met eksponente
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What is a lens?
#### Definition of a lens
##### Lenses
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##### What is in a lens?
Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.
##### Who can create a lens?
Any individual member, a community, or a respected organization.
##### What are tags?
Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.
#### In these lenses
• GETSenPhaseMaths
This module is included inLens: Siyavula: Mathematics (Gr. 7-9)
By: SiyavulaAs a part of collection: "Wiskunde Graad 9"
Collection Review Status: In Review
Click the "GETSenPhaseMaths" link to see all content selected in this lens.
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# Maak wiskunde makliker met eksponente
Module by: Siyavula Uploaders. E-mail the author
## MAAK WISKUNDE MAKLIKER MET EKSPONENTE
KLASWERK
• Onthou jy nog hoe eksponente werk? Skryf neer wat “drie tot die mag sewe” beteken. Wat is die grondtal? Wat is die eksponent? Kan jy mooi verduidelik wat ’n mag is?
• Hierdie deel het baie voorbeelde met getalle; gebruik jou sakrekenaar om hulle uit te werk sodat jy vertroue in die metodes kan ontwikkel.
1. DEFINISIE
23 = 2 × 2 × 2 en a4 = a × a × a × a en b × b × b = b3
ook
(a+b)3 = (a+b) × (a+ b) × (a+b) en 234=23×23×23×23234=23×23×23×23 size 12{ left ( { {2} over {3} } right ) rSup { size 8{4} } = left ( { {2} over {3} } right ) times left ( { {2} over {3} } right ) times left ( { {2} over {3} } right ) times left ( { {2} over {3} } right )} {}
1.1 Skryf die volgende uitdrukkings in uitgebreide vorm:
43 (p+2)5a1 (0,5)7b2 × b3
1.2 Skryf hierdie uitdrukkings as magte:
7 × 7 × 7 × 7 y × y × y × y × y –2 × –2 × –2 (x+y) × (x+y) × (x+y) × (x+y)
1.3 Antwoord sonder om dit uit te werk: Is (–7)6 dieselfde as –76 ?
• Gebruik nou ’n sakrekenaar en kyk of die twee waardes dieselfde is.
• Vergelyk ook die volgende pare deur eers te raai wat die antwoord gaan wees, en dan met jou sakrekenaar te kyk hoe goed jy geskat het.
–52 en (–5)2 –125 en (–12)5 –13 en (–1)3
• Jy behoort nou ’n goeie idee te hê hoe hakies antwoorde beïnvloed – skryf dit neer sodat jy dit sal onthou en in die toekoms kan gebruik wanneer die probleme moeiliker word.
• Ons som hierdie deel op in ’n definisie:
ar = a × a × a × a × . . . (daar moet ra’s wees, en r moet ’n natuurlike getal wees)
• Van nou af moet jy die belangrikste magte begin memoriseer:
22 = 4; 23 = 8; 24 = 16; ens. 32 = 9; 33 = 27; 34 = 81; ens. 42 = 16; 43 = 64; ens.
Die meeste eksponentsomme moet sonder ’n sakrekenaar gedoen word.
2 VERMENIGVULDIGING
• Onthou jy nog dat g3 × g8 = g11 ? Kernwoorde: vermenigvuldig; dieselfde grondtal
2.1 Vereenvoudig: (moenie uitgebreide vorm gebruik nie).
77 × 77 (–2)4 × (–2)13 ( ½ )1 × ( ½ )2 × ( ½ )3 (a+b)a × (a+b)b
• Ons vermenigvuldig magte met enerse grondtalle volgens hierdie reël:
ax × ay = ax+yook=axay=ayaxax+y=axay=ayaxax+y size 12{ size 11{a rSup { size 8{ size 7{x+y}} } } size 12{ {}=}a rSup { size 8{x} } size 12{ times }a rSup { size 8{y} } size 12{ {}=}a rSup { size 8{y} } size 12{ times }a rSup { size 8{x} } } {}, bv. 814=84×810814=84×810 size 12{8 rSup { size 8{"14"} } =8 rSup { size 8{4} } times 8 rSup { size 8{"10"} } } {}
3. DELING
• 4642=462=444642=462=44 size 12{ { {4 rSup { size 8{6} } } over {4 rSup { size 8{2} } } } =4 rSup { size 8{6 - 2} } =4 rSup { size 8{4} } } {} is hoe dit werk. Kernwoorde: deel; dieselfde grondtal
3.1 Probeer hierdie: a6aya6ay size 12{ { { size 11{a rSup { size 8{6} } }} over { size 12{a rSup { size 8{y} } } } } } {}323321323321 size 12{ { {3 rSup { size 8{"23"} } } over {3 rSup { size 8{"21"} } } } } {}a+bpa+b12a+bpa+b12 size 12{ { { left ( size 11{a+b} right ) rSup { size 8{p} } } over { size 12{ left (a+b right ) rSup { size 8{"12"} } } } } } {}a7a7a7a7 size 12{ { { size 11{a rSup { size 8{7} } }} over { size 12{a rSup { size 8{7} } } } } } {}
• Die reël wat ons gebruik vir deling van magte is: axay=axyaxay=axy size 12{ { { size 11{a rSup { size 8{x} } }} over { size 12{a rSup { size 8{y} } } } } size 12{ {}=}a rSup { size 8{x - y} } } {}.
Ookaxy=axayaxy=axay size 12{ size 11{a rSup { size 8{x - y} } } size 12{ {}= { {a rSup { size 8{x} } } over { size 12{a rSup { size 8{y} } } } } }} {}, bv. a7=a20a13a7=a20a13 size 12{ size 11{a rSup { size 8{7} } } size 12{ {}= { {a rSup { size 8{"20"} } } over { size 12{a rSup { size 8{"13"} } } } } }} {}
4. VERHEFFING VAN ’n MAG TOT ’n MAG
• bv. 324324 size 12{ left (3 rSup { size 8{2} } right ) rSup { size 8{4} } } {}= 32×432×4 size 12{3 rSup { size 8{2 times 4} } } {}= 3838 size 12{3 rSup { size 8{8} } } {}.
4.1 Doen die volgende:
• Die reël werk so: axy=axyaxy=axy size 12{ left (a rSup { size 8{x} } right ) rSup { size 8{y} } =a rSup { size 8{ ital "xy"} } } {}ookaxy=axy=ayxaxy=axy=ayx size 12{ size 11{a rSup { size 8{ bold "xy"} } } size 12{ {}= left (a rSup { size 8{x} } right ) rSup { size 8{y} } } size 12{ {}= left (a rSup { size 8{y} } right ) rSup { size 8{x} } }} {}, bv. 618=663618=663 size 12{6 rSup { size 8{"18"} } = left (6 rSup { size 8{6} } right ) rSup { size 8{3} } } {}
5. DIE MAG VAN ’n PRODUK
• So werk dit:
(2a)3 = (2a) × (2a) × (2a) = 2 × a × 2 × a × 2 × a = 2 × 2 × 2 × a × a × a = 8a3
• Dit word gewoonlik in twee stappe gedoen, nl.: (2a)3 = 23 × a3 = 8a3
5.1 Doen self hierdie: (4x)2 (ab)6 (3 × 2)4 ( ½ x)2 (a2b3)2
• Dis duidelik dat die eksponent aan elke faktor in die hakies behoort.
• Hier is die reël: (ab)x = axbxookapbp=abbapbp=abb size 12{ size 11{a rSup { size 8{p} } } size 12{ times }b rSup { size 8{p} } size 12{ {}= left ( bold "ab" right ) rSup { size 8{b} } }} {} bv. 143=2×73=2373143=2×73=2373 size 12{"14" rSup { size 8{3} } = left (2 times 7 right ) rSup { size 8{3} } =2 rSup { size 8{3} } 7 rSup { size 8{3} } } {}en32×42=3×42=12232×42=3×42=122 size 12{3 rSup { size 8{2} } times 4 rSup { size 8{2} } = left (3 times 4 right ) rSup { size 8{2} } ="12" rSup { size 8{2} } } {}
6. DIE MAG VAN ’n BREUK
• Dis baie dieselfde as die mag van ’n produk. ab3=a3b3ab3=a3b3 size 12{ left ( { { size 11{a}} over { size 11{b}} } right ) rSup { size 8{3} } size 12{ {}= { {a rSup { size 8{3} } } over { size 12{b rSup { size 8{3} } } } } }} {}
6.1 Doen hierdie, maar wees versigtig: 23p23p size 12{ left ( { {2} over {3} } right ) rSup { size 8{p} } } {}223223 size 12{ left ( { { left ( - 2 right )} over {2} } right ) rSup { size 8{3} } } {}x2y32x2y32 size 12{ left ( { { size 11{x rSup { size 8{2} } }} over { size 12{y rSup { size 8{3} } } } } right ) rSup { size 8{2} } } {}axby2axby2 size 12{ left ( { { size 11{a rSup { size 8{ - x} } }} over { size 12{b rSup { size 8{ - y} } } } } right ) rSup { size 8{ - 2} } } {}
• Weer behoort die eksponent aan beide die teller en die noemer.
• Die reël: abm=ambmabm=ambm size 12{ left ( { { size 11{a}} over { size 11{b}} } right ) rSup { size 8{m} } size 12{ {}= { {a rSup { size 8{m} } } over { size 12{b rSup { size 8{m} } } } } }} {}enambm=abmambm=abm size 12{ { { size 11{a rSup { size 8{m} } }} over { size 12{b rSup { size 8{m} } } } } size 12{ {}= left ( { {a} over { size 12{b} } } right ) rSup { size 8{m} } }} {}bv. 233=2333=827233=2333=827 size 12{ left ( { {2} over {3} } right ) rSup { size 8{3} } = { {2 rSup { size 8{3} } } over {3 rSup { size 8{3} } } } = { {8} over {"27"} } } {}ena2xbx=a2xbx=a2bxa2xbx=a2xbx=a2bx size 12{ { { size 11{a rSup { size 8{2x} } }} over { size 12{b rSup { size 8{x} } } } } = { { left ( size 11{a rSup { size 8{2} } } right ) rSup { size 8{x} } } over { size 12{b rSup { size 8{x} } } } } size 12{ {}= left ( { {a rSup { size 8{2} } } over { size 12{b} } } right ) rSup { size 8{x} } }} {}
einde van KLASWERK
TUTORIAAL
• Pas hierdie reëls saam toe om die volgende uitdrukkings te vereenvoudig — sonder ’n sakrekenaar.
1. a5a7aa8a5a7aa8 size 12{ { { size 11{a rSup { size 8{5} } } size 12{ times }a rSup { size 8{7} } } over { size 12{a size 12{ times }a rSup { size 8{8} } } } } } {}
2. x3y4x2y5x4y8x3y4x2y5x4y8 size 12{ { { size 11{x rSup { size 8{3} } } size 12{ times }y rSup { size 8{4} } size 12{ times }x rSup { size 8{2} } y rSup { size 8{5} } } over { size 12{x rSup { size 8{4} } y rSup { size 8{8} } } } } } {}
3. a2b3c2ac22bc2a2b3c2ac22bc2 size 12{ left ( size 11{a rSup { size 8{2} } b rSup { size 8{3} } c} right ) rSup { size 8{2} } size 12{ times left ( bold "ac" rSup { size 8{2} } right ) rSup { size 8{2} } } size 12{ times left ( bold "bc" right ) rSup { size 8{2} } }} {}
4. a3b2a3ab5b4ab3a3b2a3ab5b4ab3 size 12{ size 11{a rSup { size 8{3} } } size 12{ times }b rSup { size 8{2} } size 12{ times { {a rSup { size 8{3} } } over { size 12{a} } } } size 12{ times { {b rSup { size 8{5} } } over { size 12{b rSup { size 8{4} } } } } } size 12{ times left ( bold "ab" right ) rSup { size 8{3} } }} {}
5. 2xy×2x2y42x2y32xy32xy×2x2y42x2y32xy3 size 12{ left (2 size 11{ bold "xy"} right ) times left (2 size 11{x rSup { size 8{2} } y rSup { size 8{4} } } right ) rSup { size 8{2} } size 12{ times left ( { { left (x rSup { size 8{2} } y right ) rSup { size 8{3} } } over { size 12{ left (2 bold "xy" right ) rSup { size 8{3} } } } } right )}} {}
6. 23×22×278×4×8×2×823×22×278×4×8×2×8 size 12{ { {2 rSup { size 8{3} } times 2 rSup { size 8{2} } times 2 rSup { size 8{7} } } over {8 times 4 times 8 times 2 times 8} } } {}
einde van TUTORIAAL
Nog ’n paar reëls
KLASWERK
1 Beskou hierdie geval: =a53=a2a5a3=a53=a2a5a3 size 12{ { { size 11{a rSup { size 8{5} } }} over { size 12{a rSup { size 8{3} } } } } size 12{ {}=}a rSup { size 8{5 - 3} } size 12{ {}=}a rSup { size 8{2} } } {}
• Bespreek nou hierdie twee probleme en maak nog twee reëls vir hierdie gevalle.
1.1 a3a3a3a3 size 12{ { { size 11{a rSup { size 8{3} } }} over { size 12{a rSup { size 8{3} } } } } } {}
1.2 a3a5a3a5 size 12{ { { size 11{a rSup { size 8{3} } }} over { size 12{a rSup { size 8{5} } } } } } {}
2. AS DIE EKSPONENT NUL IS
• Die antwoord van 1.1 is a0 as ons die reël vir deling toepas.
• Ons weet egter goed dat die antwoord 1 moet wees, omdat ons teller en noemer dieselfde is.
• Dus kan ons sê dat enige uitdrukking met ’n eksponent wat nul is, gelyk aan 1 moet wees.
• Die reël sê: a0 = 1 ook 1 = a0 . ’n Paar voorbeelde:
30 = 1 k0 = 1 (ab2)0 = 1 (n+1)0 = 1 a3bab220=1a3bab220=1 size 12{ left ( { { size 11{a rSup { size 8{3} } b}} over { size 12{ left ( bold "ab" rSup { size 8{2} } right ) rSup { size 8{2} } } } } right ) rSup { size 8{0} } size 12{ {}=1}} {} en
1 = (enigiets)0 m.a.w. ons kan ’n 1 verander in iets wat ons pas, indien nodig!
3. AS DIE EKSPONENT NEGATIEF IS
• Kyk nou na vraag 1.2. Die antwoord is a–2 . Maar wat beteken dit?
• a3a5=aaaaaaaa=1aa=1a2a3a5=aaaaaaaa=1aa=1a2 size 12{ { { size 11{a rSup { size 8{3} } }} over { size 12{a rSup { size 8{5} } } } } size 12{ {}= { {a size 12{ times }a size 12{ times }a} over { size 12{a size 12{ times }a size 12{ times }a size 12{ times }a size 12{ times }a} } } } size 12{ {}= { {1} over {a size 12{ times }a} } } size 12{ {}= { {1} over {a rSup { size 8{2} } } } }} {}. Dus is die reël: ax=1axax=1ax size 12{ size 11{a rSup { size 8{ - x} } } size 12{ {}= { {1} over {a rSup { size 8{x} } } } }} {} en andersom.
• Van nou af probeer ons om sover moontlik antwoorde slegs met positiewe eksponente te skryf.
• Ons kan ook sê: 1ax=ax1ax=ax size 12{ { {1} over {a rSup { size 8{ - x} } } } =a rSup { size 8{x} } } {} en andersom. Hier is belangrike voorbeelde:
ab 2 c 3 = ab 2 c 3 ab 2 c 3 = ab 2 c 3 size 12{ size 11{ bold "ab" rSup { size 8{2} } c rSup { size 8{ - 3} } } size 12{ {}= { { bold "ab" rSup { size 8{2} } } over { size 12{c rSup { size 8{3} } } } } }} {}
2 x m y = 2y x m 2 x m y = 2y x m size 12{2 size 11{x rSup { size 8{ - m} } y} size 12{ {}= { {2y} over { size 12{x rSup { size 8{m} } } } } }} {}
a 2 b 5 a 3 b 5 = a 2 a 3 b 5 b 5 = a 5 b 10 a 2 b 5 a 3 b 5 = a 2 a 3 b 5 b 5 = a 5 b 10 size 12{ { { size 11{a rSup { size 8{2} } b rSup { size 8{ - 5} } }} over { size 12{a rSup { size 8{ - 3} } b rSup { size 8{5} } } } } size 12{ {}= { {a rSup { size 8{2} } a rSup { size 8{3} } } over { size 12{b rSup { size 8{5} } b rSup { size 8{5} } } } } } size 12{ {}= { {a rSup { size 8{5} } } over { size 12{b rSup { size 8{"10"} } } } } }} {}
(1)
einde van KLASWERK
HUISWERKOPDRAG
• Vereenvoudig sonder ’n sakrekenaar en laat antwoorde sonder negatiewe eksponente.
1. x3y232x2y2xy4x3y232x2y2xy4 size 12{ size 11{x rSup { size 8{3} } y rSup { size 8{2} } } size 12{ times 3 rSup { size 8{2} } } size 11{x rSup { size 8{2} } y} size 12{ times 2} bold "xy" rSup { size 8{4} } } {}
2. x43xy26x2x3y2x7y3×4x2y42yx43xy26x2x3y2x7y3×4x2y42y size 12{ { { size 11{x rSup { size 8{4} } }} over { size 12{3 bold "xy" rSup { size 8{2} } } } } size 12{ times { {6x rSup { size 8{2} } } over { size 12{x rSup { size 8{3} } y} } } } size 12{ times "2x" rSup { size 8{7} } y rSup { size 8{3} } times { {4 size 11{x rSup { size 8{2} } y rSup { size 8{4} } }} over { size 12{2y} } } }} {}
3. 5x353x5x353x size 12{ left (5 rSup { size 8{x} } right ) rSup { size 8{3} } - left (5 rSup { size 8{3} } right ) rSup { size 8{x} } } {}
4. 2a2b5c3d22abc2d34abcd322a2b5c3d22abc2d34abcd32 size 12{ left (2 size 11{a rSup { size 8{2} } b rSup { size 8{5} } c rSup { size 8{3} } d} right ) rSup { size 8{2} } size 12{ times 2}a left ( size 12{ bold "bc" rSup { size 8{2} } d} right ) rSup { size 8{3} } size 12{ times 4} bold "ab" left ( size 12{ bold "cd" rSup { size 8{3} } } right ) rSup { size 8{2} } } {}
5. 6x2y22xy33x43xy6x2y22xy33x43xy size 12{6 left ( { { size 11{x rSup { size 8{2} } }} over { size 12{y} } } right ) rSup { size 8{2} } size 12{ times left ( { {2x} over { size 12{y rSup { size 8{3} } } } } right ) rSup { size 8{3} } } size 12{ times { {x rSup { size 8{4} } } over { size 12{3 bold "xy"} } } }} {}
6. 2a23+12a308a62a23+12a308a6 size 12{ left (2 size 11{a rSup { size 8{2} } } right ) rSup { size 8{3} } size 12{+ left ("12"a rSup { size 8{3} } right ) rSup { size 8{0} } } size 12{ - 8}a rSup { size 8{6} } } {}
7. x3y431x2y132xy32x3y431x2y132xy32 size 12{ size 11{x rSup { size 8{3} } y rSup { size 8{ - 4} } } size 12{ times left (3 rSup { size 8{ - 1} } size 11{x rSup { size 8{2} } y rSup { size 8{ - 1} } } right ) rSup { size 8{ - 3} } } size 12{ times left (2 bold "xy" rSup { size 8{3} } right ) rSup { size 8{2} } }} {}
einde van HUISWERKOPDRAG
KLASWERK
• Kom ons maak net gou seker dat ons veranderlikes met getalwaardes kan vervang.
1. Om die omtrek van ’n reghoek (sylengtes 17 cm en 13,5 cm) te bereken, gebruik ons die gewone formule:
• Omtrek = 2 [ lengte + breedte ]
• Maak eers hakies vir die veranderlikes: = 2 [ ( ) + ( ) ]
• Vul nou die waardes in: = 2 [ (17) + (13,5) ]
• Verwyder hakies en vereenvoudig = 2 [ 17 + 13,5 ]volgens gewone reëls: = 2 × 20,5
• Onthou die eenhede (indien daar is): = 41 cm
2. Wat is die waarde van x3y4x2y5x4y8x3y4x2y5x4y8 size 12{ { { size 11{x rSup { size 8{3} } } size 12{ times }y rSup { size 8{4} } size 12{ times }x rSup { size 8{2} } y rSup { size 8{5} } } over { size 12{x rSup { size 8{4} } y rSup { size 8{8} } } } } } {} as x = 3 en y = 2 ?
• Daar is twee moontlikhede, vervang eers en vereenvoudig daarna of vereenvoudig eers en vervang daarna. Hier is albei metodes:
x3y4x2y5x4y8x3y4x2y5x4y8 size 12{ { { size 11{x rSup { size 8{3} } } size 12{ times }y rSup { size 8{4} } size 12{ times }x rSup { size 8{2} } y rSup { size 8{5} } } over { size 12{x rSup { size 8{4} } y rSup { size 8{8} } } } } } {} = 33×24×32×2534×2833×24×32×2534×28 size 12{ { { left (3 right ) rSup { size 8{3} } times left (2 right ) rSup { size 8{4} } times left (3 right ) rSup { size 8{2} } times left (2 right ) rSup { size 8{5} } } over { left (3 right ) rSup { size 8{4} } times left (2 right ) rSup { size 8{8} } } } } {} = 27×16×9×3281×12827×16×9×3281×128 size 12{ { {"27" times "16" times 9 times "32"} over {"81" times "128"} } } {} = 3 × 2 = 6
x3y4x2y5x4y8x3y4x2y5x4y8 size 12{ { { size 11{x rSup { size 8{3} } } size 12{ times }y rSup { size 8{4} } size 12{ times }x rSup { size 8{2} } y rSup { size 8{5} } } over { size 12{x rSup { size 8{4} } y rSup { size 8{8} } } } } } {} = x3x2y4y5x4y8x3x2y4y5x4y8 size 12{ { { size 11{x rSup { size 8{3} } } size 12{ times }x rSup { size 8{2} } size 12{ times }y rSup { size 8{4} } size 12{ times }y rSup { size 8{5} } } over { size 12{x rSup { size 8{4} } y rSup { size 8{8} } } } } } {} = x5y9x4y8x5y9x4y8 size 12{ { { size 11{x rSup { size 8{5} } } size 12{ times }y rSup { size 8{9} } } over { size 12{x rSup { size 8{4} } size 12{ times }y rSup { size 8{8} } } } } } {} = x54y98x54y98 size 12{ size 11{x rSup { size 8{5 - 4} } } size 12{ times }y rSup { size 8{9 - 8} } } {} = x × y = (3) × (2) = 6
• Sonder foute sal die antwoorde eners wees.
3.1 Watter metode is volgens jou mening die maklikste en hoekom sê jy so?
3.2 Bereken die omtrek van ’n vierkant met sylengte 6,5 cm.
3.3 Bereken die oppervlakte van ’n reghoek met sylengtes 17 cm en 13,5 cm.
3.4 As a = 5 en b = 1 en c = 2 en d = 3, bereken die waarde van: 2a2b5c3d22abc2d34abcd322a2b5c3d22abc2d34abcd32 size 12{ left (2 size 11{a rSup { size 8{2} } b rSup { size 8{5} } c rSup { size 8{3} } d} right ) rSup { size 8{2} } size 12{ times 2}a left ( size 12{ bold "bc" rSup { size 8{2} } d} right ) rSup { size 8{3} } size 12{ times 4} bold "ab" left ( size 12{ bold "cd" rSup { size 8{3} } } right ) rSup { size 8{2} } } {}.
einde van KLASWERK
## Assessering
Leeruitkomstes(LUs) LU 1 Getalle, Bewerkings en VerwantskappeDie leerder is in staat om getalle en die verwantskappe daarvan te herken, te beskryf en voor te stel, en om tydens probleemoplossing bevoeg en met selfvertroue te tel, te skat, te bereken en te kontroleer. Assesseringstandaarde(ASe) Ons weet dit as die leerder: 1.1 die historiese ontwikkeling van getallestelsels in ’n verskeidenheid historiese en kulturele kontekste (insluitend plaaslik) kan beskryf en illustreer; 1.2 rasionale getalle (insluitend baie klein getalle in wetenskaplike notasie) herken, gebruik en kan voorstel en gemaklik tussen ekwivalente vorms in geskikte kontekste kan beweeg; 1.3 probleme in konteks kan oplos, insluitend kontekste wat gebruik kan word om bewustheid by leerders te onwikkel van ander leerareas sowel as van menseregte, sosiale, ekonomiese en omgewingskwessies soos: 1.3.1 finansiële kontekste (insluitend wins en verlies, begrotings, rekeninge, lenings, enkelvoudige en saamgestelde rente, huurkoop, wisselkoers, kommissie, verhuring en die bankwese); 1.3.2 metings in die konteks van Natuurwetenskappe en Tegnologie; 1.4 probleme oor verhouding, koers en eweredigheid (direkte en omgekeerde) oplos; 1.5 skat en bereken deur geskikte bewerkings vir probleme te kies en te gebruik en die redelikheid van resultate te beoordeel (insluitend meetprobleme wat rasionale benaderings van irrasionale getalle behels); 1.6 ’n verskeidenheid tegnieke en instrumente (insluitend tegnologie) gebruik om berekeninge doeltreffend en met die nodige mate van akkuraatheid te doen, insluitende die volgende reëls en betekenisse van eksponente (leerders behoort in staat te wees om hierdie reëls en betekenisse slegs in berekeninge te gebruik): 1.6.1 xn × xm = xn + m 1.6.2 xn xm = xn – m 1.6.3 x0 = 1 1.6.4 x–n = 1xn1xn size 12{ { {1} over {x rSup { size 8{n} } } } } {} 1.7 die eienskappe van rasionale getalle herken, beskryf en gebruik. LU 2 Patrone, Funksies en AlgebraDie leerder is in staat om patrone en verwantskappe te herken, te beskryf en voor te stel, en probleme op te los deur algebraïese taal en vaardighede te gebruik. Ons weet dit as die leerder: 2.8 die eksponentwette gebruik om uitdrukkings te vereenvoudig.
## Memorandum
Eksponente
TOETS
1. Wetenskaplike Notasie
1.1 Skryf die volgende waardes as gewone getalle:
1.1.1 2,405 × 1017
1.1.2 6,55 × 10–9
1.2 Skryf die volgende getalle in wetenskaplike notasie:
1.2.1 5 330 110 000 000 000 000
1.2.2 0,000 000 000 000 000 013 104
1.3 Doen die volgende berekeninge en skryf jou antwoord in wetenskaplike notasie:
1.3.1 (6,148 × 1011) × (9 230 220 000 000 000)
1.3.2 (1,767 × 10–6) (6,553 × 10–4)
2. Eksponente
Vereenvoudig en laat antwoorde sonder negatiewe eksponente.
(Moenie ‘n sakrekenaar gebruik nie.)
2.1 3a2xy3ab2x2y33a2xy3ab2x2y3 size 12{3a rSup { size 8{2} } ital "xy" left (3 ital "ab" rSup { size 8{2} } x rSup { size 8{2} } y right ) rSup { size 8{3} } } {} 2.2 a0b0c36c2ab3c52×23a2c304abc2×18b42a3c42a0b0c36c2ab3c52×23a2c304abc2×18b42a3c42 size 12{ { { left (a rSup { size 8{0} } b rSup { size 8{0} } c right ) rSup { size 8{3} } } over {6c rSup { size 8{2} } left ( ital "ab" rSup { size 8{3} } c rSup { size 8{5} } right ) rSup { size 8{2} } } } times { {2 left (3a rSup { size 8{2} } c rSup { size 8{3} } right ) rSup { size 8{0} } } over {4 ital "abc" rSup { size 8{2} } } } times "18"b rSup { size 8{4} } left (2a rSup { size 8{3} } c rSup { size 8{4} } right ) rSup { size 8{2} } } {}
3. Substitusie
3.1 Vereenvoudig: 2x2y3 + (xy)2 – 4x
3.2 Bereken die waarde van 2x2y3 + (xy)2 – 4x as x = 4 en y = –2
4. Formules
Die formule vir die oppervlakte van ‘n sirkel is: opp. = π r2 (r is die radius).
4.1 Bereken die oppervlaktes van die volgende sirkels:
4.1.1 ‘n Sirkel met radius = 12 cm; benader antwoord tot 1 desimale plek.
4.1.2 ‘n Sirkel met ‘n deursnit van 8 m; benader tot die naaste meter.
TOETS – Memorandum
1.1.1 240 500 000 000 000 000
1.1.2 0,000 000 006 55
1.2.1 5,330 110 × 1018
1.2.2 1,3104 × 10–17
1.3.1 6,148 × 1011 × 9,23022 × 1015
= 6,148 × 9,23022 × 1011 × 1015
≈ 56,74 × 1026
= 5,674 × 1027
1.3.2 1,767×1066,553×1041,767×1066,553×104 size 12{ { {1,"767" times "10" rSup { size 8{ - 6} } } over {6,"553" times "10" rSup { size 8{ - 4} } } } } {} = 1,7676,553×106(4)1,7676,553×106(4) size 12{ { {1,"767"} over {6,"553"} } times "10" rSup { size 8{ - 6 - $$- 4$$ } } } {} ≈ 0,26 × 10–2 = 2,6 × 10–1
2.1 34a5x7y4 = 81a5x7y4
2.2 c3×2×18a6b4c86a2b6c12×4abc2c3×2×18a6b4c86a2b6c12×4abc2 size 12{ { {c rSup { size 8{3} } times 2 times "18"a rSup { size 8{6} } b rSup { size 8{4} } c rSup { size 8{8} } } over {6a rSup { size 8{2} } b rSup { size 8{6} } c rSup { size 8{"12"} } times 4 ital "abc" rSup { size 8{2} } } } } {} = 36a6b4c1124a3b7c1436a6b4c1124a3b7c14 size 12{ { {"36"a rSup { size 8{6} } b rSup { size 8{4} } c rSup { size 8{"11"} } } over {"24"a rSup { size 8{3} } b rSup { size 8{7} } c rSup { size 8{"14"} } } } } {} = 3a32b3c33a32b3c3 size 12{ { {3a rSup { size 8{3} } } over {2b rSup { size 8{3} } c rSup { size 8{3} } } } } {}
3.1 2x2y3 + x2y2 – 4x
3.2 2(4)2(–2)3 + (4)2(–2)2 – 4(4) = 2(16)(–8) + (16)(4) – 16 = –256 + 64 – 16 = – 208
4.1.1 opp = π × 122 = 452,38934… ≈ 452,4 cm2
4.1.2 opp = π × 42 = 50,26548… ≈ 50 m2
Memoranda
KLASWERK
Die leerders behoort reeds die werk in die eerste deel te ken. Diegene wat nog nie die eenvoudige eksponentwette ken nie, kan dit nou bemeester. Vir die ander dien dit as hersiening met die oog op die nuwe werk in die tweede deel.
1.1 4 × 4 × 4 (p+2) × (p+2) × (p+2) × (p+2) × (p+2) ens.
1.2 74y5 ens.
1.3 (–7)6 = 76 , dus (–7)6 ≠ –76 ens.
2.1 714 (–2)17 = –217 ens.
3.1 a6–y 32 (a+b)p–12a0
4.1 a5a ens.
TUTORIAAL
Die tutoriaal word in die klas in stilte in ‘n beperkte tyd gedoen. Aanbeveling: Sien dit onmiddelik na – dalk kan leerders mekaar se werk nasien.
Antwoorde: 1. a3 2. xy 3. a6b8c8 4. a8b6 5. 4x8y9 6. 1
KLASWERK
Nuwe werk vir die meeste leerders in graad 9.
HUISWERKOPDRAG
Antwoorde: 1. 18x6y7
2. 24x11y3
3. 0
4. 32a6b14c14d11
5. 16x10y1216x10y12 size 12{ { {"16"x rSup { size 8{"10"} } } over {y rSup { size 8{"12"} } } } } {}
6. 1 7. 108y5x108y5x size 12{ { {"108"y rSup { size 8{5} } } over {x} } } {}
KLASWERK
Substitusie veroorsaak heelwat probleme omdat dit so maklik lyk. Leerders wat stappe uitlaat (of nie neerskryf nie) maak dikwels eenvoudige foute. Verplig leerders om hakies te gebruik.
2. Hulle behoort te besluit dat vereenvoudiging eers behoort te geskied – dit is immers waarom ons hulle leer om te vereenvoudig.
3.1 26 cm
3.2 229,5 cm2
3.3 ≈ 1,45 × 1015
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# Expanding $(2y-2)^2$ by FOIL
Expanding $$(2y-2)^2$$
Isn't this same as $$\begin{gather*} (2y-2)(2y-2)\\ = 4y^2-6y+4\ ? \end{gather*}$$
This should be FOIL, shouldn't it?
• Where on earth do you get 6 from?? Commented Oct 31, 2011 at 19:15
• And how do you get $4y^2-4$? Show some intermediate results so we can see what goes wrong. It ought to be OK to ditch "FOIL" (stupid rule, addition is commutative so there is no point going around remembering a particular order of the terms -- it doesn't matter whether you do FOIL or IOLF or FLOI or FILO) -- but you still have to apply the distributive rule correctly. Commented Oct 31, 2011 at 19:17
• @Liger86: No. \begin{align*}(2y-2)(2y-2) &= 2y(2y-2) -2(2y-2)\\ &= (2y)(2y) +(2y)(-2) +(-2)(2y) +(-2)(-2)\\ &= 4y^2 -4y -4y + 4.\end{align*} Commented Oct 31, 2011 at 19:23
• Grrr... "be FOILd". I'd like to join the club and beat whoever came up with that idiotic acronym over the head with it. Commented Oct 31, 2011 at 19:24
• What on earth is FOIL??? Commented Oct 31, 2011 at 20:39
Yes, $(2y-2)^2=(2y-2)(2y-2)$.
FOILing should work, but will get you $4y^2−8y+4$, rather than $4y^2−6y+4$, as shown:
$(2y-2)(2y-2)=(2y)(2y)+(2y)(-2)+(-2)(2y)+(-2)(-2)$ $=4y^2-4y-4y+4=4y^2−8y+4$
If you want to memorize the formula which will get you the same result, it is $$(a+b)^2=a^2+2ab+b^2$$ In your example, $a=2y$ and $b=-2$, so you get
$$(2y)^2+2(2y)(-2)+(-2)^2=4y^2-8y+4$$
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# math
posted by .
mrs.berg's kitchen is shaped like a square. The area of the kitchen is 56ft^2. What is the length of her kitchen,to the nearest hundreth?
• math -
The square root of 56 is 7.483.
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# The other term used in power electronic converters for 'Constant frequency systems' in Chopper control is
This question was previously asked in
BSPHCL JE Electrical 2019 Official Paper: Batch 2 (Held on 31 Jan 2019)
View all BSPHCL JE EE Papers >
1. Pulse amplitude modulation scheme
2. Pulse width modulation scheme
3. Current limit control
4. Frequency modulation scheme
Option 2 : Pulse width modulation scheme
## Detailed Solution
The correct answer is option 2.
Output Voltage Control Techniques of Chopper
There are mainly two techniques or methods to control the output voltage of a chopper: Time Ratio Control Method and Current Limit Control.
1.) Time Ratio Control
Time Ratio Control is the method of controlling the output voltage of the chopper by changing the ON (TON) period of the chopper while keeping the chopping frequency constant or keeping the TON or TOFF constant and varying the chopping frequency.
Constant Frequency System or Pulse Width Modulation Technique:
In this technique, the time period of the chopper is maintained constant, and change in the duty cycle is achieved by changing the TON period.
Since the time period is constant, the chopping frequency will be constant, hence this scheme is called the constant frequency system.
Variable Frequency System or Frequency Modulation Scheme:
In this scheme, the chopping frequency is varied (hence, chopping time period T) either by maintaining ON time (TON) or OFF time (TOFF ).
Therefore, the duty cycle of the chopper is varied. This method of controlling duty cycle α is also called Frequency Modulation Scheme.
2.) Current Limit Control
In this control strategy, the on and off of the chopper circuit is decided by the previous set value of load current. The two set values are maximum load current and minimum load current.
When the load current reaches the upper limit, the chopper is switched off. When the load current falls below the lower limit, the chopper is switched on.
The switching frequency of the chopper can be controlled by setting a maximum and minimum level of current.
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# Discover the Simple Steps: How to Find Your BMI Formula
A body mass index (BMI) formula is a mathematical equation that calculates a person’s body fat based on their height and weight. It is commonly used to screen for weight-related health problems such as obesity and malnutrition.
BMI is a useful tool for healthcare professionals and individuals alike to assess their weight status and make informed decisions about their health. It was developed in the 19th century by Belgian mathematician and statistician Adolphe Quetelet.
This article will explore the different ways to find a BMI formula, including online calculators, mobile apps, and manual calculations. We will also discuss the limitations of BMI and how to interpret your BMI results.
## how to find bmi formula
Finding a BMI formula is an essential aspect of calculating body mass index (BMI), a measure of body fat based on height and weight. Understanding the different ways to find a BMI formula is important for healthcare professionals and individuals alike to accurately assess weight status and make informed health decisions.
• Online calculators
• Mobile apps
• Manual calculations
• Height and weight measurements
• BMI charts
• Body fat percentage
• Waist circumference
• Health risks
• Treatment options
• Public health implications
These aspects explore various dimensions related to finding a BMI formula, including methods of calculation, factors considered, health implications, and broader societal impacts. Understanding these aspects provides a comprehensive view of BMI and its role in health assessment and management.
### Online calculators
Online calculators are a convenient and accessible way to find a BMI formula. They are typically hosted on websites or mobile apps and require users to input their height and weight. The calculator then uses a mathematical formula to calculate the user’s BMI. This information can then be used to assess weight status and make informed health decisions.
Online calculators are a valuable tool for healthcare professionals and individuals alike. They provide a quick and easy way to calculate BMI, which can be used for screening purposes or to track weight loss progress. Online calculators are also relatively accurate, as they use the same mathematical formula that is used by healthcare professionals. However, it is important to note that online calculators are not a substitute for a medical evaluation. If you have any concerns about your weight or health, it is important to see a doctor.
In conclusion, online calculators are a useful tool for finding a BMI formula. They are convenient, accessible, and relatively accurate. However, it is important to note that online calculators are not a substitute for a medical evaluation.
### Mobile apps
Mobile apps are another convenient way to find a BMI formula. There are many different BMI calculators available in the app store, both free and paid. These apps typically require users to input their height and weight, and then they calculate the user’s BMI. Some apps also provide additional features, such as tracking weight loss progress or setting weight loss goals.
Mobile apps can be a valuable tool for people who want to track their weight or lose weight. They provide a quick and easy way to calculate BMI, and they can be used anywhere, anytime. Mobile apps can also be more engaging than online calculators, as they can provide users with feedback and support.
In conclusion, mobile apps are a useful tool for finding a BMI formula. They are convenient, accessible, and can provide additional features that can help users track their weight or lose weight.
### Manual calculations
Manual calculations are a critical component of how to find a BMI formula. The BMI formula is a mathematical equation that calculates a person’s body fat based on their height and weight. Manual calculations require the use of a calculator or pencil and paper to perform the mathematical operations necessary to solve the formula.
To perform a manual calculation of BMI, you will need to know your height and weight. Once you have this information, you can use the following formula to calculate your BMI:BMI = weight (kg) / (height (m))^2
For example, if you weigh 70 kg and are 1.75 m tall, your BMI would be calculated as follows:BMI = 70 kg / (1.75 m)^2 = 22.86
Manual calculations can be used to find a BMI formula in a variety of settings, including clinical settings, research studies, and personal health assessments. Manual calculations are a valuable tool for healthcare professionals and individuals alike to accurately assess weight status and make informed health decisions.
### Height and weight measurements
Height and weight measurements are critical components of how to find a BMI formula. BMI, or body mass index, is a measure of body fat based on height and weight. It is used to screen for weight-related health problems such as obesity and malnutrition. To calculate BMI, you need to know your height and weight. The formula for BMI is: BMI = weight (kg) / (height (m))^2
For example, if you weigh 70 kg and are 1.75 m tall, your BMI would be 22.86. Height and weight measurements are important because they are used to calculate BMI, which is a valuable tool for assessing weight status and making informed health decisions.
Height and weight measurements can also be used to track weight loss progress. If you are trying to lose weight, it is important to track your height and weight regularly so that you can monitor your progress. Height and weight measurements can also be used to identify health problems. For example, if you are overweight or obese, you may be at risk for developing health problems such as heart disease, stroke, and type 2 diabetes.
In conclusion, height and weight measurements are critical components of how to find a BMI formula. BMI is a valuable tool for assessing weight status and making informed health decisions. Height and weight measurements can also be used to track weight loss progress and identify health problems.
### BMI charts
BMI charts are a critical component of how to find a BMI formula. They are used to classify people into different weight categories, such as underweight, normal weight, overweight, and obese. This information can be used to assess weight status and make informed health decisions.
BMI charts are based on the BMI formula, which is a mathematical equation that calculates a person’s body fat based on their height and weight. The BMI formula is as follows:
BMI = weight (kg) / (height (m))^2
To use a BMI chart, you need to know your height and weight. Once you have this information, you can find your BMI by locating your height and weight on the chart. Your BMI will be located in the corresponding cell on the chart.
BMI charts are a valuable tool for assessing weight status and making informed health decisions. They are easy to use and can be found online or in most doctor’s offices.
### Body fat percentage
Body fat percentage is a measure of the amount of fat in the body. It is an important factor in assessing weight status and health risk. Body fat percentage can be estimated using a variety of methods, including BMI, skinfold measurements, and bioelectrical impedance analysis.
• Measurement methods
Body fat percentage can be measured using a variety of methods, each with its own advantages and disadvantages. BMI is a simple and inexpensive method, but it is not as accurate as other methods, such as skinfold measurements or bioelectrical impedance analysis.
• Health implications
Body fat percentage is an important indicator of health risk. People with high body fat percentage are at increased risk for a variety of health problems, including heart disease, stroke, type 2 diabetes, and some types of cancer.
• Weight loss
Body fat percentage can be a useful tool for tracking weight loss progress. When you lose weight, you lose both fat and muscle. However, if you are losing weight and your body fat percentage is not decreasing, it means that you are losing muscle mass instead of fat mass.
• Body composition
Body fat percentage is one component of body composition. Body composition also includes muscle mass, bone mass, and water weight. A healthy body composition includes a healthy balance of all of these components.
Body fat percentage is a complex and multifaceted concept. It is important to understand the different methods for measuring body fat percentage, the health implications of body fat percentage, and how body fat percentage can be used to track weight loss progress and assess body composition.
### Waist circumference
Waist circumference is a measure of the circumference of the waist, taken at the level of the umbilicus (navel). It is a simple and inexpensive measure that can be used to assess abdominal obesity, which is a major risk factor for several chronic diseases, such as heart disease, stroke, type 2 diabetes, and some types of cancer.
Waist circumference is closely related to BMI. In general, people with a high BMI also have a high waist circumference. However, there are some people who have a normal BMI but a high waist circumference. This is known as “normal weight obesity.” People with normal weight obesity are at an increased risk for chronic diseases, even though their BMI is in the normal range.
Waist circumference is a critical component of how to find BMI formula because it provides information about the distribution of body fat. People with a high waist circumference are more likely to have excess fat around their, which is a major risk factor for chronic diseases. This is because abdominal fat is more likely to be stored around the organs, which can lead to inflammation and insulin resistance.
Waist circumference can be used to track weight loss progress and to assess the risk of chronic diseases. If you are losing weight, it is important to measure your waist circumference regularly to ensure that you are losing fat from your abdomen. If your waist circumference is not decreasing, it means that you are losing muscle mass instead of fat mass.
In conclusion, waist circumference is a critical component of how to find BMI formula because it provides information about the distribution of body fat. People with a high waist circumference are at an increased risk for chronic diseases, even if their BMI is in the normal range.
### Health risks
Health risks are an important consideration when it comes to understanding how to find a BMI formula. BMI, or body mass index, is a measure of body fat based on height and weight. It is used to screen for weight-related health problems such as obesity and malnutrition.
• Obesity
Obesity is a major risk factor for a number of chronic diseases, including heart disease, stroke, type 2 diabetes, and some types of cancer. Obesity is defined as having a BMI of 30 or greater.
• Overweight
Overweight is defined as having a BMI between 25 and 29.9. Overweight people are at an increased risk for developing obesity and related health problems.
• Underweight
Underweight is defined as having a BMI of less than 18.5. Underweight people are at an increased risk for developing malnutrition and other health problems.
• Normal weight obesity
Normal weight obesity is a condition in which a person has a normal BMI but a high percentage of body fat. People with normal weight obesity are at an increased risk for developing chronic diseases.
It is important to be aware of the health risks associated with different BMI categories. If you are overweight or obese, it is important to take steps to lose weight and improve your overall health.
### Treatment options
Treatment options are an important consideration when it comes to understanding how to find a BMI formula. BMI, or body mass index, is a measure of body fat based on height and weight. It is used to screen for weight-related health problems such as obesity and malnutrition. Treatment options for weight-related health problems vary depending on the individual’s needs and circumstances.
• Diet
Diet is a cornerstone of treatment for weight-related health problems. A healthy diet includes plenty of fruits, vegetables, and whole grains. It also limits unhealthy fats, added sugars, and processed foods.
• Exercise
Exercise is another important part of treatment for weight-related health problems. Exercise helps to burn calories and build muscle. It also improves cardiovascular health and mood.
• Medication
Medication may be an option for people who have difficulty losing weight through diet and exercise alone. There are a number of different medications available to treat weight-related health problems.
• Surgery
Surgery is an option for people who are severely obese and have not been able to lose weight through other methods. Surgery can help to reduce the size of the stomach or intestines, which can lead to weight loss.
Treatment options for weight-related health problems are not one-size-fits-all. The best approach for one person may not be the best approach for another. It is important to talk to a healthcare provider to determine the best course of treatment.
### Public health implications
Public health implications are an important consideration when examining how to find a BMI formula. Understanding the public health implications of BMI formulas can help healthcare professionals and policymakers develop effective strategies for addressing weight-related health problems.
• Screening and surveillance
BMI formulas are used to screen for weight-related health problems such as obesity and malnutrition. Public health campaigns can encourage individuals to use these formulas to assess their BMI and take appropriate action.
• Health promotion
BMI formulas can be used to promote healthy weight management. Public health campaigns can use BMI formulas to raise awareness of the importance of maintaining a healthy weight and provide information on how to achieve and maintain a healthy weight.
• Policy development
BMI formulas can be used to develop public health policies aimed at reducing weight-related health problems. For example, BMI formulas can be used to set targets for weight loss and to evaluate the effectiveness of public health interventions.
• Resource allocation
BMI formulas can be used to allocate resources for weight-related health problems. For example, BMI formulas can be used to identify individuals who are at high risk for weight-related health problems and to target them with appropriate interventions.
Public health implications are an important consideration when examining how to find a BMI formula. By understanding the public health implications of BMI formulas, healthcare professionals and policymakers can develop effective strategies for addressing weight-related health problems.
### Frequently Asked Questions
This FAQ section addresses common questions and concerns related to finding a BMI formula, providing clear and concise answers to enhance your understanding.
Question 1: What is a BMI formula and how is it used?
A BMI formula is a mathematical equation that calculates body mass index (BMI), a measure of body fat based on height and weight. It is used to screen for weight-related health problems such as obesity and malnutrition.
Question 2: Where can I find a BMI formula?
You can find a BMI formula online, in mobile apps, or through manual calculations using the formula: BMI = weight (kg) / (height (m))^2.
Question 3: What are the different ways to calculate BMI?
You can calculate BMI using online calculators, mobile apps, or manual calculations. Each method requires inputting your height and weight.
Question 4: How do I interpret my BMI results?
BMI results are categorized as underweight, normal weight, overweight, or obese based on established cut-off points. These categories indicate your weight status and potential health risks.
Question 5: What are the limitations of BMI?
BMI does not distinguish between fat mass and muscle mass, and may not be accurate for certain individuals, such as athletes or pregnant women.
Question 6: How can I use BMI to improve my health?
Understanding your BMI can motivate you to make healthy lifestyle changes, such as adopting a balanced diet and engaging in regular physical activity.
These FAQs provide essential insights into finding and interpreting BMI formulas. To delve deeper into the topic, the next section will explore the applications and implications of BMI in various healthcare settings.
### Tips on finding and interpreting BMI formulas
Understanding how to find and interpret BMI formulas is essential for maintaining a healthy weight and managing weight-related health risks. Here are some practical tips to help you effectively utilize BMI formulas:
Tip 1: Utilize online calculators or mobile apps: Convenient tools for calculating BMI, providing quick and accessible results.
Tip 2: Perform manual calculations: Use the BMI formula (BMI = weight (kg) / (height (m))^2) for precise calculations.
Tip 3: Accurately measure height and weight: Ensure accurate BMI calculations by using standardized measuring techniques.
Tip 4: Interpret results using BMI categories: Understand your weight status based on established BMI ranges (underweight, normal, overweight, obese).
Tip 5: Consider BMI limitations: Recognize that BMI may not accurately reflect body composition for individuals with high muscle mass or pregnant women.
Tip 6: Use BMI as a screening tool: BMI is a useful indicator for identifying potential weight-related health risks, but further evaluation may be needed.
Tip 7: Consult a healthcare professional: Seek medical advice for personalized weight management strategies and to address any underlying health concerns.
Tip 8: Monitor BMI regularly: Track your BMI over time to assess progress and make necessary adjustments to your lifestyle.
Incorporating these tips can enhance your understanding and application of BMI formulas, empowering you to make informed decisions about your health and well-being.
Moving forward, the concluding section will explore the broader implications of BMI formulas in healthcare, discussing their role in clinical practice, research, and public health initiatives.
### Conclusion
This comprehensive exploration of “how to find bmi formula” has shed light on the diverse methods and implications of calculating body mass index (BMI). Key points include the accessibility of online calculators and mobile apps, the accuracy of manual calculations using the BMI formula, and the importance of interpreting results using established BMI categories.
BMI formulas serve as valuable screening tools for identifying potential weight-related health risks. However, it is essential to acknowledge their limitations and consider individual factors such as muscle mass and pregnancy. Moreover, healthcare professionals play a crucial role in providing personalized weight management strategies and addressing underlying health concerns.
Understanding how to find and interpret BMI formulas empowers individuals to make informed decisions about their health. By incorporating BMI monitoring into regular health practices, individuals can proactively assess their weight status and take appropriate steps towards a healthier lifestyle.
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latest
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http://projecteuler.net/problem=99
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crawl-data/CC-MAIN-2014-35/segments/1409535920694.0/warc/CC-MAIN-20140909054002-00425-ip-10-180-136-8.ec2.internal.warc.gz
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## Largest exponential
### Problem 99
Published on Friday, 1st July 2005, 06:00 pm; Solved by 16596
Comparing two numbers written in index form like 211 and 37 is not difficult, as any calculator would confirm that 211 = 2048 < 37 = 2187.
However, confirming that 632382518061 > 519432525806 would be much more difficult, as both numbers contain over three million digits.
Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.
NOTE: The first two lines in the file represent the numbers in the example given above.
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http://www.mathkplus.com/I-Math/Division/Division-Table.aspx
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# Division Chart, Division Table
## Division Table Also Known As Division Chart
The Division Table/Division Chart can be printed as a worksheet or used as an interactive learning tool.
Please Read! Instructions: 18 divided by 6 equals 3 (18 ÷ 6 = 3). Move the computer mouse across to column 6. Then move the mouse down to the number 18. You are now at row 3. And 3 is the answer to 18 ÷ 6. If you are using a touch screen device like a tablet or phone, you must tap the answer to see the row and column values. For example, tap 18 and column 6 and row 3 will be highlighted.
÷ 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144
After using the Division Table a.k.a Division Chart, we strongly recommend looking at solving real division problems using the First Grade Division! Also there are more interactive learning tools: Addition Table, Subtraction Table, and Multiplication Table.
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https://www.jiskha.com/similar?question=Which+value+of+r+represents+data+with+a+strong+negative+linear+correlation+between+two+variables%3F+1%29+-1.07+2%29+-0.89+3%29+-0.14+4%29+0.92&page=219
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Which value of r represents data with a strong negative linear correlation between two variables? 1) -1.07 2) -0.89 3) -0.14 4) 0.92
26,129 questions, page 219
1. Math
5.Jack participated in a 6 day bike marathon of 430 kilometers. He biked 90 kilometers on each of the first 4 days and y kilometers on the fifth day. Which equation can be used to find b, the number of kilometers Jack biked on the sixth day?
asked by Jerald on June 4, 2013
2. APUSH
Please check? 1. a 2.c 3. e 4. b 1. Before the civil war the south's slave labor force a. was a self reproducing population b. was generally unprofitable to the planter class c. was an inflexible labor system d. was employed exclusively in the cultivation
asked by Matt on March 27, 2012
3. Math
In a survey of 380 households regarding the ownership of VCRs and DVD players, the following data was obtained: 350 households own one or more VCRs. 180 households own one or more VCRs and one or more DVD players. 13 households do not own a VCR
asked by Kal Jay on October 28, 2016
4. Pre Calc
Naturalists findthat the populations of some kinds of predatory animals vary periodically. Assume that the population of foxes in a certain forext vareis sinusoidally iwth time. REcord started being kept when time t = 0 years. A minimum number, 200 foxes,
asked by Joe on December 6, 2009
5. chemistry
By using the data in this question, estimate how much oxygen a person requires on average in a life time and how far a motor can go using the same amount of oxygen. the typical daily food requirement of a person can be considered to be 1.2 kg of
asked by Anonymous on June 3, 2012
6. algebra :( help me!!
The table below shows the weekly earnings of two groups of employees working part-time at a local grocery store. GROUP A: \$245, \$224, \$218, \$295, \$214, \$312, \$284 GROUP B: \$223, \$230, \$310, \$246, \$295, \$301, \$215 PART A: What is the mean for each group?
asked by Alexandrea on April 12, 2013
7. Chemistry
By using the data in this question, estimate how much oxygen a person requires on average in a life time and how far a motor can go using the same amount of oxygen. the typical daily food requirement of a person can be considered to be 1.2 kg of
asked by Cherie on June 3, 2012
8. Physics
An object has a initial velocity of 10 m/s in the positive x direction. Under a constant acceleration it stops after undergoing a displacement of 148.3 m in the positive x direction. Because it is still under the same acceleration it then travels back in
asked by Sammy on September 4, 2014
9. Physics
An object has a initial velocity of 10 m/s in the positive x direction. Under a constant acceleration it stops after undergoing a displacement of 148.3 m in the positive x direction. Because it is still under the same acceleration it then travels back in
asked by Samantha on September 3, 2014
10. CHEMISTRY!! NEED IMMEDIATE HELP PLEASE!!!
a. Calculate the delta H, delta G, and equilibrium constant for the reaction of methane with Cl2 to give CH3Cl and HCl. Use the following information for your calculations: Bond Dissociation Energies: CH3-H (104.8kcal/mol), Cl-Cl(59.0 kcal/mol),
asked by Natasha on October 30, 2011
11. Creative Writing. ms.sue
ok . here's the 2nd && 3rd paragraph: It was when I woke up that I found out that my parents had left for the hospital. This news was much unexpected and caught me by surprise. Immediately I was filled with pleasure and enthusiasm; I had been waiting for
asked by y912f on October 13, 2009
12. algebra
A coffee merchant has coffee beans that sell for \$9 per pound and \$12 per pound. The two types are to be mixed to create 100 lb of a mixture that will sell for \$11.25 per pound. How much of each type of bean should be used in the mixture? I am trying to
asked by sandy on April 18, 2007
13. statistics
12.27 A study has been carried out to compare the United Way contributions made by clerical workers from three local corporations. A sample of clerical workers has been randomly selected from each firm, and the dollar amounts of their contributions are as
asked by jen on September 24, 2011
14. ecology
I am asked for some suggestions for investigating the distribution of limpets. There are 2 areas to be considered 'sheltered' and 'exposed' and these may be good places to carry out the investigation. They are particularly interested in the relative
asked by Maria on September 3, 2007
15. English
1. The program is biased because it covers only rich people. 2. The gradening program is useful because we can learn how to take care of trees in a garden. 3. The cooking program is informative because we can know many cookeries. 4. The animation program
asked by John on April 8, 2009
16. English
I need help with this question. All of the following elements are enemies of coherence,except which one? 1. misplaced modifiers 2. unclear antecedents 3. placing sentences in the most understandable order 4. lack of transistional words and phrases Which
asked by Thalia on November 30, 2006
17. math
A survey about the student government program at a school finds the following results. 190 students like the program 135 students think the program is unnecessary 220 students plan on running for student government next year. If a circle graph were made
asked by Anonymous on March 25, 2015
18. math
A survey about the student government program at a school finds the following results. 190 students like the program 135 students think the program is unnecessary 220 students plan on running for student government next year. If a circle graph were made
asked by anonymous on March 25, 2015
19. English
Thank you very much very much for your help. I just wanted to make sure about the grammar in the following sentences, too. 1) He found many things which he wanted to take ashore. Therefore/so, he began searching for a boat in order to transport them to
asked by Franco on March 21, 2010
20. Math
At the end of the first quarter of 2006, all the major stock market indexes had posted strong gains in the past 12 months. Mass Mutual Financial Group credited the increases to solid growth in corporate profits. The mean one-year return for stocks in the
asked by Branden on August 31, 2011
21. Algebra
Water coming out from a fountain is modeled by the function f(x) = −x2 + 8x + 2 where f(x) represents the height, in feet, of the water from the fountain at different times x, in seconds. What does the average rate of change of f(x) from x = 1 to x = 4
asked by memelord on April 14, 2017
22. Algebra HELP!!
Water coming out from a fountain is modeled by the function f(x) = -x^2 + 8x + 2 where f(x) represents the height, in feet, of the water from the fountain at different times x, in seconds. What does the average rate of change of f(x) from x = 1 to x = 4
asked by Dina M on January 18, 2015
23. Math-Algebra 1
Part 1: When writing linear equations, how do you determine which form of a line to use? Part 2: Choose 1 set of points from the choices below. Then, solve the problem and post your solution, showing your steps. Write an equation in point-slope form for
asked by Gabby on November 23, 2013
24. accounting
CAN SOMEONE TELL ME HOW MATERIAL AND CONVERSION IS BROKEN UP IN THIS PROBLEM thanks so much PROBLEM 1 Estrella Company, a company that mass produces a widely used pain medication, uses three processes: mixing, tableting, and bottling. October’s cost and
asked by amy on November 7, 2008
25. Math help
Samantha finds a bank, Star Bank, which offers a CD account with a 2% interest rate every month. Star Bank’s interest is not compounded. The interest is based on the original amount of deposit. Remember that Samantha has \$500 to start the account. She
asked by Anonymous on January 24, 2016
26. physics
Two cars are traveling along a straight-line in the same direction, the lead car at 25.0 m/s and the other car at 32.0 m/s. At the moment the cars are 40.0 m apart, the lead driver applies the brakes, causing his car to have an acceleration of -2.20 m/s2.
asked by Anonymous on October 6, 2011
27. physics
Two cars are traveling along a straight-line in the same direction, the lead car at 25.0 m/s and the other car at 32.0 m/s. At the moment the cars are 40.0 m apart, the lead driver applies the brakes, causing his car to have an acceleration of -2.20 m/s2.
asked by Anonymous on October 6, 2011
28. Maths
A police officer gathered data on the speeds, in kilometres per hour (km/h), of some cars as they drove down hill towards her car. The speed limit is 50 km/h. Analysis of the results showed that the speeds of the cars were normally distributed with a mean
asked by kate on August 23, 2007
29. statistics
It is thought that prehistoric Indians did not take their best tools, pottery, and household items when they visited higher elevations for their summer camps. It is hypothesized that archaeological sites tend to lose their cultural identity and specific
asked by jaipreet on November 16, 2010
30. health
i am doing a health project on contraceptives. where can i find the history of Lea's shield and Essure. They are both fairly new products and I could find nothing on wikipedia or anything. Also, where could I find stats on the use or non-use of
asked by maurice on June 3, 2007
31. Math
For the pair of supply-and-demand equations, where x represents the quantity demanded in units of 1,000 and p is the unit price in dollars, find the equilibrium quantity and the equilibrium price. p = - 2x + 15 and p = 5x + 1 a. equilibrium quantity 4,000
asked by Robbi on October 18, 2016
32. Statisic for managers
Brand new to statistics ,no strong math background please help. Case 6.1 Suppose that you are a contestant on Lets make a deal. Monty has just given you a free trip touring toxic waste sites around the country. He now offers you a trade: Give up the trip
asked by Tammy on February 5, 2011
33. ICT (Programming)
Based on the table below, Based on the table below, using a CASE structure to display the type of designation when a specific code is entered. CODE TYPE OF DESIGNATION L LECTURER S SENIOR LECTURER D DEAN A ADMINISTRATION (other code) WRONG CODE ENTERED! 4.
asked by Zee on April 11, 2015
34. Smartphone Help
This is an assignment in which I have to list down Requirements, advantages, and disadvantages of a smartphone. These are the requirements needed. -send and receive phone calls -send and receive texts -on screen keyboard or physical keyboard -can tether a
asked by Anonymous on February 21, 2011
35. physics
Four point charges are located at the corners of a square with sides of length a. Two of the charges are +q, and two are -q. Find the magnitude and direction of the net electric force exerted on a charge +Q, located at the center of the square, for each of
asked by need help on February 26, 2016
36. Math
A die is rolle 16 times and the number that ad for each roll was recorded. Number on die for each roll 5 2 2 6 1 4 6 1 1 3 5 2 2 3 4 4 If this data were to be displayed using a dot plot, what numbers would be displayed along the horizontal number line?
asked by Irene on August 30, 2016
37. Statistics
A die is rolled 16 times and the number that appeared for each roll was recorded. Number on die for each roll 5 2 2 6 1 4 6 1 1 3 5 2 2 3 4 4 If this data were to be displayed using a dot plot, what numbers would be displayed along the horizontal number
asked by Amethyst on August 28, 2017
38. Statistics
QUESTION 1 Use the following scenario and data for all the questions The lives of Lithium batteries used in a type of cell phones are normally distributed with an unknown standard deviation. A simple random sample of equation batteries is selected. The
asked by Neyra on October 31, 2017
39. math
No idea how to do this. I looked in my book and there usually is a data set present to get the first, second and third quartile. The lifetimes of lightbulbs of a particular type are normally distributed with a mean of 392 hours and a standard deviation of
asked by J on April 22, 2007
40. College Physics II
A steel wire and a copper wire, each of diameter 2.000 mm, are joined end to end. At 48.0°C, each has an unstretched length of 2.000 m. The wires are connected between two fixed supports 4.000 m apart on a tabletop. The steel wire extends from x =
asked by Andrew on September 3, 2013
41. physics, help! urgent!
A positive point charge q = +2.50 nC is located at x = 1.20 m and a negative charge of ƒ{2q = ƒ{5.00 nC is located at the origin. (a) Sketch the electric potential verses x for points on the x-axis in the range -1.50 m < x < 1.50 m. (b) Find a symbolic
asked by Ella on November 14, 2012
42. Physics
Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.1 kg · m2 and an angular velocity of +8.0 rad/s. Disk B is rotating with an angular velocity of -10.2 rad/s. The two disks are then linked together without the aid of any
asked by Morgan on November 21, 2014
43. Science
Consider a lead-acid battery with 500 Ah capacity and a rated voltage of 12 V. a)What is the total capacity of energy in watt-hours that can be stored in the battery? b)Assume that the battery is completely empty. The battery is now charging at a C-rate of
asked by John on June 5, 2017
44. Art
1. Many different techniques can be used to attract the viewer's attention. What element of art was used in this image to create emphasis on the girl? A. texture B. color***** C. negative space D. curved lines I think it is B because the color of the
asked by Elementory Girl on November 25, 2018
45. physics
A circular blade on a radial arm saw is turning at 257 rad/s at the instant the motor is turned off. In 17.0 s the speed of the blade is reduced to 90 rad/s. Assume the blade to be a uniform solid disk of radius 0.170 m and mass 0.400 kg. Find the net
asked by Jeannine on June 4, 2007
46. Physics
High-speed motion pictures (3500frames/second) of a jumping 280ìg flea yielded the data to plot the flea's acceleration as a function of time as shown in the figure (Figure 1). This flea was about 2 mm long and jumped at a nearly vertical takeoff angle.
asked by Jordan on October 10, 2013
47. math
For the given linear prgramming problem, write down the objective function and the constraints. Kris is trying to make his business more efficient by having a system of both large and small vans. He can spend no more than \$100,000 for both type of vans and
asked by fakaapo on April 14, 2011
48. physics
Two strong magnets on opposite sides of a small table are shown. The long-range attractive force between the magnets keeps the lower magnet in place. Suppose the weight of the table is 26.7 N, the weight of each magnet is 6.55 N, and the magnetic force on
asked by confused on October 5, 2009
49. Statistics
A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic gonadotropin (hCG) hormone. Suppose that in a given population, 6% of fetuses have a chromosome abnormality. The test correctly
asked by Johnathan on September 24, 2017
50. Statistics
A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic gonadotropin (hCG) hormone. Suppose that in a given population, 6% of fetuses have a chromosome abnormality. The test correctly
asked by Johnathan on September 25, 2017
51. Physics
A student wants to run an experiment to determine how the distribution of mass (location of mass) affects an object`s rotational velocity. For this experiment the student picks 3 rings. One is a wood ring (like a hula-hoop) with all of its mass at the
asked by Kyle on January 17, 2010
52. Probability & Statistics
Charmaine is looking at the rental car prices for Get You There Rental Car Company. The cost is based on how many miles you plan to drive the rental car. Use the list of rental prices to answer the questions 1- 5. Miles Cost 200 \$80 600 \$120 1000 \$160 1400
asked by Annie on April 30, 2013
53. Probability & Statistics
Charmaine is looking at the rental car prices for Get You There Rental Car Company. The cost is based on how many miles you plan to drive the rental car. Use the list of rental prices to answer the questions 1- 5. Miles Cost 200 \$80 600 \$120 1000 \$160 1400
asked by Annie on April 30, 2013
54. biology
please help me on this one...i need to know how a paramecia responds to certain stimuli....like fresh water, drop of dilute salt water, drop of vinegar, piece of food, air, and carbon dioxide. I need to know if it responds to a positive, negative or a no
asked by logan on August 13, 2007
55. Nice unanswered question by Maggie
"100 bushels of corn are divided among 100 men, women, and children. Men receive 3 bushels each, women 2 bushels and children 1/2 bushel each. How can the bushels be distributed? is there more than one solution? if so, find the other solutions." let the
asked by Reiny on September 16, 2009
56. Math (Statistics)
Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied. 8. Male Color Blindness: When
asked by Lexi on May 15, 2016
57. chemistry
A sample being given as a laboratory unknown is a mixture of calcium carbonate and magnesium carbonate. Both substances decompose to form calcium oxide and magnesium oxide, respectively, when heated at a high temperature for a prolonged period. An unknown
asked by anonymousokm on October 6, 2012
58. Healthcare Statistics
As a data collection specialist at the National Institutes of Health (NIH), you have been involved in a research study conducted over the past year. The study found that liver cancer rates per 100,000 males among cigarette smokers to nonsnmokers, in a
asked by Nancy on March 8, 2015
59. Math
The table shows the population of the larger U.S. cities. A. Construct a bar graph of the data. B. Describe how the population of San Diego changed from 1950 to 2000. C. Which city had the greatest percent increase from 1950 to 2000?
asked by Mikayla on March 8, 2018
60. History
Can someone please check my answer? 7) When the Articles of the Confederation were written they showed Americans early distrust of: a. The British b. States having too much power c. Strong national government d. The President My Answer~b 14) Starting in
asked by Abby on November 7, 2015
61. English
How can I revise this sentence? with the addition of a lexical and skills syllabus on biology. (therminology used in biology to describe biodiversity) 1) Do you think that the following should be included into the objectives or the results of students'
asked by Henry2 on January 28, 2012
62. math
The table shows the depth (d metres) of water in a harbour at certain times (t hours) after midnight on a particular day. Time t (hours) Depth d (m) 0 3.0 1 3.3 2 4.2 3 5.6 4 7.2 5 8.2 6 9.0 7 8.9 8 8.1 9 7.3 10 5.6 11 4.3 12 3.5 13 3.1 Use the regression
asked by grant on May 12, 2007
63. college systems analysis and design (visio 2007)
I need help on this data flow diagram pretty much i don't have a clue what i am doing. And my instructor has been no help all year so could somebody please do this for me. i don't have a clue how to use visio either. I have to do a dfd base on the JaD
asked by lacour on November 23, 2008
64. Math
Martin wants to make a scatter plot comparing the temperature outside with the number of people at the city swimming pool. Which of the following trends would most likely represent the data? A/There is no relationship between the temperature and the number
asked by Sidney on June 26, 2013
A researcher was given the assignment of predicting which of three actions would be taken by the 280 employees in the Desota plant that was going to be sold to its employees. The alternatives were to: a. Take severance pay and leave the company. b. Stay
asked by yas on October 7, 2011
A researcher was given the assignment of predicting which of three actions would be taken by the 280 employees in the Desota plant that was going to be sold to its employees. The alternatives were to: a. Take severance pay and leave the company. b. Stay
asked by yas on October 7, 2011
A researcher was given the assignment of predicting which of three actions would be taken by the 280 employees in the Desota plant that was going to be sold to its employees. The alternatives were to: a. Take severance pay and leave the company. b. Stay
asked by andrea on June 9, 2012
68. Ray Optics
Hi, this is about thin film interference in ray optics. My book presents two similar problems, but somehow uses different paths for the same variable. I'll list out the given variables: --Problem A-- n1 < nf > n2, delta phase is therefore lambda over two.
asked by Justin on December 5, 2016
69. Research
I have to write a paper about prison life on the norms and values for inmates and staff. and how do inmates meet their needs? Does it mean that inmates meet their needs by gambling and smuggling drugs in the prison? making their own alochol..sexual acts?
asked by Amanda on March 31, 2007
70. math/everything
do you know any website that can help me with all this 1.Language Arts, Writing, in two parts. •Part 1: Organization, Sentence Structure, Usage, and Mechanics •Part 2: Essay 2.Language Arts, Reading •Poetry •Drama and •Fiction •Non-fiction
asked by monica on July 22, 2010
71. math check
While writing an article on the high cost of college education, a reporter took a random sample of the cost of new textbooks for a semester. The random variable x is the cost of one book. Her sample data can be summarized by the following. (Give your
asked by tom9 on June 21, 2013
72. Math
The bus stations in Trawlwy and Fenton are 220 miles apart. A bus left Trawley for Fenton at 9:00am. Traveling at 40 miles per hour. One hour later, an express bus left Fenton for Trawley traveling along the same road at 50 miles per hour. At what time did
asked by Miranda on May 5, 2012
73. Algebra
The bus stations in Trawlwy and Fenton are 220 miles apart. A bus left Trawley for Fenton at 9:00am. Traveling at 40 miles per hour. One hour later, an express bus left Fenton for Trawley traveling along the same road at 50 miles per hour. At what time did
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# Queries to find Kth greatest character in a range [L, R] from a string with updates
Given a string str of length N, and Q queries of the following two types:
1. (1 L R K): Find the Kth greatest character (non-distinct) from the range of indices [L, R] (1-based indexing)
2. (2 J C): Replace the Jth character from the string by character C.
Examples:
Input: str = “abcddef”, Q = 3, queries[][] = {{1, 2, 5, 3}, {2, 4, g}, {1, 1, 4, 3}}
Output:
c
Explanation :
Query 1: String between indices (2, 5) is “bcdd”. The third largest character is ‘c’. Therefore, c is the required output.
Query 2: Replace S[4] by ‘g’. Therefore, S modifies to “abcgdef”.
Query 3: String between indices (1, 4) is “abcg”. The third largest character is ‘b’. Therefore, b is the required output.
Input: str=” afcdehgk”, Q = 4, queries[][] = {{1, 2, 5, 4}, {2, 5, m}, {1, 3, 7, 2}, {1, 1, 6, 4}}
Output:
c
h
d
Naive Approach: The simplest approach to solve the problem is as follows:
• For each query of type ( 1 L R K ), find the substring of S from the range of indices [L, R], and sort this substring in non-increasing order. Print the character at the Kth index in the substring.
• For each query of type ( 2 J C ), replace the Jth character in S by C.
Time Complexity: O ( Q * ( N log(N) ) ), where N logN is the computational complexity of sorting each substring.
Auxiliary Space: O(N)
The below code is the implementation of the above approach:
## C++
`// C++ Program to implement` `// the above approach` `#include "bits/stdc++.h"` `using` `namespace` `std;` `// Function to find the kth greatest` `// character from the strijng` `char` `find_kth_largest(string str, ``int` `k)` `{` ` ``// Sorting the string in` ` ``// non-increasing Order` ` ``sort(str.begin(), str.end(),` ` ``greater<``char``>());` ` ``return` `str[k - 1];` `}` `// Function to print the K-th character` `// from the substring S[l] .. S[r]` `char` `printCharacter(string str, ``int` `l,` ` ``int` `r, ``int` `k)` `{` ` ``// 0-based indexing` ` ``l = l - 1;` ` ``r = r - 1;` ` ``// Substring of str from the` ` ``// indices l to r.` ` ``string temp` ` ``= str.substr(l, r - l + 1);` ` ``// Extract kth Largest character` ` ``char` `ans` ` ``= find_kth_largest(temp, k);` ` ``return` `ans;` `}` `// Function to replace character at` `// pos of str by the character s` `void` `updateString(string str, ``int` `pos,` ` ``char` `s)` `{` ` ``// Index of S to be updated.` ` ``int` `index = pos - 1;` ` ``char` `c = s;` ` ``// Character to be replaced` ` ``// at index in S` ` ``str[index] = c;` `}` `// Driver Code` `int` `main()` `{` ` ``// Given string` ` ``string str = ``"abcddef"``;` ` ``// Count of queries` ` ``int` `Q = 3;` ` ``// Queries` ` ``cout << printCharacter(str, 1, 2, 2)` ` ``<< endl;` ` ``updateString(str, 4, ``'g'``);` ` ``cout << printCharacter(str, 1, 5, 4)` ` ``<< endl;` ` ``return` `0;` `}`
## Java
`// Java Program to implement` `// the above approach` `//include "bits/stdJava.h"` `import` `java.util.*;` `class` `GFG{` `// Function to find the kth greatest` `// character from the strijng` `static` `char` `find_kth_largest(``char` `[]str, ` ` ``int` `k)` `{` ` ``// Sorting the String in` ` ``// non-increasing Order` ` ``Arrays.sort(str);` ` ``reverse(str);` ` ``return` `str[k - ``1``];` `}` ` ``static` `char``[] reverse(``char` `a[]) ` ` ``{` ` ``int` `i, n = a.length;` ` ``char` `t;` ` ``for` `(i = ``0``; i < n / ``2``; i++) ` ` ``{` ` ``t = a[i];` ` ``a[i] = a[n - i - ``1``];` ` ``a[n - i - ``1``] = t;` ` ``}` ` ``return` `a;` `}` ` ` `// Function to print the K-th character` `// from the subString S[l] .. S[r]` `static` `char` `printCharacter(String str, ``int` `l,` ` ``int` `r, ``int` `k)` `{` ` ``// 0-based indexing` ` ``l = l - ``1``;` ` ``r = r - ``1``;` ` ``// SubString of str from the` ` ``// indices l to r.` ` ``String temp = str.substring(l, r - l + ``1``);` ` ``// Extract kth Largest character` ` ``char` `ans = ` ` ``find_kth_largest(temp.toCharArray(), k);` ` ``return` `ans;` `}` `// Function to replace character at` `// pos of str by the character s` `static` `void` `updateString(``char` `[]str, ` ` ``int` `pos, ``char` `s)` `{` ` ``// Index of S to be updated.` ` ``int` `index = pos - ``1``;` ` ``char` `c = s;` ` ``// Character to be replaced` ` ``// at index in S` ` ``str[index] = c;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ``// Given String` ` ``String str = ``"abcddef"``;` ` ``// Count of queries` ` ``int` `Q = ``3``;` ` ``// Queries` ` ``System.out.print(printCharacter(str, ``1``, ` ` ``2``, ``2``) + ``"\n"``);` ` ``updateString(str.toCharArray(), ``4``, ``'g'``);` ` ``System.out.print(printCharacter(str, ``1``, ` ` ``5``, ``4``) + ``"\n"``);` `}` `}` `// This code is contributed by shikhasingrajput`
## Python3
`# Python3 Program to implement` `# the above approach` `# Function to find the kth greatest` `# character from the strrijng` `def` `find_kth_largest(strr, k):` ` ``# Sorting the in` ` ``# non-increasing Order` ` ``strr ``=` `sorted``(strr)` ` ``strr ``=` `strr[:: ``-``1``]` ` ``return` `strr[k ``-` `1``]` `# Function to prthe K-th character` `# from the subS[l] .. S[r]` `def` `printCharacter(strr, l, r, k):` ` ` ` ``#0-based indexing` ` ``l ``=` `l ``-` `1` ` ``r ``=` `r ``-` `1` ` ``# Subof strr from the` ` ``# indices l to r.` ` ``temp``=` `strr[l: r ``-` `l ``+` `1``]` ` ``#Extract kth Largest character` ` ``ans ``=` `find_kth_largest(temp, k)` ` ``return` `ans` `# Function to replace character at` `# pos of strr by the character s` `def` `updateString(strr, pos, s):` ` ``# Index of S to be updated.` ` ``index ``=` `pos ``-` `1` ` ``c ``=` `s` ` ``# Character to be replaced` ` ``# at index in S` ` ``strr[index] ``=` `c` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ` ` ``# Given strring` ` ``strr ``=` `"abcddef"` ` ``strr``=``[i ``for` `i ``in` `strr]` ` ``# Count of queries` ` ``Q ``=` `3` ` ``# Queries` ` ``print``(printCharacter(strr, ``1``, ``2``, ``2``))` ` ``updateString(strr, ``4``, ``'g'``)` ` ``print``(printCharacter(strr, ``1``, ``5``, ``4``))` `# This code is contributed by Mohit Kumar 29`
## C#
`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` `// Function to find the kth greatest` `// character from the strijng` `static` `char` `find_kth_largest(``char` `[]str, ` ` ``int` `k)` `{` ` ` ` ``// Sorting the String in` ` ``// non-increasing Order` ` ``Array.Sort(str);` ` ``reverse(str);` ` ` ` ``return` `str[k - 1];` `}` `static` `char``[] reverse(``char` `[]a) ` `{` ` ``int` `i, n = a.Length;` ` ``char` `t;` ` ` ` ``for``(i = 0; i < n / 2; i++) ` ` ``{` ` ``t = a[i];` ` ``a[i] = a[n - i - 1];` ` ``a[n - i - 1] = t;` ` ``}` ` ``return` `a;` `}` `// Function to print the K-th character` `// from the subString S[l] .. S[r]` `static` `char` `printchar(String str, ``int` `l,` ` ``int` `r, ``int` `k)` `{` ` ` ` ``// 0-based indexing` ` ``l = l - 1;` ` ``r = r - 1;` ` ``// SubString of str from the` ` ``// indices l to r.` ` ``String temp = str.Substring(l, r - l + 1);` ` ``// Extract kth Largest character` ` ``char` `ans = find_kth_largest(` ` ``temp.ToCharArray(), k);` ` ``return` `ans;` `}` `// Function to replace character at` `// pos of str by the character s` `static` `void` `updateString(``char` `[]str, ` ` ``int` `pos, ``char` `s)` `{` ` ` ` ``// Index of S to be updated.` ` ``int` `index = pos - 1;` ` ``char` `c = s;` ` ``// char to be replaced` ` ``// at index in S` ` ``str[index] = c;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ``// Given String` ` ``String str = ``"abcddef"``;` ` ``// Count of queries` ` ``//int Q = 3;` ` ``// Queries` ` ``Console.Write(printchar(str, 1, 2, 2) + ``"\n"``);` ` ``updateString(str.ToCharArray(), 4, ``'g'``);` ` ` ` ``Console.Write(printchar(str, 1, 5, 4) + ``"\n"``);` `}` `}` `// This code is contributed by Amit Katiyar`
Output:
```a
b
```
Efficient Approach: The above approach can be optimized by precomputing the count of all the characters which are greater than or equal to character C ( ‘a’ ≤ C ≤ ‘z’ ) efficiently using a Fenwick Tree.
Follow the steps below to solve the problem:
• Create a Fenwick Tree to store the frequencies of all characters from ‘a’ to ‘z
• For every query of type 1, check for each character from ‘z’ to ‘a’, whether it is the Kth the greatest character.
• In order to perform this, traverse from ‘z’ to ‘a’ and for each character, check if the count of all the characters traversed becomes ≥ K or not. Print the character for which the count becomes ≥ K.
Below is the implementation of the above approach:
## C++
`// C++ Program to implement` `// the above approach` `#include "bits/stdc++.h"` `using` `namespace` `std;` `// Maximum Size of a String` `const` `int` `maxn = 100005;` `// Fenwick Tree to store the` `// frequencies of 26 alphabets` `int` `BITree[26][maxn];` `// Size of the String.` `int` `N;` `// Function to update Fenwick Tree for` `// Character c at index val` `void` `update_BITree(``int` `index, ``char` `C,` ` ``int` `val)` `{` ` ``while` `(index <= N) {` ` ``// Add val to current node` ` ``// Fenwick Tree` ` ``BITree[C - ``'a'``][index]` ` ``+= val;` ` ``// Move index to parent node` ` ``// in update View` ` ``index += (index & -index);` ` ``}` `}` `// Function to get sum of frequencies` `// of character c till index` `int` `sum_BITree(``int` `index, ``char` `C)` `{` ` ``// Stores the sum` ` ``int` `s = 0;` ` ``while` `(index) {` ` ``// Add current element of` ` ``// Fenwick tree to sum` ` ``s += BITree[C - ``'a'``][index];` ` ``// Move index to parent node` ` ``// in getSum View` ` ``index -= (index & -index);` ` ``}` ` ``return` `s;` `}` `// Function to create the Fenwick tree` `void` `buildTree(string str)` `{` ` ``for` `(``int` `i = 1; i <= N; i++) {` ` ``update_BITree(i, str[i], 1);` ` ``}` ` ``cout << endl;` `}` `// Function to print the kth largest` `// character in the range of l to r` `char` `printCharacter(string str, ``int` `l,` ` ``int` `r, ``int` `k)` `{` ` ``// Stores the count of` ` ``// charcters` ` ``int` `count = 0;` ` ``// Stores the required` ` ``// character` ` ``char` `ans;` ` ``for` `(``char` `C = ``'z'``; C >= ``'a'``; C--) {` ` ``// Calculate frequency of` ` ``// C in the given range` ` ``int` `times = sum_BITree(r, C)` ` ``- sum_BITree(l - 1, C);` ` ``// Increase count` ` ``count += times;` ` ``// If count exceeds K` ` ``if` `(count >= k) {` ` ``// Required character` ` ``// found` ` ``ans = C;` ` ``break``;` ` ``}` ` ``}` ` ``return` `ans;` `}` `// Function to update character` `// at pos by character s` `void` `updateTree(string str, ``int` `pos,` ` ``char` `s)` `{` ` ``// 0 based index system` ` ``int` `index = pos;` ` ``update_BITree(index, str[index], -1);` ` ``str[index] = s;` ` ``update_BITree(index, s, 1);` `}` `// Driver Code` `int` `main()` `{` ` ``string str = ``"abcddef"``;` ` ``N = str.size();` ` ``// Makes the string 1-based indexed` ` ``str = ``'#'` `+ str;` ` ``// Number of queries` ` ``int` `Q = 3;` ` ``// Construct the Fenwick Tree` ` ``buildTree(str);` ` ``cout << printCharacter(str, 1, 2, 2)` ` ``<< endl;` ` ``updateTree(str, 4, ``'g'``);` ` ``cout << printCharacter(str, 1, 5, 4)` ` ``<< endl;` ` ``return` `0;` `}`
## Java
`// Java Program to implement` `// the above approach` `//include "bits/stdJava.h"` `import` `java.util.*;` `class` `GFG{` `// Maximum Size of a String` `static` `int` `maxn = ``100005``;` `// Fenwick Tree to store the` `// frequencies of 26 alphabets` `static` `int` `[][]BITree = ``new` `int``[``26``][maxn];` `// Size of the String.` `static` `int` `N;` `// Function to update Fenwick Tree for` `// Character c at index val` `static` `void` `update_BITree(``int` `index, ` ` ``char` `C, ``int` `val)` `{` ` ``while` `(index <= N) ` ` ``{` ` ``// Add val to current node` ` ``// Fenwick Tree` ` ``BITree[C - ``'a'``][index] += val;` ` ``// Move index to parent node` ` ``// in update View` ` ``index += (index & -index);` ` ``}` `}` `// Function to get sum of frequencies` `// of character c till index` `static` `int` `sum_BITree(``int` `index, ``char` `C)` `{` ` ``// Stores the sum` ` ``int` `s = ``0``;` ` ``while` `(index > ``0``) ` ` ``{` ` ``// Add current element of` ` ``// Fenwick tree to sum` ` ``s += BITree[C - ``'a'``][index];` ` ``// Move index to parent node` ` ``// in getSum View` ` ``index -= (index & -index);` ` ``}` ` ``return` `s;` `}` `// Function to create the Fenwick tree` `static` `void` `buildTree(String str)` `{` ` ``for` `(``int` `i = ``1``; i <= N; i++) ` ` ``{` ` ``update_BITree(i, str.charAt(i), ``1``);` ` ``}` ` ``System.out.println();` `}` `// Function to print the kth largest` `// character in the range of l to r` `static` `char` `printCharacter(String str, ``int` `l,` ` ``int` `r, ``int` `k)` `{` ` ``// Stores the count of` ` ``// charcters` ` ``int` `count = ``0``;` ` ``// Stores the required` ` ``// character` ` ``char` `ans = ``0``;` ` ``for` `(``char` `C = ``'z'``; C >= ``'a'``; C--) ` ` ``{` ` ``// Calculate frequency of` ` ``// C in the given range` ` ``int` `times = sum_BITree(r, C) - ` ` ``sum_BITree(l - ``1``, C);` ` ``// Increase count` ` ``count += times;` ` ``// If count exceeds K` ` ``if` `(count >= k) ` ` ``{` ` ``// Required character` ` ``// found` ` ``ans = C;` ` ``break``;` ` ``}` ` ``}` ` ``return` `ans;` `}` `// Function to update character` `// at pos by character s` `static` `void` `updateTree(String str, ` ` ``int` `pos, ``char` `s)` `{` ` ``// 0 based index system` ` ``int` `index = pos;` ` ``update_BITree(index, ` ` ``str.charAt(index), -``1``);` ` ``str = str.substring(``0``, index) + s + ` ` ``str.substring(index + ``1``);` ` ``update_BITree(index, s, ``1``);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ``String str = ``"abcddef"``;` ` ``N = str.length();` ` ``// Makes the String 1-based indexed` ` ``str = ``'/'` `+ str;` ` ``// Number of queries` ` ``int` `Q = ``3``;` ` ``// Conthe Fenwick Tree` ` ``buildTree(str);` ` ``System.out.print(printCharacter(str, ``1``, ` ` ``2``, ``2``) + ``"\n"``);` ` ``updateTree(str, ``4``, ``'g'``);` ` ``System.out.print(printCharacter(str, ``1``, ` ` ``5``, ``4``) + ``"\n"``);` `}` `}` `// This code is contributed by shikhasingrajput`
## Python3
`# Python3 Program to implement` `# the above approach` `# Maximum Size of a String` `maxn ``=` `100005` `# Fenwick Tree to store the` `# frequencies of 26 alphabets` `BITree ``=` `[[``0` `for` `x ``in` `range``(maxn)]` ` ``for` `y ``in` `range``(``26``)]` `# Size of the String.` `N ``=` `0` `# Function to update Fenwick Tree for` `# Character c at index val` `def` `update_BITree(index, C, val):` ` ``while` `(index <``=` `N):` ` ``# Add val to current node` ` ``# Fenwick Tree` ` ``BITree[``ord``(C) ``-` ` ``ord``(``'a'``)][index]``+``=` `val` ` ``# Move index to parent node` ` ``# in update View` ` ``index ``+``=` `(index & ``-``index)` `# Function to get sum of ` `# frequencies of character ` `# c till index` `def` `sum_BITree(index, C):` ` ``# Stores the sum` ` ``s ``=` `0` ` ``while` `(index):` ` ``# Add current element of` ` ``# Fenwick tree to sum` ` ``s ``+``=` `BITree[``ord``(C) ``-` ` ``ord``(``'a'``)][index]` ` ``# Move index to parent node` ` ``# in getSum View` ` ``index ``-``=` `(index & ``-``index)` ` ``return` `s` `# Function to create ` `# the Fenwick tree` `def` `buildTree(st):` ` ` ` ``for` `i ``in` `range``(``1``,` ` ``N ``+` `1``):` ` ``update_BITree(i, ` ` ``st[i], ``1``)` ` ` ` ``print``()` `# Function to print the ` `# kth largest character ` `# in the range of l to r` `def` `printCharacter(st, l, ` ` ``r, k):` ` ` ` ``# Stores the count of` ` ``# charcters` ` ``count ``=` `0` ` ``for` `C ``in` `range``(``ord``(``'z'``), ` ` ``ord``(``'a'``) ``-` ` ``1``, ``-``1``):` ` ``# Calculate frequency of` ` ``# C in the given range` ` ``times ``=` `(sum_BITree(r, ``chr``(C)) ``-` ` ``sum_BITree(l ``-` `1``, ``chr``(C)))` ` ``# Increase count` ` ``count ``+``=` `times` ` ``# If count exceeds K` ` ``if` `(count >``=` `k):` ` ``# Required character` ` ``# found` ` ``ans ``=` `chr``( C)` ` ``break` ` ` ` ``return` `ans` `# Function to update character` `# at pos by character s` `def` `updateTree(st, pos, s):` ` ` ` ``# 0 based index system` ` ``index ``=` `pos;` ` ``update_BITree(index, ` ` ``st[index], ``-``1``)` ` ``st.replace(st[index], s, ``1``)` ` ``update_BITree(index, s, ``1``)` `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` ` ` ` ``st ``=` `"abcddef"` ` ``N ``=` `len``(st)` ` ``# Makes the string ` ` ``# 1-based indexed` ` ``st ``=` `'#'` `+` `st` ` ``# Number of queries` ` ``Q ``=` `3` ` ``# Construct the Fenwick Tree` ` ``buildTree(st)` ` ``print` `(printCharacter(st, ``1``, ` ` ``2``, ``2``))` ` ``updateTree(st, ``4``, ``'g'``)` ` ``print` `(printCharacter(st, ``1``, ` ` ``5``, ``4``))` `# This code is contributed by Chitranayal`
## C#
`// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{` `// Maximum Size of a String` `static` `int` `maxn = 100005;` `// Fenwick Tree to store the` `// frequencies of 26 alphabets` `static` `int` `[,]BITree = ``new` `int``[26, maxn];` `// Size of the String.` `static` `int` `N;` `// Function to update Fenwick Tree for` `// char c at index val` `static` `void` `update_BITree(``int` `index, ` ` ``char` `C, ``int` `val)` `{` ` ``while` `(index <= N) ` ` ``{` ` ``// Add val to current node` ` ``// Fenwick Tree` ` ``BITree[C - ``'a'``, index] += val;` ` ``// Move index to parent node` ` ``// in update View` ` ``index += (index & -index);` ` ``}` `}` `// Function to get sum of frequencies` `// of character c till index` `static` `int` `sum_BITree(``int` `index, ``char` `C)` `{` ` ``// Stores the sum` ` ``int` `s = 0;` ` ``while` `(index > 0) ` ` ``{` ` ``// Add current element of` ` ``// Fenwick tree to sum` ` ``s += BITree[C - ``'a'``, index];` ` ``// Move index to parent node` ` ``// in getSum View` ` ``index -= (index & -index);` ` ``}` ` ``return` `s;` `}` `// Function to create the Fenwick tree` `static` `void` `buildTree(String str)` `{` ` ``for` `(``int` `i = 1; i <= N; i++) ` ` ``{` ` ``update_BITree(i, str[i], 1);` ` ``}` ` ``Console.WriteLine();` `}` `// Function to print the kth largest` `// character in the range of l to r` `static` `char` `printchar(String str, ``int` `l,` ` ``int` `r, ``int` `k)` `{` ` ``// Stores the count of` ` ``// charcters` ` ``int` `count = 0;` ` ``// Stores the required` ` ``// character` ` ``char` `ans = (``char``)0;` ` ``for` `(``char` `C = ``'z'``; C >= ``'a'``; C--) ` ` ``{` ` ``// Calculate frequency of` ` ``// C in the given range` ` ``int` `times = sum_BITree(r, C) - ` ` ``sum_BITree(l - 1, C);` ` ``// Increase count` ` ``count += times;` ` ``// If count exceeds K` ` ``if` `(count >= k) ` ` ``{` ` ``// Required character` ` ``// found` ` ``ans = C;` ` ``break``;` ` ``}` ` ``}` ` ``return` `ans;` `}` `// Function to update character` `// at pos by character s` `static` `void` `updateTree(String str, ` ` ``int` `pos, ``char` `s)` `{` ` ``// 0 based index system` ` ``int` `index = pos;` ` ``update_BITree(index, ` ` ``str[index], -1);` ` ``str = str.Substring(0, index) + s + ` ` ``str.Substring(index + 1);` ` ``update_BITree(index, s, 1);` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ``String str = ``"abcddef"``;` ` ``N = str.Length;` ` ``// Makes the String 1-based indexed` ` ``str = ``'/'` `+ str;` ` ``// Number of queries` ` ``int` `Q = 3;` ` ``// Conthe Fenwick Tree` ` ``buildTree(str);` ` ``Console.Write(printchar(str, 1, 2, 2) + ``"\n"``);` ` ``updateTree(str, 4, ``'g'``);` ` ``Console.Write(printchar(str, 1, 5, 4) + ``"\n"``);` `}` `}` `// This code is contributed by Rajput-Ji`
Output:
```a
b
```
Time Complexity: O( QlogN + NlogN )
Auxiliary Space: O(26 * maxn), where maxn denotes the maximum possible length of the string.
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# BookRiff
If you don’t like to read, you haven’t found the right book
## What is chi square test of hypothesis testing?
Chi-square test is a nonparametric test used for two specific purpose: (a) To test the hypothesis of no association between two or more groups, population or criteria (i.e. to check independence between two variables); (b) and to test how likely the observed distribution of data fits with the distribution that is …
## What is the null hypothesis for a x2 test?
The null hypothesis of the Chi-Square test is that no relationship exists on the categorical variables in the population; they are independent.
What does X squared mean in statistics?
A chi-square (χ2) statistic is a measure of the difference between the observed and expected frequencies of the outcomes of a set of events or variables. χ2 depends on the size of the difference between actual and observed values, the degrees of freedom, and the samples size.
Where chi-square test is used?
The Chi-Square test is a statistical procedure used by researchers to examine the differences between categorical variables in the same population. For example, imagine that a research group is interested in whether or not education level and marital status are related for all people in the U.S.
### What is difference between chi-square and t test?
A t-test tests a null hypothesis about two means; most often, it tests the hypothesis that two means are equal, or that the difference between them is zero. A chi-square test tests a null hypothesis about the relationship between two variables.
### How does a chi-square test work?
The chi-square test of independence works by comparing the categorically coded data that you have collected (known as the observed frequencies) with the frequencies that you would expect to get in each cell of a table by chance alone (known as the expected frequencies).
What is chi-square test what are its applications explain with examples?
For example, you might use the Chi-Square test to compare the incidence PONV between patients that received ondansetron, patients that received droperidol, and patients that received a placebo. The Chi square test is used to compare a group with a value, or to compare two or more groups, always using categorical data.
Should I use chi-square or t-test?
a t-test is to simply look at the types of variables you are working with. If you have two variables that are both categorical, i.e. they can be placed in categories like male, female and republican, democrat, independent, then you should use a chi-square test.
#### How is χ 2 used in hypothesis testing?
When we conduct a χ 2 test, we compare the observed frequencies in each response category to the frequencies we would expect if the null hypothesis were true. These expected frequencies are determined by allocating the sample to the response categories according to the distribution specified in H 0.
#### Which is the null hypothesis of the chi squared test?
The psychiatrist wants to investigate whether the distribution of the patients by social class differed in these two units. She therefore erects the null hypothesis that there is no difference between the two distributions. This is what is tested by the chi squared (χ²) test (pronounced with a hard ch as in “sky”).
When do you use the chi squared test?
A chi-squared test, also written as χ2 test, is a statistical hypothesis test that is valid to perform when the test statistic is chi-squared distributed under the null hypothesis, specifically Pearson’s chi-squared test and variants thereof.
Which is the best test for hypothesis testing?
The test of hypothesis with a discrete outcome measured in a single sample, where the goal is to assess whether the distribution of responses follows a known distribution, is called the χ 2 goodness-of-fit test.
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# Thread: Sin X Squared & Center of gravity
1. ## Sin X Squared & Center of gravity
Hi, I'm New to this forum so hello everyone!
Ive been asked to find the center of gravity of Sin x squared, between the limits of 0 -180 degrees.
I am a bit lost, once plotted how could i find the center of gravity between 0-180?
Thanks
2. Could you tell us more about the problem? A function, such as " $sin^2(x)$" or " $sin(x^2)$" does not have a "center of gravity". That requires a physical object. A geometric object may have a "center" or "centroid" which can be calculated in the same way as a center of gravity and so called (but incorrectly in my opinion) "center of gravity". But you don't even have a geometric figure. Do you mean the center of gravity of two dimensional figure with boundary the graph of the given function and the x-axis? Or the center of gravity of the one dimensional curve? Or some other figure?
So please: do you mean the center of
1) The curve $y= sin^2(x)$
2) The curve $y= sin(x^2)$
3) The area between the graph of $y= sin^2(x)$ and the x- axis
4) The area between the graph of $y= sin(x^2)$ and the x- axis
or something else?
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# C++ Program To Find Sum of Fibonacci Numbers at Even Indexes Upto N Terms
• Last Updated : 27 Jan, 2023
Given a positive integer N, the task is to find the value of F2 + F4 + F6 +………+ F2n upto N terms where Fi denotes the i-th Fibonacci number.
The Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……
Examples:
Input: n = 5
Output: 88
N = 5, So the fibonacci series will be generated from 0th term upto 10th term:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55
Sum of elements at even indexes = 0 + 1 + 3 + 8 + 21 + 55
Input: n = 8
Output: 1596
0 + 1 + 3 + 8 + 21 + 55 + 144 + 377 + 987 = 1596.
Method-1: This method includes solving the problem directly by finding all Fibonacci numbers till 2n and adding up the only the even indices. But this will require O(n) time complexity.
Below is the implementation of the above approach:
## C++
`// C++ Program to find sum of``// even-indiced Fibonacci numbers``#include ``using` `namespace` `std;` `// Computes value of first``// fibonacci numbers and``// stores the even-indexed sum``int` `calculateEvenSum(``int` `n)``{`` ``if` `(n <= 0)`` ``return` `0;` ` ``int` `fibo[2 * n + 1];`` ``fibo[0] = 0, fibo[1] = 1;` ` ``// Initialize result`` ``int` `sum = 0;` ` ``// Add remaining terms`` ``for` `(``int` `i = 2; i <= 2 * n; i++)`` ``{`` ``fibo[i] = fibo[i - 1] + fibo[i - 2];` ` ``// For even indices`` ``if` `(i % 2 == 0)`` ``sum += fibo[i];`` ``}` ` ``// Return the alternating sum`` ``return` `sum;``}` `// Driver code``int` `main()``{` ` ``// Get n`` ``int` `n = 8;` ` ``// Find the even-indiced sum`` ``cout << ``"Even indexed Fibonacci Sum upto "` `<<`` ``n << ``" terms: "` `<< calculateEvenSum(n) <<`` ``endl;` ` ``return` `0;``}`
Output:
`Even indexed Fibonacci Sum upto 8 terms: 1596`
Time complexity: O(N).
Auxiliary space: O(N).
Method-2:
It can be clearly seen that the required sum can be obtained thus:
2 ( F2 + F4 + F6 +………+ F2n ) = (F1 + F2 + F3 + F4 +………+ F2n) – (F1 – F2 + F3 – F4 +………+ F2n)
Now the first term can be obtained if we put 2n instead of n in the formula given here.
Thus F1 + F2 + F3 + F4 +………+ F2n = F2n+2 – 1.
The second term can also be found if we put 2n instead of n in the formula given here
Thus, F1 – F2 + F3 – F4 +………- F2n = 1 + (-1)2n+1F2n-1 = 1 – F2n-1.
So, 2 ( F2 + F4 + F6 +………+ F2n
= F2n+2 – 1 – 1 + F2n-1
= F2n+2 + F2n-1 – 2
= F2n + F2n+1 + F2n+1 – F2n – 2
= 2 ( F2n+1 -1)
Hence, ( F2 + F4 + F6 +………+ F2n) = F2n+1 -1 .
So in order to find the required sum, the task is to find only F2n+1 which requires O(log n) time.( Refer to method 5 or method 6 in this article.
Below is the implementation of the above approach:
## C++
`// C++ Program to find even indexed``// Fibonacci Sum in O(Log n) time.``#include ``using` `namespace` `std;` `const` `int` `MAX = 1000;` `// Create an array for memoization``int` `f[MAX] = { 0 };` `// Returns n'th Fibonacci number``// using table f[]``int` `fib(``int` `n)``{`` ``// Base cases`` ``if` `(n == 0)`` ``return` `0;`` ``if` `(n == 1 || n == 2)`` ``return` `(f[n] = 1);` ` ``// If fib(n) is already computed`` ``if` `(f[n])`` ``return` `f[n];` ` ``int` `k = ((n & 1) ? (n + 1) / 2 :`` ``n / 2);` ` ``// Applying above formula [Note value`` ``// n&1 is 1 if n is odd, else 0].`` ``f[n] = (n & 1) ? (fib(k) * fib(k) +`` ``fib(k - 1) * fib(k - 1)) :`` ``(2 * fib(k - 1) + fib(k)) *`` ``fib(k);` ` ``return` `f[n];``}` `// Computes value of even-indexed ``// Fibonacci Sum``int` `calculateEvenSum(``int` `n)``{`` ``return` `(fib(2 * n + 1) - 1);``}` `// Driver code``int` `main()``{`` ``// Get n`` ``int` `n = 8;` ` ``// Find the alternating sum`` ``cout << ``"Even indexed Fibonacci Sum upto "` `<<`` ``n << ``" terms: "` `<< calculateEvenSum(n) <<`` ``endl;` ` ``return` `0;``}`
Output:
`Even indexed Fibonacci Sum upto 8 terms: 1596`
Time complexity: O(log(N)).
Auxiliary Space: O(N).
My Personal Notes arrow_drop_up
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# $$\displaystyle\int_{−\pi/2}^{\pi/2}$$(ecos x sin x + esin x cos x)dx किसके बराबर है ?
This question was previously asked in
NDA 02/2022 Mathematics Official Paper (Held On 04 Sep 2022)
View all NDA Papers >
1. $$\frac{e^2−1}{e}$$
2. $$\frac{e^2+1}{e}$$
3. $$\frac{1−e^2}{e}$$
4. 0
Option 1 : $$\frac{e^2−1}{e}$$
Free
NDA Mathematics Full Mock Test
8.2 K Users
120 Questions 300 Marks 150 Mins
## Detailed Solution
संकल्पना:
$$\displaystyle\int_{−a}^{a}$$f(x) dx = 0, यदि f(x) एक विषम फलन है।
गणना:
दिया गया है, I =$$\displaystyle\int_{−\pi/2}^{\pi/2}$$(ecos x sin x + esin x cos x)dx
⇒ I = $$\displaystyle\int_{−\pi/2}^{\pi/2}$$ecos x sin x dx + $$\displaystyle\int_{−\pi/2}^{\pi/2}$$esin x cos x dx
⇒ I = I1 + I2
जहाँ I1 = $$\displaystyle\int_{−\pi/2}^{\pi/2}$$ecos x sin x dx और I2 = $$\displaystyle\int_{−\pi/2}^{\pi/2}$$esin x cos x dx
Iको हल करने पर
माना f(x) = ecos x sin x
⇒ f(-x) = ecos (-x) sin (-x) = - ecos x sin x
⇒ f(-x) = f(x), अतः f(x) एक विषम फलन है।
$$\displaystyle\int_{−\pi/2}^{\pi/2}$$f(x) dx = 0,
अर्थात, I1 = $$\displaystyle\int_{−\pi/2}^{\pi/2}$$ecos x sin x dx = 0
इसलिए, I = $$\displaystyle\int_{−\pi/2}^{\pi/2}$$esin x cos x dx
माना sin x = t → cos x dx = dt
जब x = -$${\pi \over 2}$$, t = -1 और जब x = $${\pi \over 2}$$, t = 1
⇒ I = $$\displaystyle\int_{−1}^{1}$$et dt
⇒ I = et $$|_{-1}^1$$ = e - e-1
⇒ I = $$\frac{e^2−1}{e}$$
सही विकल्प (1) है।
Last updated on May 15, 2024
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# Python Bustlings
If all else fails, and you really do want to edit your source code, you’ll need to edit `sys.path`. `sys.path` is a list of locations where Python will look for code.
``````import sys
sys.path.append("/path/to/your/tweepy/directory")
import tweepy
Creating a class in python
``````
Code Example 6 – the Shape class
```class Shape:
def __init__(self,x,y):
self.x = x
self.y = y
description = "This shape has not been described yet"
author = "Nobody has claimed to make this shape yet"
def area(self):
return self.x * self.y
def perimeter(self):
return 2 * self.x + 2 * self.y
def describe(self,text):
self.description = text
def authorName(self,text):
self.author = text
def scaleSize(self,scale):
self.x = self.x * scale
self.y = self.y * scale
source : http://sthurlow.com/python/lesson08/
Making own library in python
`` ``
# My Readings 1 – Stochastic Processes Sheldon Ross – Preliminaries
Three axioms of probability
In probability theory, Boole’s inequality, also known as the union bound, says that for any finite or countable set of events, the probability that at least one of the events happens is no greater than the sum of the probabilities of the individual events.
# Computational Investing Wiki
http://wiki.quantsoftware.org/index.php?title=Computational_Investing_I#Week_1
Passing by reference : This means that if we do in python
fred = squareArray
and then fred[1,1] = 0
Python does not make a copy of squareArray, rather its just passed as a reference, hence by changing fred[1,1], squareArray also gets changed.
Whereas if we want to pass by value something like
fred = squareArray.copy() should be done.
the general formula for the variance of returns for a portfolio is:
σ²(port) = ΣΣw(i)w(j)σ(i)σ(j)ρ(i,j)
where the first sum is taken over all is, and the second over all js. Thus, for a 5-asset portfolio, the formula would be:
σ²(port) = w1²σ1² + w2²σ2² + w3²σ3² + w3²σ3² + w5²σ5²
+ 2w1w2σ1σ2ρ(1,2) + 2w1w3σ1σ3ρ(1,3) + 2w1w4σ1σ4ρ(1,4) + 2w1w5σ1σ5ρ(1,5)
+ 2w2w3σ2σ3ρ(2,3) + 2w2w4σ2σ4ρ(2,4) + 2w2w5σ2σ5ρ(2,5)
+ 2w3w4σ3σ4ρ(3,4) + 2w3w5σ3σ5ρ(3,5)
+ 2w4w5σ4σ5ρ(4,5)
ρ is covariance
https://meitham.com/2012/11/17/highest-sharpe-ratio/
http://www.investopedia.com/terms/c/capm.asp#ixzz3YFEJisis
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# Lesson 7 Drawing various shapes by ashrafp
VIEWS: 11 PAGES: 3
• pg 1
``` High School Curriculum
Lesson 7: Drawing squares, rectangles, circles, ellipses, and
polygons
OBJECTIVES
1. Demonstrate how to draw squares, rectangles, circles, ellipses, and polygons.
QCC: ...........................................................................................................486-502_
TIME: 90 minutes
REFERENCES
Landscape Program: Design Imaging Group, “Planscape.” 320 Long Island Ave,
Holtsville, NY 11742. 631-654-1600. www.designimaginggroup.com
EQUIPMENT, SUPPLIES, MATERIALS
Computers
TEACHING PROCEDURE
Introduction and Mental Set
How are shapes drawn?
Discussion
1. Drawing squares
a. On the draw menu, select square.
OR
b. Select square shape from the tool bar.
c. Click on screen.
d. Drag the mouse.
e. Release the left mouse button.
f. Press the space bar to set the dimensions.
g. A text box will appear for side length and heading (angle).
h. Type 25’ for length and 270 for heading.
i. Press right mouse button to end command.
Georgia High School Agricultural Education Curriculum, Unit Lesson 7 1
Georgia Agricultural Education Curriculum Office
Updated July 2003
2. Drawing rectangles
a. On the draw menu, select rectangle.
OR
a. Select rectangle shape from the tool bar.
b. Click on screen.
c. Drag the mouse.
d. Release the left mouse.
e. Press the space bar to set the dimensions.
f. A text box will appear for length and width.
g. Type 15’ for height and 20’ for width.
h. Press right mouse button to end command.
3. Drawing circles
a. On the draw menu, select circle.
OR
a. Select circle shape from the tool bar.
b. Click on screen.
c. Drag the mouse.
d. Release the left mouse.
e. Press the space bar to set the dimensions.
f. A text box will appear for radius.
g. Type 8’ for radius.
h. Press right mouse button to end command.
4. Drawing ellipses (oval shape)
a. On the draw menu, select ellipse.
OR
a. Select ellipse shape from the tool bar.
b. Click on screen.
c. Drag the mouse.
d. Release the left mouse.
e. Press the space bar to set the dimensions.
f. A text box will appear for width and height.
g. Type 7’ for length and 22’ for width.
h. Press right mouse button to end command.
5. Drawing ngons (multiple sided shape)
a. On the draw menu, select ngon.
b. Text box will appear for number of sides.
c. Click on screen and drag the mouse.
d. Release the left mouse.
e. Press the space bar to set the dimensions.
f. A text box will appear for radius.
g. Type 10’ for radius.
h. Type 45 for heading.
Georgia High School Agricultural Education Curriculum, Unit Lesson 7 2
Georgia Agricultural Education Curriculum Office
Updated July 2003
i. Press right mouse button to end command.
OR
j. Select octagon shape from the tool bar.
If the octagon shape is selected from the tool bar, the number of sides cannot
be altered.
Steps c through i remain the same.
Assignment: A. Save as: shapes1.prj
B. Delete previous shapes from computer screen.
C. Draw the following shapes without overlapping. Label shapes by
placing the question number beside the shape.
1. 8’ x 8’ square with a heading of 0.
2. 10’5” x 20’ rectangle
3. circle with a radius of 6’
4. 14’ x 7’2” ellipse
5. 6 sided shape (hexagon) with a radius of 13’ and heading of 0
6. circle with a radius of 9’6”
7. 12’ x 12’ square with a heading of 45
8. 13’ x 8’ ellipse
9. 20’ x 12’6” rectangle
10. 9 sided shape (nonagon) with a radius of 10’11” and heading of 0
D. Type shape name and number using annotation.
E. Save as: shapes1.prj
SUMMARY
Evaluation
Save document shapes1.prj on disk.
Teacher check disks or printed copy.
Georgia High School Agricultural Education Curriculum, Unit Lesson 7 3
Georgia Agricultural Education Curriculum Office
Updated July 2003
```
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Decision values in Libsvm
I'm new to SVM. I used Libsvm for Matlab, and after a prediction phase I've got a decision values array. From SVM theory: each test record z is assigned as positive if
f(z)=1
where f(z) is defined as
f(z)=sign(w*z+b)
So how can I relate the decision value from the array for an instance z with f(z)? Is the prediction based on decision value so: if dec_value>0 then z is positive otherwise z is negative?
Yes, you are correct, if f(z) is positive, then the instance belongs to class +1, if its negative it belongs to class -1. The value of f(z) is not interpretable.
While the function:
f(z) = sign(w*z+b)
looks like an equation for a hyperplane, it differs in that w is not a normal vector - its length is not 1, so the value of f(z) is not the distance from the hyperplane, this is why it is specified as sign(..), to make it clear the value is only used to determine which side of the hyperplane the instance falls on.
Some background:
The goal is to find the hyperplane which gives the maximum margin between the two classes:
So, the purpose is to maximize the margin, which is , therefore minimizing . Remember, usually when w is used to represent a hyperplane as the normal vector, is 1. That isn't the case here clearly, as there would be no optimization problem. Instead of keeping = 1 and varying the width of the margin, we've fixed the width of the margin to 2 and are allowing to vary in size instead.
This gives us the primal optimization problem (with soft margins):
This seems to be what you are referring to. However, this equation comes from basic soft maximum margin classifier, which is foundation of SVM. True SVM is formulated as a Lagrangian dual to allow the use of kernels. The neat thing about SVM is that when the above problem (and its constraints) are formulated in the Lagrangian, all the variables except for the lagrangian multipliers drop out, leaving us with the following problem:
Notice there is no w. The training points x (y are the labels, 1 or -1), now only appear together as a dot product, allowing us to employ the kernel trick to obtain a non-linear model.
But if we don't have w what is our decision function? It becomes a function of our support vectors and the lagrangian multipliers we found.
This is what libsvm produces and what it stores as the model you have trained. It stores the support vectors and the associated alphas. For linear SVM, you can obtain the primal w, this is explained here in the LibSVM FAQ, but it is not going to be what you get back automatically from LibSVM, and this can only be done for the linear kernel.
The value of the SVM decision function based on the lagrangian multipliers and support vectors should only be interpreted by its sign as well.
Reading the docs tells me that:
The third [return value] is a matrix containing decision values or probability estimates (if '-b 1' is specified). If k is the number of classes in training data, for decision values, each row includes results of predicting k(k-1)/2 binary-class SVM's.
So for a two-class problems, what you get is a vector containing the decision values f(z) , so this means everything belonging to the first class has d<0 and everything belonging to the second class has d>0.
In general: libsvm considers its first class to be the first label it gets and so on. So for having reliable results you'd need to sort your data first.
In the binary case this holds too: whatever labels you feed svmtrain, it will take the first it encounters as class 1, and the second as class -1. This is trivially verifiable by feeding it a trivial dataset:
```Y = [-ones(100,1);ones(100,1)];
m = svmtrain(Y,Y); % train with all labels as data (never do this in practices, not the "all" part, not the training on labels part ;)
[a,b,c] = svmpredict(Y,Y,m); % of course this will give 100% accuracy.
b' % you can see that the first label will always have an internal representation of 1.
```
For multi-class classification this is different: it then contains k(k-1)/2 entries, corresponding to all one-against-all class scenarios for each pixel. This means for eg. a 4 class problem you'd have 4*3/2 = 6 values for each sample:
```[ f12(z) f13(z) f14(z) f23(z) f24(z) f34(z)]
```
Now how these function values map to classes through one-against-all I could not really deduce easily from looking at the code... But I guess you where mostly interested in the 2 class case anyhow, no?
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https://www.wyzant.com/resources/answers/500541/sales_of_26_900_grow_to_78_000_in_11_years_calculate_the_percentage_growth_p_a
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-1
Sales of \$26,900 grow to \$78,000 in 11 years. Calculate the percentage growth p.a.
i don't know the formula for this question and the teacher didn't explain it very well so I don't understand it very well
Walter B. | Success-Based Tutor Specializing in Your StudentSuccess-Based Tutor Specializing in Your...
4.9 4.9 (540 lesson ratings) (540)
0
We are looking for the growth rate that will grow \$26,900 into \$78,000 in 11 years.
The formula is:
Future Value = Present Value * (1+growth rate)11
Rearranging gives us
(Future Value/Present Value) = (1+growth rate)11
Taking the 11th root of both sides yields
(Future Value/Present Value)^(1/11) = (1+growth rate) = 1.1016
(1+growth rate) = 1.1016 therefore growth rate = 1.1016 - 1 = .1016 or 10.16% for the growth rate that is needed to increase \$26,900 to \$78,000 in 11 years.
Hope this helps
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=145&t=12627&p=31115
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## Rate = k[A][B] vs rate = -1/a d[A]/dt
$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$
204635822
Posts: 25
Joined: Fri Sep 25, 2015 3:00 am
### Rate = k[A][B] vs rate = -1/a d[A]/dt
What is the difference between the two equations. Rate = k[A][B] vs rate = -1/a d[A]/dt. I understand that the second one means that reactants go down as reaction progresses because they are used up, but do we ever use that to solve any problems, or is it more of a conceptual thing?
YeChan Lee
Posts: 14
Joined: Fri Sep 25, 2015 3:00 am
### Re: Rate = k[A][B] vs rate = -1/a d[A]/dt
Yes we do use that equation. Just take a look at the practice quiz and the first problem uses those two equations.
Nataliya Karashchuk
Posts: 13
Joined: Fri Oct 02, 2015 3:00 am
### Re: Rate = k[A][B] vs rate = -1/a d[A]/dt
Going off of this, how is Rate=k[A][B] derived from the Rate=-1/a d[A]/dt and is that beyond the scope of this class? I don't really understand the link between the equations
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http://math.stackexchange.com/questions/39371/where-does-this-1-come-from-when-balancing-this-integral-equation
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# Where does this 1 come from when balancing this integral equation?
$$\int e^{ax}\cos(bx)\,\mathrm dx = \frac1{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,\mathrm dx$$
$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\cos(bx)\,\mathrm dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C$$
Where does the $1$ in $\displaystyle \left(1 + \frac{b^2}{a^2}\right)$ come from?
-
Yeah so they have taken the quantity $-b^{2}/a^{2}$ to the Left hand side. So you have the LHS as $$1 \cdot \int e^{ax}\cos(bx) \rm{dx} + \frac{b^{2}}{a^{2}} \int e^{ax}\cos(bx) \ \text{dx} = \Bigl(1+\frac{b^{2}}{a^{2}}\Bigr)\int e^{ax}\cos(bx) \ \text{dx}$$
If $a \neq 0$ and you multiply $a$ with $1$ you get $a$ itself. – user9413 May 16 '11 at 8:53
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#### What is 43.5 percent of 420,000?
How much is 43.5 percent of 420000? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 43.5% of 420000 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 43.5% of 420,000 = 182700
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating fourty-three point five of four hundred and twenty thousand How to calculate 43.5% of 420000? Simply divide the percent by 100 and multiply by the number. For example, 43.5 /100 x 420000 = 182700 or 0.435 x 420000 = 182700
#### How much is 43.5 percent of the following numbers?
43.5 percent of 420000.01 = 18270000.435 43.5 percent of 420000.02 = 18270000.87 43.5 percent of 420000.03 = 18270001.305 43.5 percent of 420000.04 = 18270001.74 43.5 percent of 420000.05 = 18270002.175 43.5 percent of 420000.06 = 18270002.61 43.5 percent of 420000.07 = 18270003.045 43.5 percent of 420000.08 = 18270003.48 43.5 percent of 420000.09 = 18270003.915 43.5 percent of 420000.1 = 18270004.35 43.5 percent of 420000.11 = 18270004.785 43.5 percent of 420000.12 = 18270005.22 43.5 percent of 420000.13 = 18270005.655 43.5 percent of 420000.14 = 18270006.09 43.5 percent of 420000.15 = 18270006.525 43.5 percent of 420000.16 = 18270006.96 43.5 percent of 420000.17 = 18270007.395 43.5 percent of 420000.18 = 18270007.83 43.5 percent of 420000.19 = 18270008.265 43.5 percent of 420000.2 = 18270008.7 43.5 percent of 420000.21 = 18270009.135 43.5 percent of 420000.22 = 18270009.57 43.5 percent of 420000.23 = 18270010.005 43.5 percent of 420000.24 = 18270010.44 43.5 percent of 420000.25 = 18270010.875
43.5 percent of 420000.26 = 18270011.31 43.5 percent of 420000.27 = 18270011.745 43.5 percent of 420000.28 = 18270012.18 43.5 percent of 420000.29 = 18270012.615 43.5 percent of 420000.3 = 18270013.05 43.5 percent of 420000.31 = 18270013.485 43.5 percent of 420000.32 = 18270013.92 43.5 percent of 420000.33 = 18270014.355 43.5 percent of 420000.34 = 18270014.79 43.5 percent of 420000.35 = 18270015.225 43.5 percent of 420000.36 = 18270015.66 43.5 percent of 420000.37 = 18270016.095 43.5 percent of 420000.38 = 18270016.53 43.5 percent of 420000.39 = 18270016.965 43.5 percent of 420000.4 = 18270017.4 43.5 percent of 420000.41 = 18270017.835 43.5 percent of 420000.42 = 18270018.27 43.5 percent of 420000.43 = 18270018.705 43.5 percent of 420000.44 = 18270019.14 43.5 percent of 420000.45 = 18270019.575 43.5 percent of 420000.46 = 18270020.01 43.5 percent of 420000.47 = 18270020.445 43.5 percent of 420000.48 = 18270020.88 43.5 percent of 420000.49 = 18270021.315 43.5 percent of 420000.5 = 18270021.75
43.5 percent of 420000.51 = 18270022.185 43.5 percent of 420000.52 = 18270022.62 43.5 percent of 420000.53 = 18270023.055 43.5 percent of 420000.54 = 18270023.49 43.5 percent of 420000.55 = 18270023.925 43.5 percent of 420000.56 = 18270024.36 43.5 percent of 420000.57 = 18270024.795 43.5 percent of 420000.58 = 18270025.23 43.5 percent of 420000.59 = 18270025.665 43.5 percent of 420000.6 = 18270026.1 43.5 percent of 420000.61 = 18270026.535 43.5 percent of 420000.62 = 18270026.97 43.5 percent of 420000.63 = 18270027.405 43.5 percent of 420000.64 = 18270027.84 43.5 percent of 420000.65 = 18270028.275 43.5 percent of 420000.66 = 18270028.71 43.5 percent of 420000.67 = 18270029.145 43.5 percent of 420000.68 = 18270029.58 43.5 percent of 420000.69 = 18270030.015 43.5 percent of 420000.7 = 18270030.45 43.5 percent of 420000.71 = 18270030.885 43.5 percent of 420000.72 = 18270031.32 43.5 percent of 420000.73 = 18270031.755 43.5 percent of 420000.74 = 18270032.19 43.5 percent of 420000.75 = 18270032.625
43.5 percent of 420000.76 = 18270033.06 43.5 percent of 420000.77 = 18270033.495 43.5 percent of 420000.78 = 18270033.93 43.5 percent of 420000.79 = 18270034.365 43.5 percent of 420000.8 = 18270034.8 43.5 percent of 420000.81 = 18270035.235 43.5 percent of 420000.82 = 18270035.67 43.5 percent of 420000.83 = 18270036.105 43.5 percent of 420000.84 = 18270036.54 43.5 percent of 420000.85 = 18270036.975 43.5 percent of 420000.86 = 18270037.41 43.5 percent of 420000.87 = 18270037.845 43.5 percent of 420000.88 = 18270038.28 43.5 percent of 420000.89 = 18270038.715 43.5 percent of 420000.9 = 18270039.15 43.5 percent of 420000.91 = 18270039.585 43.5 percent of 420000.92 = 18270040.02 43.5 percent of 420000.93 = 18270040.455 43.5 percent of 420000.94 = 18270040.89 43.5 percent of 420000.95 = 18270041.325 43.5 percent of 420000.96 = 18270041.76 43.5 percent of 420000.97 = 18270042.195 43.5 percent of 420000.98 = 18270042.63 43.5 percent of 420000.99 = 18270043.065 43.5 percent of 420001 = 18270043.5
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# Input argument "" is undefined
12 views (last 30 days)
Anita on 30 Nov 2011
My problem arises when I run the following piece of code:
theta=[psi;s_u;b];
[theta_to , f_to , cov_to] = maximize(@p_11,theta,2,1,0);
I get this error message:
??? Undefined function or method 'p_11'
for input arguments of type 'double'.
I have made sure that all my functions are in the same library. When I try to call p_ll, I get the following error message:
??? Input argument "theta" is
undefined.
Error in ==> p_ll at 3
psi=theta(1)
The p_ll function looks like this:
function ll=p_ll(theta)
global R N T x y u
psi=theta(1);
s_u=theta(2);
b=theta(3:end);
ll=zeros(N,1);
[S_AR, A]=p_S_AR(T, psi, s_u);
for i=1:N
xb_i=x((i-1)*T+1:i*T,:)*b;
y_i=y((i-1)*T+1:i*T);
ll(i)=log(p_lli(T, R, y_i, xb_i, A, u(:,:,i)));
end
It may be trivial but I cannot find my mistake(s). Does anyone have an idea?
Hin Kwan Wong on 30 Nov 2011
function ll=p_ll(theta)
is p_ll with letter L not number ONE
maximize(@p_11,theta,2,1,0);
you put it as p_11 with number ONE
Anita on 30 Nov 2011
Thank you! It works now.
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https://github.com/QuantConnect/Tutorials/blob/master/06%20Introduction%20to%20Options%5B%5D/06%20The%20Greek%20Letters/03%20gamma.html
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# QuantConnect/Tutorials
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Gamma is the rate of change of the portfolio's delta with respect to the underlying asset's price. It represents the second-order sensitivity of the option to a movement in the underlying asset’s price.
Long options, either calls or puts, always yield positive Gamma. Gamma is higher for options that are at-the-money and closer to expiration because the Delta of the near term options move toward either 0 or 1.00 is imminent. Deeper-in-the-money or farther-out-of-the-money options have lower Gamma as their Deltas already approached 0 or 1.00 (or 0 or -1.00 for puts) and will not change as quickly with movement in the underlying. For European options:
$gamma(call)=gamma(put)=\frac{N^{'}(d_1)e^{-q(T-t)}}{S\sigma\sqrt{(T-t)}}$
def gamma(self):
d1 = self.d1()
dn1 = self.dn(d1)
return dn1 * exp(-self.q * self.T) / (self.s * self.sigma * sqrt(self.T))
z = gamma
fig = plt.figure(figsize=(20,11))
ax = fig.add_subplot(111, projection='3d')
ax.view_init(12,320)
ax.plot_wireframe(s, T, z, rstride=1, cstride=1)
ax.plot_surface(s, T, z, facecolors=cm.jet(delta),linewidth=0.001, rstride=1, cstride=1, alpha = 0.75)
ax.set_zlim3d(0, z.max())
ax.set_xlabel('stock price')
ax.set_ylabel('Time to Expiration')
ax.set_zlabel('gamma')
m = cm.ScalarMappable(cmap=cm.jet)
m.set_array(delta)
cbar = plt.colorbar(m)
The color of the graph above represents delta value.
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https://www.scribd.com/document/272210798/C1-Sequences-and-Series-Questions
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You are on page 1of 8
# Dr.
Faisal Rana
C1
1
2
www.biochemtuition.com
faisal.rana@biochemtuition.com
Worksheet A
SEQUENCES AND SERIES
Write down the first five terms of the sequences with nth terms, un , given for n ≥ 1 by
a un = 4n + 5
b un = (n + 1)2
c un = 2n
e un = n3 − 2n
f un = 1 −
g un = 1 −
1
3
n
d un =
1
2n
n
n +1
h un = 32 × ( 12 )n
The nth term of each of the following sequences is given by un = an + b, for n ≥ 1.
Find the values of the constants a and b in each case.
3
a 4, 7, 10, 13, 16, …
b 0, 7, 14, 21, 28, …
c 16, 14, 12, 10, 8, …
d 0.4, 1.7, 3.0, 4.3, 5.6, …
e 100, 83, 66, 49, 32, …
f −13, −5, 3, 11, 19, …
Find a possible expression for the nth term of each of the following sequences.
a 1, 6, 11, 16, 21, …
d
1
2
, 1, 2, 4, 8, …
g 4, 7, 12, 19, 28, …
4
b 3, 9, 27, 81, 243, …
c 2, 8, 18, 32, 50, …
e 22, 11, 0, −11, −22, …
f 0, 1, 8, 27, 64, …
h
1
3
,
2
5
,
3
7
,
4
9
,
5
,
11
i
1, 3, 7, 15, 31, …
The nth term of a sequence, un , is given by
un = c + 3n − 2.
Given that u3 = 11,
a find the value of the constant c,
b find the value of u6.
5
The nth term of a sequence, un , is given by
un = n(2n + k).
Given that u6 = 2u4 − 2,
a find the value of the constant k,
b prove that for all values of n, un − un − 1 = 4n + 3.
6
The nth term of a sequence, un , is given by
un = kn − 3.
Given that u1 + u2 = 0,
a find the two possible values of the constant k.
b For each value of k found in part a, find the corresponding value of u5.
7
Write down the first four terms of each sequence.
a un = un–1 + 4, n > 1, u1 = 3
b un = 3un–1 + 1, n > 1, u1 = 2
c un+1 = 2un + 5, n > 0, u1 = –2
d un = 7 – un–1, n ≥ 2, u1 = 5
e un = 2(5 – 2un–1), n > 1, u1 = –1
f un =
g un+1 = 1 –
1
3
un, n ≥ 1, u1 = 6
1
10
h un+1 =
Solomon Press
(un–1 + 20), n ≥ 2, u1 = 10
1
2 + un
, n ≥ 1, u1 = 0
a – 4. 15 A sequence of terms {tn} is defined. n > 0. u3 = –0. 60. Given that u3 = 7. a find expressions for u2 and u3 in terms of k. Faisal Rana www. find u4 and u1. u3 = 10 b un+1 = c un+1 = 0.5. 2 13 . 13. … Given that the following sequences can be defined by recurrence relations of the form un = aun−1 + b. 27. 39. 9. n > 1.Dr.rana@biochemtuition. where k is a constant. 12 13 For the sequences given by the following recurrence relations. 11. –1. find expressions for u2 and u3 in terms of the constant k. u1 = 2. by the recurrence relation tn = ktn–1 + 2. 3.com C1 SEQUENCES AND SERIES 8 In each case. 3. u3 = 5 un −1 . a un = 3un–1 – 2. 4. 30. Given that u5 = 30. u1 = 2. u3 . 8. 5 12 . write down a recurrence relation that would produce the given sequence. n > 1. for n > 1. 171. 26. … c 62. 5. u1 = 4 f un+1 = 3 3 61k 3 + un . n > 1. a find expressions for t2 and t3 in terms of k. u1 = k 3 3 A sequence is given by the recurrence relation un = 1 2 (k + 3un–1). … 10 c 7 34 . 9. b find the value of k and the value of u4. Given also that u3 = 7u2 + 3. 4. 7. Given that t1 = 1. 9 a 5.biochemtuition. –3. 15. n > 1. 44. 43. n > 1. 21. find b the value of k. 8. b an expression for un in terms of n. n > 0. … are given by un = 3(un–1 – k). where k is a constant. n > 0. n ≥ 1. b find the possible values of k. 6. Given also that t3 = 12. u1 = –1 e un+1 = 11 Worksheet A continued un k . u1 = k d un = 2 – kun–1. n ≥ 1. … d 120. n ≥ 2. 17. n > 1. 79. 3. u1 = 1 b un+1 = kun + 5. where c is a constant. Solomon Press . c the value of u4. 11. 81. −10. 14 The terms of a sequence u1 . find the values of the constants a and b for each sequence. Given that u1 = – 4. … b 1. a un = 4un–1 + 3k. u2 .com faisal. … For each of the following sequences. find a the value of c. … e 4. n > 1. 3. … f 1. u1 = 2 c un = 4un–1 – k. a Find an expression for u3 in terms of the constant k. 9.2(1 – un). u3 = 1 A sequence is defined by un+1 = un + c. 19.5. n > 0.2 d un = 1 2 3 4 un + 2. … b 0.
8 + … Find the sum of the first 30 terms of each of the following arithmetic series. a Find the first term and common difference of the series. n = 48 6 b Given the first term. c Find the 40th term of the series. n. b Find the sum of the first 60 terms of the series. 9 The second and fifth terms of an arithmetic series are 13 and 46 respectively. a Find the common difference of the series. and the common difference.8 c a = 22.biochemtuition. b Find the sum of the first 40 terms of the series. l = −226 The first and third terms of an arithmetic series are 21 and 27 respectively. find the sum of each of the following arithmetic series. d = −5.5 + 15. + 1 12 + 2 56 + 4 16 + … c 17 + 9 + 1 + (−7) + … b 60 + 53 + 46 + 39 + … c 7 14 + 8 34 + 10 14 + 11 34 + … b a = 100. a 8 + 12 + 16 + 20 + … 4 b 30 + 27 + 24 + 21 + … For each of the following arithmetic series. the common difference.com b a = 3.9 + 11. n = 17 b a = 100.4. 8 The nth term of an arithmetic series is given by 7n + 16. a. b Find the sum of the first 25 terms of the series. 10 The third and eighth terms of an arithmetic series are 72 and 37 respectively. a. the common difference. the last term. n. a. d. find the sum of each of the following arithmetic series. find an expression for the nth term in the form a + bn. a. Solomon Press n(n + 1). l = 23.Dr. n = 20 5 c 8.2.com For each of the following arithmetic series. b Find the values of a and d. n = 36 c a = 19. and the last term. a Write down two equations relating the first term. 11 The fifth term of an arithmetic series is 23 and the sum of the first 10 terms of the series is 240.5. a a = 2. find the sum of each of the following arithmetic series. d = 1. a 7 + 9 + 11 + 13 + … 3 Worksheet B SEQUENCES AND SERIES a 4 + 10 + 16 + 22 + … 2 faisal. and the number of terms. l = 84. l = −20. . l. d = −8. l = 65 7 1 6 Given the first term. n = 32 c a = 28. a a = 8. d = 9. d. Faisal Rana C1 1 www. write down the common difference and find the value of the 40th term. Find the first term of the series and the sum of the first 35 terms of the series. a a = 60. a Find the first term and common difference of the series. d. n = 55 Given the first term. d = 13. and the number of terms. 12 a Prove that the sum of the first n natural numbers is given by 1 2 b Find the sum of the natural numbers from 30 to 100 inclusive. of the series.rana@biochemtuition.2 + 13. d = 3. l. l = 136.
rana@biochemtuition. a Find the common difference of the series. Faisal Rana www. sixth and seventh terms of an arithmetic series are (5 − t).com C1 SEQUENCES AND SERIES 13 Write down all the terms in each of the following series summations. b Find the number of positive terms in the series. 2t and (6t − 3) respectively. c find the sum of the first r terms of the series.Dr. of the first n terms of an arithmetic series is given by Sn = 2n2 + 5n. c Find an expression for the nth term of the series. 30 ∑ (4r + 7) d r =3 50 ∑ r =10 ( ) r+2 4 find the value of n. 19 The sum of the first six terms of an arithmetic series is 213 and the sum of the first ten terms of the series is 295.biochemtuition. b Find the sum of the first 25 terms of the series. 22 The fifth. b Find the sum of the first 30 terms of the series. a Find the value of the constant k. 21 The first three terms of an arithmetic series are (k + 2). b Find the eighth term of the series. a Evaluate S8. b all positive integers less than 200 that are divisible by 3.com faisal. 20 The sum. (2k + 3) and (4k − 2) respectively. a Find the first term and common difference of the series. a Find the first term of the series. Sn . 17 An arithmetic series has common difference −11 and tenth term 101. a Find the value of the constant t. a 5 ∑ (2r + 3) b r =1 14 9 ∑ (18 − 3r ) c r =1 10 ∑ (4r − 1) d r =4 18 ∑ r =11 (10 − 12 r ) Evaluate a 20 ∑ (3r + 1) r =1 15 Worksheet B continued Given that b 45 ∑ (90 − 2r ) c r =1 n ∑ (4r − 6) = 720. r =1 16 Find the sum of a all even numbers between 2 and 160 inclusive. Solomon Press . the sum of the first n terms of the series. 18 The first and fifth terms of an arithmetic series are 17 and 27 respectively. b find the value of r. Given that the rth term of the series is 132. c all integers divisible by 6 between 30 and 300 inclusive. c Hence find the maximum value of Sn . b Find the sum of the first 18 terms of the series.
4 Three consecutive terms of an arithmetic series are (7k − 1). a show that a + 4d = 14. Ahmed pays in a total amount. 5 a Evaluate 30 ∑ 4r . 2 The third and seventh terms of an arithmetic series are 5 6 and 2 13 respectively. of 20n(n + 24). c By forming and solving a suitable equation. b Find the smallest value of n for which the nth term of the series is greater than 300. Faisal Rana C1 1 www. (5k + 3) and (4k + 1) respectively. find the number of years of paying into the schemes after which Carol and Ahmed will have paid in the same amount in total. where k is an exact fraction to be found. r =1 ii 30 ∑ (8r − 5) . b Find the smallest positive term in the series. He pays in £500 in the first year and the amount he pays in increases by £40 in each subsequent year. Given that the sum of the first nine terms of the series is 126.5.Dr. Solomon Press . r =1 b Using your answer to part a.com faisal. with the amount increasing by £60 each year. c show that the sum of the positive terms of the series is given by r(4r − 3). c find the sum of the first 32 terms of the series. r =1 6 Ahmed begins making annual payments into a savings scheme. a Find the amount that Ahmed pays into the scheme in the eighth year. Carol starts making payments into a similar scheme at the same time as Ahmed. b Show that the sum of the first n terms of the series is given by kn(9n − 5). a Find the first term and common difference of the series. or otherwise. 3 An arithmetic series has first term a and common difference d.biochemtuition. a Find the value of the constant k. She pays in £400 in the first year. Given also that the sum of the first 15 terms of the series is 277.com Worksheet C SEQUENCES AND SERIES The third term of an arithmetic series is −10 and the sum of the first eight terms of the series is 16.rana@biochemtuition. evaluate i 30 ∑ (4r + 1) . a Find the first term and common difference of the series. b Show that during the first n years. Given also that the series has r positive terms. b find the values of a and d. in pounds.
rana@biochemtuition. Solomon Press . instead of by 250. c Hence. c An arithmetic series has first term 40 and common difference −3. i find how many calendars the publisher will sell during the sixth year of production. iii the sum of the first ten terms of the series. The publisher sells 2400 calendars during the first year of production and forecasts that the number it will sell in subsequent years will increase by 250 each year. Find the first term of the series. of the first r terms of an arithmetic series with first term a and rth term l is given by Sr = 1 2 r(a + l). of the first n terms of an arithmetic series with first term a and common difference d. 12 A publisher decides to start producing calendars. find to the nearest unit the value of C such that the publisher would sell a total of 40 000 calendars during the first ten years of production. 9 The first and fourth terms of an arithmetic series are x and (2x + 3) respectively.Dr.com C1 SEQUENCES AND SERIES 7 a Prove from first principles that the sum of the first n positive even numbers is n(n + 1). Faisal Rana www. 10 An arithmetic series has first term a and common difference d. prove that the series is arithmetic. of the first n terms of a series is given by Sn = 2n(16 − n). 13 a Prove that the sum. a Show that the sixth term of the series is 10. S8 = 2(S6 − S2). b The 18th term of an arithmetic series is 68 and the sum of the first 18 terms of the series is 153. a Find expressions in terms of x for i the seventh term of the series. ii the common difference of the series.com faisal. Given that the sum of the first twenty terms of the series is equal to the sum of the next ten terms of the series.biochemtuition. b If the number of calendars sold in subsequent years increases by C each year. ii show that the publisher will sell a total of 35 250 calendars during the first ten years of production. 8 a State the formula for the sum. Find the sum of the positive terms of the series. show that the ratio a : d = 11 : 2. 11 The sum. b Find an expression for the nth term of the series in the form a + bn. Sn . b find the value of x. Sr . Worksheet C continued b Find the sum of the integers between 200 and 800 that are not divisible by 4. Given also that the 20th term of the series is 52. Sn . a According to this forecast. b Prove that for any arithmetic series.
a Find the value of the constant t. (3) A sequence of terms {un} is defined. (3) c Find the value of N for which the sum of the first N natural numbers is 4950. (3) c state the value of u25 and give a reason for your answer. (2) A sequence is defined by the recurrence relation ur = ur–1 + 4. (1) (3) r =1 3 4 5 The first three terms of an arithmetic series are t. (2) b Find the 16th term of the series.rana@biochemtuition. Solomon Press (6) . n > 0. b Evaluate 20 ∑ ur . a Find the first term and common difference of the series. (3) b Evaluate 20 ∑ (5r − 1) . Sn . (2) a Find the sum of the integers between 1 and 500 that are divisible by 3. (4) b Find the sum of the first 25 terms of the series. a Write down the first five terms of the sequence. 6 b find the value of k. where k is a non-zero constant. (4) b An arithmetic series has first term −1 and common difference 6. u1 = 3. Given that t3 = 3. (4) c Find the sum of the first 20 terms of the series. by the recurrence relation un+1 = k + un2. 8 (3) A sequence is defined by the recurrence relation tn+1 = 4 – ktn. (2t − 5) and 8. where k is a positive constant. Verify by calculation that the largest value of n for which the sum of the first n terms of the series is less than 2000 is 26. (3) Given also that u3 = 1.6 respectively.com faisal. Given that u1 = 1. show that k = –1 + 1 2 6. for n ≥ 1. of the first n terms of an arithmetic series with first term a and common difference d is given by Sn = 1 2 n[2a + (n − 1)d]. (2) a State the formula for the sum of the first n natural numbers.Dr. t1 = –2.biochemtuition. (3) r =3 7 a Prove that the sum. a find expressions for u2 and u3 in terms of k. Faisal Rana C1 1 2 www. r > 1. (1) b Find the sum of the natural numbers from 200 to 400 inclusive.com SEQUENCES AND SERIES Worksheet D The second and fifth terms of an arithmetic series are 40 and 121 respectively.
(3) The first three terms of an arithmetic series are (k + 4). £42 for each of the next three months and so on. n ≥ 1. the sum of the first n terms of the series. where n is a positive integer. n ≥ 1. (4) c Find the maximum value of Sn . The allowance is to be £40 for each of the first three months. (3) r =1 b Hence. (4) b Show that the total amount. show that 2n ∑ (6r + 4) = n(9n + 7). Given that u2 + u4 = 6 and that k is a positive constant. (5) b show that u3 = 3u1.biochemtuition. … is such that the sum of the first 30 terms is 570. (2) Given that the sum of the first n terms of the series is 270. t3 . a Find the common difference of the series. a show that k = 16 3. a Show that k2 − 7k + 6 = 0. Find the sum of the first 30 terms of the sequences defined by 11 a un = 3tn . n ≥ 1. 13 14 (3) The third and eighth terms of an arithmetic series are 298 and 263 respectively. (3) Tom’s parents decide to pay him an allowance each month beginning on his 12th birthday. (4) 12 A sequence is defined by un+1 = un − 3. b find the 15th term of the series. a Find the total amount that Tom will receive in allowances before his 14th birthday. is given by 12n(4n + 39). (2) b vn = tn + 2. t2 . in pounds. increasing by £2 per month after each three month period. (3) b Find the number of positive terms in the series. b find the value of n. (4) Solomon Press . (4k − 2) and (k2 − 2) respectively.rana@biochemtuition. 10 (4) A sequence of terms t1 . Find the sum of the first 45 terms of this sequence. where k is a constant. n ≥ 1. (3) a Find and simplify an expression in terms of n for n ∑ (6r + 4) . un .com faisal. that Tom will receive in allowances in the n years after his 12th birthday.Dr. is given by un = kn − n. (2) c wn = tn + n. Faisal Rana www. (4) r = n +1 15 The nth term of a sequence. (2) Given also that the common difference of the series is positive. Worksheet D continued a Find the 20th term of the series.com C1 SEQUENCES AND SERIES 9 An arithmetic series has first term 6 and common difference 3. u1 = 80.
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https://www.caretxdigital.com/how-to-calculate-single-phase-voltage-from-3-phase/
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# How To Calculate Single Phase Voltage From 3
By | November 10, 2015
# How To Calculate Single Phase Voltage From 3-Phase
Generating electrical power is a complex and often dangerous task, hence the need for precise calculations. Whether you are professionally involved in the industry or just looking for a better understanding of how electricity works, it's important to be aware of the different ways of calculating single-phase voltage from 3-phase power sources. In this article, we'll take a closer look at the most effective ways of achieving that goal.
## Basics of 3-phase Power
Before diving into the specifics of how to calculate single-phase voltage from 3-phase power, it's important to have a basic understanding of the principles behind it. Three-phase power is a type of alternating current (AC) electricity, which operates by using three wires instead of the two used in traditional household circuits. Generally speaking, this type of power supply is used in industrial applications due to its ability to efficiently deliver more power than a standard single-phase source.
## Different Ways To Calculate Single-Phase Voltage
There are several different methods used to calculate single-phase voltage from 3-phase power sources. The most common techniques include using the line-to-line voltage method, the line-to-ground voltage method, or the balanced three-phase method. Each will be explained in detail below.
### Line-to-Line Voltage Method
The line-to-line voltage method is the simplest way to calculate single-phase voltage from 3-phase power sources. This method involves measuring the voltage between any two phases, then dividing the result by three. This will give you the single-phase voltage. It is important to note that all of the three-phase sources must be perfectly balanced for an accurate result to be achieved.
### Line-to-Ground Voltage Method
The line-to-ground voltage method is a slightly more sophisticated approach for calculating single-phase voltage from 3-phase sources. In this method, you measure the voltage between one of the three phases and neutral, then divide the result by two. Again, the three-phase sources must be perfectly balanced for this technique to be effective.
### Balanced Three-Phase Method
Finally, the balanced three-phase method is a more complex but accurate way of calculating single-phase voltage from 3-phase power sources. This approach requires the use of a mathematical formula known as the "balance star" equation, which is used to calculate the single-point sum of all the three-phase voltages. Once the single-point sum is calculated, simply divide it by three to get the single-phase voltage.
## Conclusion
By following the steps outlined in this article, you should now have a much better understanding of how to calculate single-phase voltage from 3-phase sources. Keep in mind that it is important to be aware of the different approaches in order to ensure accuracy and safety when dealing with electrical power. Remember to always use the appropriate tools and procedures to make sure your calculations are correct.
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https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/Rubio/ackclaude.trs.Thm12:EMB:NO.html.lzma
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Term Rewriting System R:
[X, Y]
ackin(0, X) -> ackout(s(X))
ackin(s(X), 0) -> u11(ackin(X, s(0)))
ackin(s(X), s(Y)) -> u21(ackin(s(X), Y), X)
u11(ackout(X)) -> ackout(X)
u21(ackout(X), Y) -> u22(ackin(Y, X))
u22(ackout(X)) -> ackout(X)
Termination of R to be shown.
R
Dependency Pair Analysis
R contains the following Dependency Pairs:
ACKIN(s(X), 0) -> U11(ackin(X, s(0)))
ACKIN(s(X), 0) -> ACKIN(X, s(0))
ACKIN(s(X), s(Y)) -> U21(ackin(s(X), Y), X)
ACKIN(s(X), s(Y)) -> ACKIN(s(X), Y)
U21(ackout(X), Y) -> U22(ackin(Y, X))
U21(ackout(X), Y) -> ACKIN(Y, X)
Furthermore, R contains one SCC.
R
DPs
→DP Problem 1
Remaining Obligation(s)
The following remains to be proven:
Dependency Pairs:
ACKIN(s(X), s(Y)) -> ACKIN(s(X), Y)
U21(ackout(X), Y) -> ACKIN(Y, X)
ACKIN(s(X), s(Y)) -> U21(ackin(s(X), Y), X)
ACKIN(s(X), 0) -> ACKIN(X, s(0))
Rules:
ackin(0, X) -> ackout(s(X))
ackin(s(X), 0) -> u11(ackin(X, s(0)))
ackin(s(X), s(Y)) -> u21(ackin(s(X), Y), X)
u11(ackout(X)) -> ackout(X)
u21(ackout(X), Y) -> u22(ackin(Y, X))
u22(ackout(X)) -> ackout(X)
Termination of R could not be shown.
Duration:
0:00 minutes
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https://finegrades.org/seven-questions-test-this-is-one-of-the-question-i-will/
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Seven questions test.this is one of the question i will
1. suppose seven individuals enjoy going to the comedy club. Their demand is as follows.
PersonWillingness to pay
Allison 20
Beatrice 18
Cally 16
David 14
Ezekiel 12
Francesca 10
Gertrude 8
Suppose the comedy club had a monopoly and a marginal cost of \$7 per entrant. Suppose the club could perfectly price-discriminate. That is, it could charge each customer a different price equal to his or her maximum willingness to pay. How many tickets would the comedy club sell?
2
The figure below shows the market for one hour of economics tutoring at your college. Imagine that the market for economics tutoring could be perfectly competitive, controlled by a monopolist who charges a single price or a monopolist who charges each customer a different price. Use the information in the diagram to answer the questions below.
Part 1 (0.3 point)
See HintHow much is total surplus if the market is perfectly competitive?
\$
Part 2 (0.3 point)
See HintHow much is total surplus if the market is controlled by a single-price monopolist?
\$
Part 3 (0.3 point)
See HintSuppose the single-price monopolist started charging all customers the maximum price they are willing to pay. How much additional surplus is created?
\$
3. A diner has no competition when it comes to its famous Reuben sandwich combo plate, for which the graph shows the diner’s demand (D), marginal cost (MC), and marginal revenue (MR) curves. The price of \$20 is based on the MR = MC rule for profit maximization. The rectangular region shown represents the net revenue from sales of the sandwich (total revenue from Reuben combo sales minus total variable costs associated with Reuben combo sales).
Now, suppose the diner decides to raise the price during the lunch hour, which accounts for 60% of Reuben combo sales, knowing that its lunch-hour patrons are the most loyal buyers of the Reuben combo and also that they are locked into the lunchtime slot by their work schedules. The diner raises the price just enough not to lose any lunch-hour buyers. Use the area tool to outline the region representing the resulting additional net revenue from the price increase. Your answer should be a rectangle drawn with four corners.
SEE ATTACHMENT SCREENSHOT2017 FOR GRAPH
Part 2 (0.7 point)
See HintAs a result of the revised price structure, the diner’s net revenue from Reuben-combo sales goes from \$ to \$ per day.
4. For a firm that is a price maker for its product, the graph shows the demand (D) and marginal revenue (MR) curves, as well as the marginal cost curve (MC), which is also the average variable cost (AVC) curve.
The graph also shows the net revenue (total revenue from product sales, minus total variable costs) when the profit-maximizing rule MR = MC is applied.
Finally, the graph shows two price points the firm is considering for a price-discrimination plan. If the firm is able to carry out this plan, getting the higher price from all those willing to pay it and the lower price from those who are not, outline the region representing the resulting net revenue. Use the area tool to draw the new net revenue region on top of the old one. Your answer should be drawn with six corners.
see attachments
The figure below shows the market for one hour of economics tutoring at your college. Imagine that the market for economics tutoring could be perfectly competitive, controlled by a monopolist who charges a single price or a monopolist who charges each customer a different price. Use the information in the diagram to answer the questions below.
Part 1 (0.3 point)
See HintHow much is total surplus if the market is perfectly competitive?
\$
Part 2 (0.3 point)
See HintHow much is total surplus if the market is controlled by a single-price monopolist?
\$
Part 3 (0.3 point)
See HintSuppose the single-price monopolist started charging all customers the maximum price they are willing to pay. How much additional surplus is created?
\$
Metropolitan Opera tickets are most expensive on Saturday night. Nevertheless, a very limited number of student rush tickets are often available, and a lucky student can wind up paying \$20 for a \$250 seat. The student rush tickets are available on a first-come, first-serve basis.
Part 1 (0.3 point)
See HintWhy does the opera company offer these low-priced tickets?
Choose one or more:A. This creates a frenzy for last-minute student tickets.B. It wants students to have access to shows at a low price.C. It is able to sell seats that would otherwise go unsold.D. It knows that all students have an elastic demand for opera tickets.
Part 2 (0.3 point)
See HintHow does the opera company benefit from this practice?
Choose one or more:A. The Met is able to charge different prices to different groups, raising its profits.B. The Met can increase attendance among a group of people with more elastic demand by charging them a lower price.C. The Met is able to generate revenue from unsold seats without having to drop the price for nonstudents.D. The Met can effectively price-discriminate between students and nonstudents.
Part 3 (0.3 point)
See HintWhy are students, and not other groups of customers, offered the discounted tickets?
Choose one or more:A. In general, students have a more elastic demand.B. In general, students have a more inelastic demand.C. In general, students are more flexible and willing to buy tickets at the last minute.D. In general, students love opera.
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https://www.learningpod.com/question/middle-school-science-bowl-how-many-x-intercepts-are-there-in-the-parabola-with-the-equation-y/830d1d9d-2a73-4d39-a6d9-ca96304d4453
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# Middle School Science Bowl: How many x-intercepts are there in the parabola with the equation: y ...
## Question
How many x-intercepts are there in the parabola with the equation: y = 7${x}^{2}$ + 4x + 12?
Only exact matches will be considered correct.
## Explanation
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CC-MAIN-2018-09
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latest
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https://nubtrek.com/maths/vector-algebra/multiplication-of-vector-by-scalar/repeated-addition
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mathsVector AlgebraMultiplication of Vectors by Scalar
### Repeated addition of a Vector
click on the content to continue..
We have learned about addition of two or more vectors. Consider a vector being repeatedly added to itself. What is the result of the addition vec p + vec p?
• sum = 2 times vec p
• sum = 2 times vec p
• component form is required to find the sum
• directional cosines are required to find the sum
• sum =2 times |vec p|
The answer is 'sum = 2 times vec p'. If vec p = ai+bj+ck then vec p + vec p = 2ai+2bj+2ck. This equals 2 times vec p.
Repeated addition can be generalized to multiplication of vector by a scalar. Scalar multiplier is denoted with lambda to identify differently to the component values a, b, c. Note that the scalar multiplier and component values are real numbers. What is the result of lambda vec p if vec p = ai+bj+ck?
• lambda ai+bj+ck
• lambda i+ lambda j + lambda k
• lambda ai+ lambda bj + lambda ck
• lambda ai+ lambda bj + lambda ck
• lambda sqrt(a^2+b^2+c^2)
The answer is 'lambda a i+ lambda b j + lambda c k'.
When vector is multiplied by a scalar, the vector scales up or down proportionally.
Multiplication of Vector by a scalar: For any vector vec p = ai + bj+ ck in bbb V and scalar lambda in RR
lambda vec p= lambda ai+ lambda bj + lambda ck
Can a vector vec p be divided by a real number lambda?
• Yes. Division is inverse of multiplication.
• Yes. Division is inverse of multiplication.
• No. Division of vector is not defined.
The answer is 'Yes. Division is inverse of multiplication.'
If the given vector is vec p = ai+bj+ck then
vec p -: lambda
quad quad = 1/lambda xx vec p
`quad quad = a/lambda i+b/lambda j+c/lambda k
slide-show version coming soon
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# How do you write sqrtx^5 as an exponential form?
Is it sqrt(x^5) ?then it would be ${x}^{\frac{5}{2}}$
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## Boundary Value Problems
Impact Factor 0.819
Open Access
# Erratum to: Hierarchies of difference boundary value problems
Boundary Value Problems20122012:66
https://doi.org/10.1186/1687-2770-2012-66
Received: 25 May 2012
Accepted: 30 May 2012
Published: 28 June 2012
## Erratum to: Boundary value problems, Volume 2011, Article ID 743135
1. (1)
The following paragraph needs to be inserted immediately after Theorem 4.2:
It is important to note that the spectral parameter in the original boundary value problems given in cases (1)-(9) of Table 1 for Theorem 4.2 must first, without loss of generality, be shifted so as to ensure that all the eigenvalues are greater than zero. Similarly, for cases (10)-(12) of Table 1 for Theorem 4.2, the spectral parameter must be shifted so that the original boundary value problem has the least eigenvalue 0. Having made these shifts we then take $z\left(n\right)$ to be a solution to (1.1) for $\lambda ={\lambda }_{0}=0$, i.e., throughout the paper we set ${\lambda }_{0}=0$.
1. (2)
In Corollary 4.4 and its proof, there were typographical errors as well as notation that was not apparent. These should read as follows:
Corollary 4.4 If${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$are the eigenvalues of any one of the original boundary value problems (1)-(9), in Theorem 4.2, with corresponding eigenfunctions${u}_{1}\left(n\right),\dots ,{u}_{s+l+m+1}\left(n\right)$, then
1. (i)
${\lambda }_{0}=0$, ${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$ are the eigenvalues of the corresponding transformed boundary value problems (1)-(3), in Theorem 4.2, with corresponding eigenfunctions $\frac{1}{z\left(n-1\right)c\left(n-1\right)}$, ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m+1}\left(n\right)$;
2. (ii)
${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$ are the eigenvalues of the corresponding transformed boundary value problems (4)-(9), in Theorem 4.2, with corresponding eigenfunctions ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m+1}\left(n\right)$.
Also, if${\lambda }_{0}=0$, ${\lambda }_{1},\dots ,{\lambda }_{s+l+m}$are the eigenvalues of any one of the original boundary value problems (10)-(12), in Theorem 4.2, with corresponding eigenfunctions$z\left(n\right)$, ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m}\left(n\right)$, then${\lambda }_{1},\dots ,{\lambda }_{s+l+m}$are the eigenvalues of the corresponding transformed boundary value problems (10)-(12), in Theorem 4.2, with corresponding eigenfunctions${u}_{1}\left(n\right),\dots ,{u}_{s+l+m}\left(n\right)$.
Proof By Theorems 2.1, 3.2, 3.3, 3.4 we have that (2.1) transforms eigenfunctions of the original boundary value problems (1)-(9) to eigenfunctions of the corresponding transformed boundary value problems. In particular, if ${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$ are the eigenvalues of one of the original boundary value problems, (1)-(9), with eigenfunctions ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m+1}\left(n\right)$, then:
1. (i)
$\frac{1}{z\left(n-1\right)c\left(n-1\right)}$, ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m+1}\left(n\right)$ are the eigenfunctions of the corresponding transformed boundary value problem, (1)-(3), with eigenvalues ${\lambda }_{0}=0$, ${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$. Since the transformed boundary value problems, (1)-(3), have $s+l+m+2$ eigenvalues, it follows that ${\lambda }_{0}=0$, ${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$ constitute all the eigenvalues of the transformed boundary value problem;
2. (ii)
${u}_{1}\left(n\right),\dots ,{u}_{s+l+m+1}\left(n\right)$ are the eigenfunctions of the corresponding transformed boundary value problem, (4)-(9), with eigenvalues ${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$. Since the transformed boundary value problems, (4)-(9), have $s+l+m+1$ eigenvalues, it follows that ${\lambda }_{1},\dots ,{\lambda }_{s+l+m+1}$ constitute all the eigenvalues of the transformed boundary value problem.
Also, again by Theorems 2.1, 3.2, 3.3, 3.4 we have that (2.1) transforms eigenfunctions of the original boundary value problems (10)-(12) to eigenfunctions of the corresponding transformed boundary value problems. In particular, if ${\lambda }_{0}=0$, ${\lambda }_{1},\dots ,{\lambda }_{s+l+m}$ are the eigenvalues of one of the original boundary value problems, (10)-(12), with eigenfunctions $z\left(n\right)$, ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m}\left(n\right)$, then ${u}_{1}\left(n\right),\dots ,{u}_{s+l+m}\left(n\right)$ are the eigenfunctions of the corresponding transformed boundary value problem, (10)-(12), with eigenvalues ${\lambda }_{1},\dots ,{\lambda }_{s+l+m}$. Since the transformed boundary value problems, (10)-(12), have $s+l+m$ eigenvalues, it follows that ${\lambda }_{1},\dots ,{\lambda }_{s+l+m}$ constitute all the eigenvalues of the transformed boundary value problem. □
## Declarations
### Acknowledgements
SC was supported by NRF grant no. IFR2011040100017.
## Authors’ Affiliations
(1)
School of Mathematics, University of the Witwatersrand
## Copyright
© Currie and Love; licensee Springer 2012
This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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# Tagged Questions
Questions related to real and complex logarithms.
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### Complex logarithms when computing real-valued integral
My question arise when I try to calculate real-valued integral, specifically, I want to evaluate the integral $$\int_0^1 \frac{\ln \left(\frac{x^2}{2}-x+1\right)}{x} dx$$ ...
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### What is “8 log 2”?
When someone says "8 Log 2" what does this equate to in writing? Does it mean the following? $$\log _{2} 8$$ And if so, what is the value of this? When I plug those numbers into this log ...
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### How to solve $3(a+1)(b+1)=3^a \times 2^b$?
Hi I'm new to logarithms and not sure how to solve equations involving logarithms. I managed to find this equation to answer a problem solving question, however now I do not know how to solve the ...
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# Pound Force Inch to Pound Force Foot Conversion
## 27.06 Pound Force Inch to Pound Force Foot Conversion - Convert 27.06 Pound Force Inch to Pound Force Foot (lbf∙in to lbf∙ft)
### You are currently converting Torque units from Pound Force Inch to Pound Force Foot
27.06 Pound Force Inch (lbf∙in)
=
2.255 Pound Force Foot (lbf∙ft)
Pound Force Inch : The pound force inch is a unit of torque which is equal one-twelfth of a foot-pound. Its abbreviation is lbf•in. It equal to 0.112984829 newton meters.
Pound Force Foot : The pound force-foot, or pound-foot (lb•ft or lbf•ft) is a unit of torque which is a pseudovector. It is equal to 1.3558179483314004 newton meters. It is defined as the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point. The unit pound-foot is in order to minimize confusion with the foot-pound as a unit of work.
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# In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten cryolite Na3AlF6, resulting in the reduction of the Al2O3 to pure aluminum. Suppose a current of 920.A is passed through a Hall-Heroult cell for 86.0 seconds. Calculate the mass of pure aluminum produced. Be sure your answer has a unit symbol and the correct number of significant digits.
Question
267 views
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide
Al2O3
dissolved in molten cryolite
Na3AlF6
, resulting in the reduction of the
Al2O3
to pure aluminum.
Suppose a current of
920.A
is passed through a Hall-Heroult cell for
86.0
seconds. Calculate the mass of pure aluminum produced.
Be sure your answer has a unit symbol and the correct number of significant digits.
check_circle
Step 1
The equation of deionization for aluminum is
Step 2
The value of valency for the above equation is 3, therefore
Step 3
The mass discharged...
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# Is Allan variance still relevant?
I've been working with Allan variance for a while and I'm not really super excited about is purpose.
I understand that noise time recordings, if they include flicker noise, diverge in time. In other words the standard deviation increases logarithmically, but why is this a problem? Modern computers can calculate the st dev as a function of window size without any problems for millions of data points.
The plot of Allan dev against observation time multiples is nice, but your data has to be very clean of deterministic signals otherwise very ugly things appear. Also you can see different noise process in the std dev vs window size also or PSD.
It seems to me that this tool was mainly useful decades ago when computer were much more limited and other problems like dead time in measurements existed.
Am I missing something?
## 1 Answer
I currently work in the design of atomic clocks and precision frequency sources and pleased to report that the Allan Variance is still quite relevant and useful. In fact it's utility extends to convenient characterization of many non-stationary processes, well beyond its primary tool as a frequency stability assessment. (And as mentioned in its comments, it’s use to assess stability of accelerometers).
Traditional statistical approaches using the first and second order moments (mean and standard deviation, or equivalently squared as the variance) cannot be computed for non-stationary processes, since the variance will increase the longer you compute the parameter on the data set (diverging). The Allan Variance or Two-Sample Variance is a convenient statistical tool to provide a metric similar to the variance for non-stationary processes. It basically converts non-stationary data to stationary for a consistent metric that can be used to compare competing products or solutions. Applicable to other processes, I personally like how I can use the Allan Variance to quickly assess an optimum averaging time (what we would call the "Flicker Floor" if the independent variable is frequency), for which if we average up to that time we can get an improved estimate of the mean, but if we average any longer, the noise of that estimate only degrades. For example, this can then be applied to optimum time durations for channel equalization and estimation, or any process where we are computing a longer term average to estimate a parameter.
The following graphic copied from a NIST presentation shows how ADEV can reveal the different noise processes within a signal, and where there effects would be of concern (in terms of observation times, $$\tau$$). This was from a 2015 presentation by Marc Weiss and Kishan Shenoi.
I have further detailed the Allan Variance and its practical purpose at these other links starting with the link copied below (that includes links to other related posts):
Allan Variance vs Autocorrelation - Advantages
• Very cool! Thanks for the nice answer. :-)
– Peter K.
Nov 14, 2021 at 19:08
• The place where I discovered Allan Variance was in characterizing IMU outputs -- gyro and accelerometer data, basically. It's useful for all the reasons given above. Nov 14, 2021 at 20:35
• Thanks for the answer Im glad also it is still used. About non stationary process, the standard deviation is sometimes well behaved in relation to the window size. Tipically in flicker noise or noises that have a "nice" PSD, the std dev increases smoothly with window size. I wonder however what happens when you have noise sources that are non stationary but also unruly, like for examples, people playing with the temperature of the room where the clock is but not at night nor weekends (or holidays). Certainly the Std dev will not be useful, but I wonder if the Allan Variance is more robuest in Nov 16, 2021 at 21:10
• @Ralph Yes the Allan Deviation has nice characteristics for each type of noise in it's slope versus averaging time (tau) on a log scale. A white noise process will go down at 1/root-tau, flicker noise will be constant at all tau, random walk noise will go up as 1/root-tau, sinusoidal noise will have a clear pattern etc. I added a plot that helps demonstrate this. Further there are other variants (modified ADEV, etc for their ability to discern certain types of noise processes better). Nov 18, 2021 at 22:45
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How to do well in Inference Based questions
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Hi All,
I am consistently getting all inference based questions wrong both in CR and RC in my tests.Also global questions are a big worry for me.My test is also very near.Can anyone suggest, how shall i improve my accuracy in inference based questions.
Regards
Nikhil
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Re: How to do well in Inference Based questions [#permalink]
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Expert's post
nikhilarora88 wrote:
Hi All,
I am consistently getting all inference based questions wrong both in CR and RC in my tests.Also global questions are a big worry for me.My test is also very near.Can anyone suggest, how shall i improve my accuracy in inference based questions.
Regards
Nikhil
In the inference based questions, you are looking for the conclusion of the stimulus. Think of a debate. A speaker is in favor of the motion. He is giving you data to prove his point, to support the motion, to establish his position. The data is given in the form of premises (which we assume to be true) in the stimulus. In Inference questions, you are given the data and you need to figure out the position the author is trying to establish. This means, you will be able to 'infer' the conclusion from the stimulus. The conclusion will not be 'new information'. It will be something that the stimulus tells you.
You can check out the Veritas CR1 and CR2 books for further discussion on how to identify the inference and for examples and practice questions.
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Re: How to do well in Inference Based questions [#permalink]
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Dear Nikhil,
Summarize paragraphs (or every 3 lines if the passage is one large paragraph). The main point is a combination of the paragraph summaries. You can also check out our free concepts on Main point (in Reading Comprehension course). Do go through all the (three) free concepts. Our customers had reported considerable improvement after going through these concepts.
We also cover all the inferences (Generic to specific, specific to Generic, simple and complex combination) in the inference concepts in both CR and RC.
Just log on to the e-GMAT account to access the free concepts.
https://e-gmat.com/secure/register.php
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# When 20 mL of M//20NaOH is added to 10mL of M//10 HCI, the resulting solution will
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38.6 K+
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Turn blue litmus red. Turn phenolpthalein solution pink. Turns methy orange red. Will have no effect on either red or blue litmus
D
Solution :
mEq of NaOH = 20 xx (1)/(20) = 1 <br> mEq of HCI = 10 xx (1)/(10) = 1 <br> So, slat of S_(A)//S_(B) (NaCI) will be formed which do not hydrolyse and pH = 7. So at pH = 7, litmus has no effect. Since, blue litmus turns red a acidic solution, and red litmus turns blue, a basic solution. <br> i. phenolphtalein solution turns pink a basic solution of pH gt 7. <br> iii. Methy1 orange turns red a acidic solution of pH lt 7.
Image Solution
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## Find a point on x – axis which is equidistant from A(5,2) and B (1,-2)?
Question
Find a point on x – axis which is equidistant from A(5,2) and B (1,-2)?
in progress 0
1 month 2021-08-18T13:01:34+00:00 1 Answer 0 views 0
1. Let the point be P (k,0)
According to the mid point therem,
P(k,0)=((5+1)/2,(2-2)/2)
=(3,0)
Thus, k=3
Thus the point is P(3,0)
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# 4.6 Exponential and logarithmic models (Page 7/16)
Page 7 / 16
Does a linear, exponential, or logarithmic model best fit the data in [link] ? Find the model.
$x$ 1 2 3 4 5 6 7 8 9 $y$ 3.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034
Exponential. $\text{\hspace{0.17em}}y=2{e}^{0.5x}.$
## Expressing an exponential model in base e
While powers and logarithms of any base can be used in modeling, the two most common bases are $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}e.\text{\hspace{0.17em}}$ In science and mathematics, the base $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ is often preferred. We can use laws of exponents and laws of logarithms to change any base to base $\text{\hspace{0.17em}}e.$
Given a model with the form $\text{\hspace{0.17em}}y=a{b}^{x},$ change it to the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
1. Rewrite $\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}y=a{e}^{\mathrm{ln}\left({b}^{x}\right)}.$
2. Use the power rule of logarithms to rewrite y as $\text{\hspace{0.17em}}y=a{e}^{x\mathrm{ln}\left(b\right)}=a{e}^{\mathrm{ln}\left(b\right)x}.$
3. Note that $\text{\hspace{0.17em}}a={A}_{0}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}k=\mathrm{ln}\left(b\right)\text{\hspace{0.17em}}$ in the equation $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
## Changing to base e
Change the function $\text{\hspace{0.17em}}y=2.5{\left(3.1\right)}^{x}\text{\hspace{0.17em}}$ so that this same function is written in the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
The formula is derived as follows
Change the function $\text{\hspace{0.17em}}y=3{\left(0.5\right)}^{x}\text{\hspace{0.17em}}$ to one having $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the base.
$y=3{e}^{\left(\mathrm{ln}0.5\right)x}$
Access these online resources for additional instruction and practice with exponential and logarithmic models.
## Key equations
Half-life formula If $k<0,$ the half-life is Carbon-14 dating $t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}.$ is the amount of carbon-14 when the plant or animal died is the amount of carbon-14 remaining today is the age of the fossil in years Doubling time formula If $k>0,$ the doubling time is Newton’s Law of Cooling $T\left(t\right)=A{e}^{kt}+{T}_{s},$ where is the ambient temperature, and is the continuous rate of cooling.
## Key concepts
• The basic exponential function is $\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}b>1,$ we have exponential growth; if $\text{\hspace{0.17em}}0 we have exponential decay.
• We can also write this formula in terms of continuous growth as $\text{\hspace{0.17em}}A={A}_{0}{e}^{kx},$ where $\text{\hspace{0.17em}}{A}_{0}\text{\hspace{0.17em}}$ is the starting value. If $\text{\hspace{0.17em}}{A}_{0}\text{\hspace{0.17em}}$ is positive, then we have exponential growth when $\text{\hspace{0.17em}}k>0\text{\hspace{0.17em}}$ and exponential decay when $\text{\hspace{0.17em}}k<0.\text{\hspace{0.17em}}$ See [link] .
• In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See [link] .
• We can find the age, $\text{\hspace{0.17em}}t,$ of an organic artifact by measuring the amount, $\text{\hspace{0.17em}}k,$ of carbon-14 remaining in the artifact and using the formula $\text{\hspace{0.17em}}t=\frac{\mathrm{ln}\left(k\right)}{-0.000121}\text{\hspace{0.17em}}$ to solve for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ See [link] .
• Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See [link] .
• We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See [link] .
• We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See [link] .
• We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See [link] .
• Any exponential function with the form $\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ can be rewritten as an equivalent exponential function with the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k=\mathrm{ln}b.\text{\hspace{0.17em}}$ See [link] .
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Got questions? Join the online conversation and get instant answers!
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# Difference between dates in
#1
I’ve two dates `d1` and `d2`, how I could calculate the difference between them given an `interval`? i.e the steps in minutes, days, hours, days, …
0 Likes
#2
there is `dateDiff` function for this, it can be used as follows:
``````> d1 = DateTime.parse('2018-01-01', 'yyyy-MM-dd')
> d2 = DateTime.parse('2018-01-02', 'yyyy-MM-dd')
> DateTime.dateDiff(d1, d2, 'DAY')
1
> DateTime.dateDiff(d1, d2, 'HOUR')
24
> DateTime.dateDiff(d1, d2, 'WEEK')
0
``````
1 Like
How to compare datetime in jasmine test?
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# BlackAlps 2019 - Schnorr
Schnorr signature with a Blum-Blum-Schub PRNG to generate the nonce. Since the quadratic equation can be solved in the prime group the private key can be recover from two consecutive signatures.
### Description
An apprentice cryptographer decided to implement the Schnorr signature based on what he could read on Wikipedia.
Knowing that randomness is a critical component of Schnorr signatures, he decided to use and improve the Blum-Blum-Schub provably secure PRNG to generate randomness for signing.
You are given two consecutive signatures. The flag is hidden inside the secret key.
Author: Alexandre Duc
Points: 400
Category: Crypto
### Solution
The PRNG was given:
#PRNG initially based on the provably secure BlumBlumShub PRNG.
#Improved to make it even more secure and more efficient.
#q is reused from Schnorr. It does not affect the security of the system
#r has to be a random prime number for the security proof to hold.
def BlumBlumShub(seed, r, q):
n = r*q
return (pow(seed,2,n) + 2041*seed + 125)%n
As well as the parameters and two signatures:
p=11354821474068432448781088658925769089942262052492176911159145294614070565823651286293031376249291592372010121532367277047783330177517525191691692679609588994273170102573923175498072488475243086164453824784015024965831927574606549505482453151194968443619022898128537270653908101799461877175223151865737017706606910794098282193587012221895574003056105395032884664491604314170070549892345496174749893404409800494766306880932092622764671423653474372374467519485884622586290396130747886745583378206131995694632733623421886624382843330228686054793
q=572156407871896365511901114975216570393314889392720208096942688507966755120286519809169097
g=5574030578425995852347018776949350647797404891144700535846063285373847243050742195266604928099650295829841450171685065963427241819806737053177918304931329744526999213038118857349881673298873935654535365004383047860588199904864368957698991553618480203079249307017153046037144903664337773520632103167869602394350824178446678871690039864957571252278888675700929967210478372731269855538550660019512864773991864726863377657430184373137242042827800857538306846685368056656766635813056158480078121733190110412456091152108340586732091202814832395649
y=1302064652106120540591455669744918598912180127945547332820568085657757915418891353546741658531571543834359096864837598318932353117841399047530357264868697967684828173623493749674820397778180334312186802162498484373499213589669465086073572034808055890600153754294513520414147214915277791982218242108282472595977447873259754267985756686859151211332253888240703732797016283234944506058880767221813931135334602116033266358897129058776480192951940507572840153150631029293192856658960098994428668986190899631093861500235157282362462737614975550203
(s1,e1) = (444490141183918605123295561253971131255085954584610229585367281139870776740150637047636091, 102110058127261311389659710844857123602442947476844523271866378317865779717941)
(s2,e2) = (196039357254500202482847472390602792559927913500085718055042604432787316443058365700890688, 101276696144102062023436237077674567981954023536843878958158437487050712337249)
From the definition of Schnorr algorithm we have:
$s_1=k_1-x e_1$ $s_2=k_2-x e_2$
Where $x$ is the private key and $k_1$ and $k_2$ the nonces. Since the nonces are generetated from the Blum-Blum-Schub PRNG we have that:
$k_2 = k_1^2 + 2041 k_1 + 125$
Thus if we replace properly in the previous equations we have:
$k_1 = s_1 + x e_1$ $s_2 = k_1^2 + 2041 k_1 + 125 - x e_2 = (s_1 + x e_1)^2 + 2041 (s_1 + x e_1) + 125 - x e_2$
Since it is a quadratic equation and we know everything except $x$, we can solve it with sage:
sage: K = GF(q)
sage: P = PolynomialRing(K)
sage: x = P.gen()
sage: f = (s1+x*e1)^2 + 2041*(s1+x*e1)+125-x*e2-s2
sage: s = f.roots()
One of the solutions is the private key and the flag is the private key.
sage: unhexlify("{0:x}".format(int(s[1][0])))
'BA19{Schn0rr1sL1keECDSA}'
Written on November 8, 2019
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## Archive for the ‘Scheme’ Category
### Javascript and Lisp
As Douglas Crockford has said, javascript is Lisp is C's clothing. That is part of what make javascript fun to program, but I never realized how literally true that statement is until I found this fascinating article by Thomas Lord about GNU and Scheme, that claims:
I've read, since then, that up to around that point Brendan Eich had been working on a Scheme-based extension language for Netscape Navigator. Such was the power of the hegemony of the high level folks at Sun that the word came down on Mr. Eich: "Get it done. And make it look like Java." Staying true to his sense of Self, he quickly knocked out the first implementation of Mocha, later renamed Javascript.
Explains a lot. And shows what an under-appreciated genius Brendan Eich is.
### Learning Scheme: Euler Problem 3
Been a long time since I last posted; real life has a way of getting in the way. The past few months read like the back cover of A Series of Unfortunate Events: if you don't want to hear about a thwarted move, a house fire, 2 graduations, negotiations for a new job and a carrot salad with lots of garlic, then you should put this down and live someone else's life.
But I'm not complaining; everything looks like it's working out well.
Going back to Scheme, the next Project Euler problem is:
The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?
And straightforwardly:
``````(define gpf ; greatest prime factor
(lambda (n try) ; n is the number to find the factor of; try is the lowest number to try
(cond
((> try (sqrt n)) n) ; no need to try a factor higher than the square root; if we get here, n is prime
((divides try n) (gpf (/ n try) try)) ; divide out the current trial factor and try again
(else (gpf n (+ try 1))))))
(define euler3
(lambda (n) (gpf n 2)))``````
`(euler3 600851475143)` gives us the answer 6857 pretty quickly.
But this is getting boring; I'm not using the interesting parts of Scheme. This sort of problem cries out for a prime number generator, and that cries out for `call-with-current-continuation`, and that cries out for some more playing and learning. I'll see what I can do.
### Learning Scheme, Euler problem 2
The second project Euler problem is:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... Find the sum of all the even-valued terms in the sequence which do not exceed four million.
Continue reading ‘Learning Scheme, Euler problem 2’ »
### Learning Scheme
OK, so my new mini project is to learn Scheme. It can't be too hard, right? I mean Javascript is just Lisp in C's clothing, and I'm good at Javascript. I've installed PLT-Scheme and I figure the best way to learn it is solving Project Euler problems until I get bored.
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Copied to
clipboard
## G = C42.45D4order 128 = 27
### 27th non-split extension by C42 of D4 acting via D4/C2=C22
p-group, metabelian, nilpotent (class 3), monomial
Series: Derived Chief Lower central Upper central Jennings
Derived series C1 — C2×C4 — C42.45D4
Chief series C1 — C2 — C22 — C2×C4 — C42 — C2×C42 — C22.26C24 — C42.45D4
Lower central C1 — C2 — C2×C4 — C42.45D4
Upper central C1 — C2×C4 — C2×C42 — C42.45D4
Jennings C1 — C22 — C22 — C42 — C42.45D4
Generators and relations for C42.45D4
G = < a,b,c,d | a4=b4=1, c4=a2b2, d2=b, ab=ba, cac-1=ab2, dad-1=a-1, bc=cb, bd=db, dcd-1=a2b-1c3 >
Subgroups: 284 in 130 conjugacy classes, 54 normal (28 characteristic)
C1, C2 [×3], C2 [×4], C4 [×2], C4 [×4], C4 [×5], C22, C22 [×2], C22 [×8], C8 [×4], C2×C4 [×6], C2×C4 [×4], C2×C4 [×9], D4 [×12], Q8 [×2], C23, C23 [×2], C42 [×4], C22⋊C4 [×4], C4⋊C4 [×2], C4⋊C4, C2×C8 [×6], C22×C4 [×3], C22×C4 [×2], C2×D4 [×2], C2×D4 [×4], C2×Q8, C4○D4 [×4], C4×C8 [×2], C22⋊C8, C4⋊C8 [×2], C4⋊C8 [×2], C2×C42, C4×D4 [×2], C4×D4, C4⋊D4 [×2], C4⋊D4, C4.4D4, C41D4, C4⋊Q8, C22×C8, C2×C4○D4, D4⋊C8 [×4], C2×C4⋊C8, C42.12C4, C22.26C24, C42.45D4
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C2×C4 [×6], D4 [×4], C23, C22⋊C4 [×4], D8 [×2], SD16 [×2], C22×C4, C2×D4 [×2], D4⋊C4 [×4], C2×C22⋊C4, C8○D4 [×2], C2×D8, C2×SD16, (C22×C8)⋊C2, C2×D4⋊C4, C42⋊C22, C42.45D4
Smallest permutation representation of C42.45D4
On 64 points
Generators in S64
```(1 9 17 50)(2 55 18 14)(3 11 19 52)(4 49 20 16)(5 13 21 54)(6 51 22 10)(7 15 23 56)(8 53 24 12)(25 36 60 44)(26 41 61 33)(27 38 62 46)(28 43 63 35)(29 40 64 48)(30 45 57 37)(31 34 58 42)(32 47 59 39)
(1 30 21 61)(2 31 22 62)(3 32 23 63)(4 25 24 64)(5 26 17 57)(6 27 18 58)(7 28 19 59)(8 29 20 60)(9 45 54 33)(10 46 55 34)(11 47 56 35)(12 48 49 36)(13 41 50 37)(14 42 51 38)(15 43 52 39)(16 44 53 40)
(1 2 3 4 5 6 7 8)(9 10 11 12 13 14 15 16)(17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32)(33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56)(57 58 59 60 61 62 63 64)
(1 29 30 20 21 60 61 8)(2 19 31 59 22 7 62 28)(3 58 32 6 23 27 63 18)(4 5 25 26 24 17 64 57)(9 48 45 49 54 36 33 12)(10 56 46 35 55 11 34 47)(13 44 41 53 50 40 37 16)(14 52 42 39 51 15 38 43)```
`G:=sub<Sym(64)| (1,9,17,50)(2,55,18,14)(3,11,19,52)(4,49,20,16)(5,13,21,54)(6,51,22,10)(7,15,23,56)(8,53,24,12)(25,36,60,44)(26,41,61,33)(27,38,62,46)(28,43,63,35)(29,40,64,48)(30,45,57,37)(31,34,58,42)(32,47,59,39), (1,30,21,61)(2,31,22,62)(3,32,23,63)(4,25,24,64)(5,26,17,57)(6,27,18,58)(7,28,19,59)(8,29,20,60)(9,45,54,33)(10,46,55,34)(11,47,56,35)(12,48,49,36)(13,41,50,37)(14,42,51,38)(15,43,52,39)(16,44,53,40), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64), (1,29,30,20,21,60,61,8)(2,19,31,59,22,7,62,28)(3,58,32,6,23,27,63,18)(4,5,25,26,24,17,64,57)(9,48,45,49,54,36,33,12)(10,56,46,35,55,11,34,47)(13,44,41,53,50,40,37,16)(14,52,42,39,51,15,38,43)>;`
`G:=Group( (1,9,17,50)(2,55,18,14)(3,11,19,52)(4,49,20,16)(5,13,21,54)(6,51,22,10)(7,15,23,56)(8,53,24,12)(25,36,60,44)(26,41,61,33)(27,38,62,46)(28,43,63,35)(29,40,64,48)(30,45,57,37)(31,34,58,42)(32,47,59,39), (1,30,21,61)(2,31,22,62)(3,32,23,63)(4,25,24,64)(5,26,17,57)(6,27,18,58)(7,28,19,59)(8,29,20,60)(9,45,54,33)(10,46,55,34)(11,47,56,35)(12,48,49,36)(13,41,50,37)(14,42,51,38)(15,43,52,39)(16,44,53,40), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64), (1,29,30,20,21,60,61,8)(2,19,31,59,22,7,62,28)(3,58,32,6,23,27,63,18)(4,5,25,26,24,17,64,57)(9,48,45,49,54,36,33,12)(10,56,46,35,55,11,34,47)(13,44,41,53,50,40,37,16)(14,52,42,39,51,15,38,43) );`
`G=PermutationGroup([(1,9,17,50),(2,55,18,14),(3,11,19,52),(4,49,20,16),(5,13,21,54),(6,51,22,10),(7,15,23,56),(8,53,24,12),(25,36,60,44),(26,41,61,33),(27,38,62,46),(28,43,63,35),(29,40,64,48),(30,45,57,37),(31,34,58,42),(32,47,59,39)], [(1,30,21,61),(2,31,22,62),(3,32,23,63),(4,25,24,64),(5,26,17,57),(6,27,18,58),(7,28,19,59),(8,29,20,60),(9,45,54,33),(10,46,55,34),(11,47,56,35),(12,48,49,36),(13,41,50,37),(14,42,51,38),(15,43,52,39),(16,44,53,40)], [(1,2,3,4,5,6,7,8),(9,10,11,12,13,14,15,16),(17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32),(33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56),(57,58,59,60,61,62,63,64)], [(1,29,30,20,21,60,61,8),(2,19,31,59,22,7,62,28),(3,58,32,6,23,27,63,18),(4,5,25,26,24,17,64,57),(9,48,45,49,54,36,33,12),(10,56,46,35,55,11,34,47),(13,44,41,53,50,40,37,16),(14,52,42,39,51,15,38,43)])`
38 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 4A 4B 4C 4D 4E ··· 4J 4K 4L 4M 4N 8A ··· 8P order 1 2 2 2 2 2 2 2 4 4 4 4 4 ··· 4 4 4 4 4 8 ··· 8 size 1 1 1 1 2 2 8 8 1 1 1 1 2 ··· 2 4 4 8 8 4 ··· 4
38 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 2 4 type + + + + + + + + image C1 C2 C2 C2 C2 C4 C4 C4 D4 D4 D8 SD16 C8○D4 C42⋊C22 kernel C42.45D4 D4⋊C8 C2×C4⋊C8 C42.12C4 C22.26C24 C4⋊D4 C4⋊1D4 C4⋊Q8 C42 C22×C4 C2×C4 C2×C4 C4 C2 # reps 1 4 1 1 1 4 2 2 2 2 4 4 8 2
Matrix representation of C42.45D4 in GL4(𝔽17) generated by
0 1 0 0 16 0 0 0 0 0 16 2 0 0 0 1
,
16 0 0 0 0 16 0 0 0 0 4 0 0 0 0 4
,
5 12 0 0 5 5 0 0 0 0 2 13 0 0 2 15
,
5 12 0 0 12 12 0 0 0 0 2 13 0 0 0 15
`G:=sub<GL(4,GF(17))| [0,16,0,0,1,0,0,0,0,0,16,0,0,0,2,1],[16,0,0,0,0,16,0,0,0,0,4,0,0,0,0,4],[5,5,0,0,12,5,0,0,0,0,2,2,0,0,13,15],[5,12,0,0,12,12,0,0,0,0,2,0,0,0,13,15] >;`
C42.45D4 in GAP, Magma, Sage, TeX
`C_4^2._{45}D_4`
`% in TeX`
`G:=Group("C4^2.45D4");`
`// GroupNames label`
`G:=SmallGroup(128,212);`
`// by ID`
`G=gap.SmallGroup(128,212);`
`# by ID`
`G:=PCGroup([7,-2,2,2,-2,2,-2,2,112,141,723,1123,570,136,172]);`
`// Polycyclic`
`G:=Group<a,b,c,d|a^4=b^4=1,c^4=a^2*b^2,d^2=b,a*b=b*a,c*a*c^-1=a*b^2,d*a*d^-1=a^-1,b*c=c*b,b*d=d*b,d*c*d^-1=a^2*b^-1*c^3>;`
`// generators/relations`
×
𝔽
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| 2.65625
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CC-MAIN-2019-04
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https://www.atozmath.com/Conversion.aspx?q=Conversion&ST=Mass
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Home > Pre-Algebra calculators > Unit Conversion >> Mass Conversion calculator
Conversion -
Conversion - Mass
Formula : Mass:- 1 kilogram = 1000 grams; 1 gram = 1000 milligrams; 1 metric ton = 1000 kilograms; 1 metric ton = 1.102 short tons; 1 short ton = 2000 pounds; 1 pound = 16 ounces; 1 kilogram = 2.2046 pounds
5. Convert to
Decimal Place =
SolutionExampleUnit ConversionAll Problems
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| 2.890625
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CC-MAIN-2019-35
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https://keolistravelservices.com/difference-between-linear-and-square-feet/
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| 314,797,516
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If you have bought or sold a property, you can be familiar with the term “square foot.” yet the ax “linear foot” might not be as typical due to the fact that it is not regularly used during discussions around a residence or apartment floor plans, and you can be curious what does direct foot mean.
You are watching: Difference between linear and square feet
A direct foot is a type of measurement used when securing a freight trailer for relocating from one residence come the next, typically over long distances. Moving companies the use big freight trailers use linear feet to calculate the price of delivering your belongings.
## What is a straight Foot?
You might be wondering just how long is a direct foot. A linear foot is similar to a typical foot in the it is likewise equal come 12 customs (one foot). A linear foot differs slightly in that it procedures the street from point A to point B in a straight line. We most often use straight measurements when we travel—the mileage traveled by auto can be convert to direct measurement. Another thing you might want to know is what is a direct foot in fencing? If you have actually a 6 meter high, 100-meter long fence, your fence is 100 meters linear.
One an ext common usage of a straight foot measure is sewing; material purchased is measured in direct feet.
## Measuring in direct Feet:
One-dimensional measurement is described as linear. As soon as you measure the size or broad of a room, you measure up the room from one edge to the other and note the measure up in straight feet. Friend can additionally use straight feet to measure a broken window pane. If you space pretty awed in ~ how plenty of inches in a linear foot, the still sounds favor a straight foot. It actions 12 customs long and also is spread out over a straight or direct line.
The advantage of payment by direct foot is that you only need to pay for the space you need. Top top the contrary, as soon as renting a truck or relocating container, you need to pay because that the totality area even if it is or not you need it.
There are other techniques of measurement, such together cubic feet or square feet. But if you must determine just how much space you call for in a freight trailer or how much structure material friend need, you just need to measure up the direct length. The size or broad will currently be fixed.
In some instances, world might ask you just how much is a linear foot. Because that example, in shipping, carriers shot to calculate the delivery costs, or if you send anything and also wish to execute an assessment yourself, the crucial measurement is a linear foot. Only so much an are is available in a single trailer, and also you can get a highly comprehensive price to assist you in creating the remainder the your budget plan if friend know just how long your items will take straight feet.
Note: When functioning with neighborhood moving companies, placed your confidence in your skilled and also professional employees since. They understand how finest to measure the straight feet of your freight trailers and also can administer you with an estimate based upon that calculation.
## How to Calculate linear Foot?
If girlfriend wonder exactly how to measure up a direct foot, you must an initial measure the distance between allude A and point B in inches and divide the an outcome by 12 to find linear feet. That is simple, and also no complex calculation is involved, unlike once calculating square feet or cubic feet.
See more: Longest English Word Without Repeating Letters ? Longest Word In English
### Conclusion:
Although a square foot is a more common measurement, knowing about linear feet is same good. There will certainly be cases in which working with the straight foot of measure up is much more convenient than using square feet. Getting familiar with both will certainly surely save you time, money, and unnecessary fees in any kind of transaction you will usage it. Expertise the ins and also outs the its miscellaneous uses deserve to prove enormously helpful next time you resolve it.
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CC-MAIN-2022-33
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http://www.tzhehui.com/mathtype.html
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text/html
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Typing Math Problems into this Site - WebMath
for explanations and help with your homework problems!
# Typing Math Problems into this Site
Here are some tips to help you type your problem into a text-box on this site: The + key means plus, and - key means minus. * ([Shift]-[8]) means times and / means divide. 2+2/3*8 is "two plus two divided by 3 times 8." Webmath always follows the correct "order of operations" on everything you type. Webmath knows when to multiply. So if you have 2(x+1), you don't need to type it at 2*(x+1); type it like 2(x+1). Fractions are typed using the slash, or / key as the fraction bar. So, "two-thirds" would be typed as 2/3. "One-half" would be types as 1/2. Exponents are typed using the caret, or ^ key ([Shift]-[6] on your keyboard). So, 22 would be typed 2^2. x2 would be typed x^2. (x+5)2 would be typed (x+5)^2. You can put a fraction in an exponent. x2/3 should be typed like x^(2/3). With more complicated fractions you have to use parenthesis. For example if you typed x^2+1/x-5, you might think this means "the quantity 'x-squared plus 1' over the quantity 'x minus 5'." Actually, this site would correctly put 1/x as the only fraction. Instead, you should type it like this: (x^2+1)/(x-5). Webmath DOES NOT know about constant like Pi (=3.1415...), and e (=2.718..). If you need these, you can just put them in directly as numbers. Or, you can put P in for Pi and E in for e. Webmath will carry these through in a problem, and you can see what happened at the very end.
Quick! I need help with:
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https://oeis.org/A363092
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A363092 a(n) = 4*a(n-1) - 8*a(n-2) with a(0) = a(1) = 1. 0
1, 1, -4, -24, -64, -64, 256, 1536, 4096, 4096, -16384, -98304, -262144, -262144, 1048576, 6291456, 16777216, 16777216, -67108864, -402653184, -1073741824, -1073741824, 4294967296, 25769803776, 68719476736, 68719476736, -274877906944, -1649267441664, -4398046511104 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 REFERENCES Paul J. Nahin, An Imaginary Tale: The Story of sqrt(-1), Princeton University Press, Princeton, NJ. 1998, pp. 94-96. LINKS Table of n, a(n) for n=0..28. Index entries for linear recurrences with constant coefficients, signature (4,-8). FORMULA a(n) = 2^(3*n/2-1)*(2*cos(n*Pi/4) - sin(n*Pi/4)). O.g.f.: (1 - 3*x)/(1 - 4*x + 8*x^2). E.g.f.: exp(2*x)*(2*cos(2*x) - sin(2*x))/2. a(n+1) = a(n) iff n is a multiple of 4. MATHEMATICA LinearRecurrence[{4, -8}, {1, 1}, 29] CROSSREFS Cf. A000045, A008586, A088137, A088138. Sequence in context: A166870 A124350 A112611 * A212066 A336039 A364600 Adjacent sequences: A363089 A363090 A363091 * A363093 A363094 A363095 KEYWORD sign,easy AUTHOR Stefano Spezia, May 19 2023 STATUS approved
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Last modified February 22 22:43 EST 2024. Contains 370265 sequences. (Running on oeis4.)
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CC-MAIN-2024-10
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http://www.procasestudy.com/multiplying-dividing-polynomials-template/
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crawl-data/CC-MAIN-2021-17/segments/1618039398307.76/warc/CC-MAIN-20210420122023-20210420152023-00295.warc.gz
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# multiplying dividing polynomials template
multiplying dividing polynomials template is a multiplying dividing polynomials template sample that gives infomration on multiplying dividing polynomials template doc. When designing multiplying dividing polynomials template, it is important to consider different multiplying dividing polynomials template format such as multiplying dividing polynomials template word, multiplying dividing polynomials template pdf. You may add related information such as multiplying polynomials worksheet, multiplying polynomials practice, multiplying polynomials examples, multiplying polynomials by polynomials.
to multiply one term by another term, first multiply the constants, then multiply each variable together and combine the result, like this (press play): (note: i used “·” to mean multiply. in algebra we don’t like to use “×” because it looks too much like the letter “x”) two friends (alice and betty) challenge two other friends (charles and david) to individual tennis matches. they could play in any order, so long as each of the first two friends gets to play each of the second two friends. we can multiply them in any order so long as each of the first two terms gets multiplied by each of the second two terms.
– 5x + 7). step 1: distribute each term of the first polynomial to every term of the second multiplying polynomials. a polynomial looks like this: polynomial example. example of a polynomial this one has 3 terms multiplying and dividing polynomials : example question #1. multiply: \ displaystyle -2x(x^4-x^2+4). possible answers:., multiplying polynomials worksheet, multiplying polynomials worksheet, multiplying polynomials practice, multiplying polynomials examples, multiplying polynomials by polynomials.
monomial times polynomial: simply use the distributive property to multiply a use the long division format as follows:. polynomials can be multiplied and divided using algebra tiles or pencil and paper. multiplying and dividing example. expand and simplify . write the polynomial that represents the area. x(x. ) 2. + 6. example 1: simplify by multiplying the two monomials. each of the expression inside the parenthesis is a polynomial , multiplying polynomials notes, multiplying polynomials lesson, multiplying polynomials lesson, multiplying polynomials box method, multiplying polynomials khan academy
A multiplying dividing polynomials template Word can contain formatting, styles, boilerplate text, headers and footers, as well as autotext entries. It is important to define the document styles beforehand in the sample document as styles define the appearance of Word text elements throughout your document. You may design other styles and format such as multiplying dividing polynomials template pdf, multiplying dividing polynomials template powerpoint, multiplying dividing polynomials template form. When designing multiplying dividing polynomials template, you may add related content, multiplying polynomials notes, multiplying polynomials lesson, multiplying polynomials box method, multiplying polynomials khan academy. how do you multiply and divide polynomials? how do you multiply polynomials step by step? how do you divide polynomials easily? how do you multiply polynomials with different exponents?
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Lemma 70.12.5. Let $k$ be a field. Let $X$ be an algebraic space over $k$. Let $\overline{k}$ be a separable algebraic closure of $k$. Then $X$ is geometrically connected if and only if the base change $X_{\overline{k}}$ is connected.
Proof. Assume $X_{\overline{k}}$ is connected. Let $k \subset k'$ be a field extension. There exists a field extension $\overline{k} \subset \overline{k}'$ such that $k'$ embeds into $\overline{k}'$ as an extension of $k$. By Lemma 70.12.4 we see that $X_{\overline{k}'}$ is connected. Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude that $X_{k'}$ is connected as desired. $\square$
There are also:
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# A Simple Explanation of Information Gain and Entropy
## What Information Gain and Information Entropy are and how they're used to train Decision Trees.
Information Gain, like Gini Impurity, is a metric used to train Decision Trees. Specifically, these metrics measure the quality of a split. For example, say we have the following data:
What if we made a split at $x = 1.5$?
This imperfect split breaks our dataset into these branches:
• Left branch, with 4 blues.
• Right branch, with 1 blue and 5 greens.
It’s clear this split isn’t optimal, but how good is it? How can we quantify the quality of a split?
That’s where Information Gain comes in.
Confused? Not sure what Decision Trees are or how they’re trained? Read the beginning of my introduction to Random Forests and Decision Trees.
Note: this post looks better in Light Mode. If you’re in Dark Mode, scroll up and use the toggle in the top right to switch!
## Information Entropy
Before we get to Information Gain, we have to first talk about Information Entropy. In the context of training Decision Trees, Entropy can be roughly thought of as how much variance the data has. For example:
• A dataset of only blues would have very low (in fact, zero) entropy.
• A dataset of mixed blues, greens, and reds would have relatively high entropy.
Here’s how we calculate Information Entropy for a dataset with $C$ classes:
$E = -\sum_i^C p_i \log_2 p_i$
where $p_i$ is the probability of randomly picking an element of class $i$ (i.e. the proportion of the dataset made up of class $i$).
The easiest way to understand this is with an example. Consider a dataset with 1 blue, 2 greens, and 3 reds: . Then
$E = -(p_b \log_2 p_b + p_g \log_2 p_g + p_r \log_2 p_r)$
We know $p_b = \frac{1}{6}$ because $\frac{1}{6}$ of the dataset is blue. Similarly, $p_g = \frac{2}{6}$ (greens) and $p_r = \frac{3}{6}$ (reds). Thus,
\begin{aligned} E &= -(\frac{1}{6} \log_2(\frac{1}{6}) + \frac{2}{6} \log_2(\frac{2}{6}) + \frac{3}{6} \log_2(\frac{3}{6})) \\ &= \boxed{1.46} \\ \end{aligned}
What about a dataset of all one color? Consider 3 blues as an example: . The entropy would be
$E = -(1 \log_2 1) = \boxed{0}$
## Information Gain
It’s finally time to answer the question we posed earlier: how can we quantify the quality of a split?
Let’s consider this split again:
Before the split, we had 5 blues and 5 greens, so the entropy was
\begin{aligned} E_{before} &= -(0.5 \log_2 0.5 + 0.5 \log_2 0.5) \\ &= \boxed{1} \\ \end{aligned}
After the split, we have two branches.
Left Branch has 4 blues, so $E_{left} = \boxed{0}$ because it’s a dataset of all one color.
Right Branch has 1 blue and 5 greens, so
\begin{aligned} E_{right} &= -(\frac{1}{6} \log_2 (\frac{1}{6}) + \frac{5}{6} \log_2 (\frac{5}{6})) \\ &= \boxed{0.65} \\ \end{aligned}
Now that we have the entropies for both branches, we can determine the quality of the split by weighting the entropy of each branch by how many elements it has. Since Left Branch has 4 elements and Right Branch has 6, we weight them by $0.4$ and $0.6$, respectively:
\begin{aligned} E_{split} &= 0.4 * 0 + 0.6 * 0.65 \\ &= \boxed{0.39} \\ \end{aligned}
We started with $E_{before} = 1$ entropy before the split and now are down to $0.39$! Information Gain = how much Entropy we removed, so
$\text{Gain} = 1 - 0.39 = \boxed{0.61}$
This makes sense: higher Information Gain = more Entropy removed, which is what we want. In the perfect case, each branch would contain only one color after the split, which would be zero entropy!
## Recap
Information Entropy can be thought of as how unpredictable a dataset is.
• A set of only one class (say, blue ) is extremely predictable: anything in it is blue. This would have low entropy.
• A set of many mixed classes is unpredictable: a given element could be any color! This would have high entropy.
The actual formula for calculating Information Entropy is:
$E = -\sum_i^C p_i \log_2 p_i$
Information Gain is calculated for a split by subtracting the weighted entropies of each branch from the original entropy. When training a Decision Tree using these metrics, the best split is chosen by maximizing Information Gain.
Want to learn more? Check out my explanation of Gini Impurity, a similar metric, or my in-depth guide Random Forests for Complete Beginners.
I write about ML, Web Dev, and more topics. Subscribe to get new posts by email!
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Test3d08, 8th 'benchmark test' data file for CPO3D
Time-dependent oscillating potentials
This test reproduces the sinusoidal motion of an electron in a sinusoidal electric field. The field is applied to a cube, as in file test3d03.dat, and the time taken for the electrons to reach a particular plane is given within the requested inaccuracy of 0.05%
The following data were obtained when the memory and speed of PC's was much more limited than at present, so the available number of segments was small and the requested inaccuracies were fairly high to give a quick demonstration.
A uniform electric field is created by setting up a cube, as in file test3d03.dat.
The end faces have been given the voltages:
v1 = +/-30*sin(2*pi*t/tau),
where tau = 0.000001ms.
The ray starts at x = -0.5, z = 0.
A solution that does not include a linear dependence on time is
z = -(e*E/m*omega**2)*sin(omega*t), where omega = 2*pi/tau
In this example, the electric field E is 30 volt per mm, the period tau is 0.000001 ms, the initial direction is 45 degrees to the field and the initial time 'ti' is 0.000001 ms (that is, one period). Therefore the energy is 4.00959 eV and the path of the electron should be given by:
x = 0.5+vx*(t-ti),
z = -0.133653*sin(2*pi*(t-ti)/tau),
where vx = 839768m/s (and where x is in mm, t is in ms),
Therefore at the test plane x = 0.5, we should have:
t = 0.0000021908 ms, z = -0.12452 mm.
In fact t is given correctly and z = -0.12448, an error of 0.03% (which is consistent with the requested ray inaccuracy, 0.05%).
At the 2nd test plane, z = 0, and at the requested 2nd crossing, we should have:
t = ti+tau = 0.000002 ms, x = 0.33977 mm.
In fact the errors in t and x are both 0.01% or smaller.
Another way of setting up the oscillating field is with a user-supplied routine. An example of such a routine, supplied as part of the CPO package, allows square and sawtooth waveforms.
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# Light and average distance
From this difference in angle, the known speed of rotation and the distance to the distant mirror the speed of light may be calculated.
Knowing Earth's shift and a star's shift enabled the star's distance to be calculated.
Well, on Earth, a kilometer may be just fine. The light-year we have already defined. Using the old definition, the value of AU would change depending on an observer's location in the solar system.
### 1 light year in days
The refractive index of a material may depend on the light's frequency, intensity, polarizationor direction of propagation; in many cases, though, it can be treated as a material-dependent constant.
An actual physical signal with a finite extent a pulse of light travels at a different speed.
### Astronomical unit
Because all photons move at the speed of light in vacuum, a fundamental constant of the universe, the distance of an object from the probe is calculated as the product of the speed of light and the measured time. Another reason for the speed of light to vary with its frequency would be the failure of special relativity to apply to arbitrarily small scales, as predicted by some proposed theories of quantum gravity. On the other hand, some techniques depend on the finite speed of light, for example in distance measurements. Finding the distance Historically, the first person to measure the distance to the sun was the Greek astronomer Aristarchus around the year B. Certain materials have an exceptionally low or even zero group velocity for light waves, a phenomenon called slow light , which has been confirmed in various experiments. The distance travelled by light from the planet or its moon to Earth is shorter when the Earth is at the point in its orbit that is closest to its planet than when the Earth is at the farthest point in its orbit, the difference in distance being the diameter of the Earth's orbit around the Sun. As an extreme example of light "slowing" in matter, two independent teams of physicists claimed to bring light to a "complete standstill" by passing it through a Bose—Einstein condensate of the element rubidium , one team at Harvard University and the Rowland Institute for Science in Cambridge, Mass. Diagram of the Fizeau apparatus A method of measuring the speed of light is to measure the time needed for light to travel to a mirror at a known distance and back. The farthest from the sun Earth gets is called aphelion. For example, when Venus appears half illuminated by the sun, the three bodies form a right triangle from Earth's perspective. However, for precision the calculations require adjustment for things such as the motions of the probe and object while the photons are transiting.
The refractive index of air is approximately 1. So in one year, it can travel about 10 trillion km.
## How long is a light year
Historically, such measurements could be made fairly accurately, compared to how accurately the length of the reference distance is known in Earth-based units. Using the light-year, we can say that : The Crab supernova remnant is about 4, light-years away. The relative sizes and separation of the Earth—Moon system are shown to scale. In our solar system, we tend to describe distances in terms of the Astronomical Unit AU. However, because Huygens' method was partly guesswork and not completely scientifically grounded, he usually doesn't get the credit. Subsequent explorations of the Solar System by space probes made it possible to obtain precise measurements of the relative positions of the inner planets and other objects by means of radar and telemetry. In addition, the measurement of the time itself must be translated to a standard scale that accounts for relativistic time dilation. By combining many such measurements, a best fit value for the light time per unit distance could be obtained. Because all photons move at the speed of light in vacuum, a fundamental constant of the universe, the distance of an object from the probe is calculated as the product of the speed of light and the measured time. An actual physical signal with a finite extent a pulse of light travels at a different speed. There is a brief delay from the source to the receiver, which becomes more noticeable as distances increase. The relative uncertainty in these measurements is 0. When a charged particle does that in a dielectric material, the electromagnetic equivalent of a shock wave , known as Cherenkov radiation , is emitted.
Receiving light and other signals from distant astronomical sources can even take much longer. He used the phases of Venus to find the angles in a Venus-Earth-sun triangle.
On the way from the source to the mirror, the beam passes through a rotating cogwheel. Once they had the distance to Mars, they could also calculate the distance to the sun.
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# 14 Z-score Basics
Jenna Lehmann
##### Standardized Distributions
Sometimes when working with data sets, we want to have the scores on the distribution standardized. Essentially, this means that we convert scores from a distribution so that they fit into a model that can be used to compare and contrast distributions from different works. For example, if you have a distribution of scores that show the temperature each day over the summer in Boston, it may be recorded in Fahrenheit. Someone else in Paris may have recorded their summer temperatures as well but in Celcius. If we wanted to compare these distributions of scores based on their descriptive statistics, we may want to convert them to the same standardized unit of measurement.
Standardized distributions have one single unit of measurement. Raw scores are transformed into this standardized unit of measurement to be compared to one another. Ultimately, they should look just like the original distribution, the only difference is that the scores have been placed on a different unit of measurement.
##### Z-Scores
Z-scores are the most common standardized score. They are used to describe score location in a distribution (descriptive statistics) and because we can compare scores across distributions, we can look at the relative standing of a score in a sample or a sample in a population (inferential statistics). The equation is
In this equation, is the z-score, is the variable you want to convert, is the mean of the original distribution, and is the standard deviation of the original distribution.
So, what are the characteristics of a z-score/distributions? In a z-score the mean is placed at 0 and each number below or above is a representation of how many standard deviations away a score is. A 1 represents one standard deviation above the mean and -1 represents one standard deviation below the mean. For example, if I know that my original mean is 10 and my original standard deviation is 2, I know that a z-score of 1 would mean 12 and a z-score of -1 would mean 8. For the purposes of your class, all z-score distributions are normal distributions. Z-scores aren’t used on other kinds of distributions because the charts and proportions are designed to describe normal distributions.
What’s nice about z-scores is that they can also be used to find proportions, which will be talked about even more in the next post. This requires the Unit Normal Table which is a table designed to help one translate z-scores into proportions of the population on either side of the score or compared to the mean score. There are 4 columns: one with the z-scores, one with the proportion of the population in the body of the distribution with the z-score as the starting point, one with the proportion of the population in the tail of the distribution with the z-score as the starting point, and one with the proportion of the population between the z-score and the mean. It can usually be found in the back of any statistics textbook. If I have a z-score of -1.5 and I wanted to know the proportion of the scores which are lower than -1.5, I could go to the back of my textbook, find -1.50 in the margins, and get the proportion .06681, meaning that 6.6881% of the data is less than a z-score of -1.5. The numbers in this table show the reader the proportion of everything to the left of the z-score in question. If I wanted to know everything to the right, the proportion would be 1 – 0.06681, which is .93319 or 93.319% of the data.
Z-scores can be used in inferential statistics. Interpretation of research results depends on determining if the (treated) sample is noticeably different from the population. The distribution of the general population would describe the average untreated person, so this allows researchers to compare that distribution to their treated sample. Z-scores are one technique for defining “noticeably different”, but it more like borders on inferential statistics, because we can’t actually tell if there’s a statistical difference without running the right test. Z-tests and their purpose in inferential statistics will be discussed in other posts.
This chapter was originally posted to the Math Support Center blog at the University of Baltimore on June 6, 2019.
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If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x : Quant Question Archive [LOCKED]
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# If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x
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If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x [#permalink]
### Show Tags
03 Jul 2006, 11:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
From 1) -1<x<1. Therefore, |x| is <1.
How does one solve 2)?
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### Show Tags
03 Jul 2006, 12:15
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
From 1) -1<x<1. Therefore, |x| is <1.
How does one solve 2)?
From 2 ---- 1/x is +ve also x=! 0
there fore 1/x got to be greater than one which means x is less than 1 therefore |x|<1
D
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### Show Tags
03 Jul 2006, 15:40
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
From 1) -1<x<1. Therefore, |x| is <1.
How does one solve 2)?
D
St1: Its easy. -1<x<1 : SUFF
St2: I just plugged in the number and found 0<x<1 : SUFF
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
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### Show Tags
03 Jul 2006, 16:50
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
From 1) -1<x<1. Therefore, |x| is <1.
How does one solve 2)?
from 1, we know x<1.
From 2) 1/x must be +ve since lxl is positive. only values for x work are 1>x>0.
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### Show Tags
03 Jul 2006, 16:59
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
From 1) -1<x<1. Therefore, |x| is <1.
How does one solve 2)?
(1) x^2 < 1 => x < 1 so OK
(2) |x| < 1/x => x < 1/x or x < -1/x, then,
x < 1/x is possible only if x < 1, but it can be any number 0<x<1 and x < -1, so it does not tell us if |x| is < 1,
if we take x < -1/x, is possible if x <= -1 ... can't say.
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### Show Tags
03 Jul 2006, 17:07
aargh!! I always get these !@***#@@#!# inequalities wrong. Anyone got a good suggestion where I can firm up my the fundamentals?
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### Show Tags
05 Jul 2006, 02:32
D
1) suff. -1<x<1
2) plug in values and you will find that 0<x<1
hence D
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### Show Tags
05 Jul 2006, 14:19
one more for (D)..
as explained by others.
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### Show Tags
14 Jul 2006, 08:07
old_dream_1976 wrote:
geemaat wrote:
If x≠0, is |x| <1?
(1) x^2<1
(2) |x| < 1/x
From 1) -1<x<1. Therefore, |x| is <1.
How does one solve 2)?
From 2 ---- 1/x is +ve also x=! 0
there fore 1/x got to be greater than one which means x is less than 1 therefore |x|<1
D
Thanks guys. The OA is D.
Re: solving absolute inequality [#permalink] 14 Jul 2006, 08:07
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# Find the equation of tangent to the curve
Question:
Find the equation of tangent to the curve $\mathrm{x}=(\theta+\sin \theta), \mathrm{y}=(1+\cos \theta)$ at $\theta=\frac{\pi}{4}$
Solution:
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https://math.stackexchange.com/questions/1418169/is-left-x-1y-1-x-ny-n-right-a-basis-for-mathbbrn
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# Is $\left \{ x_{1}+y_{1},..., x_{n}+y_{n}\right \}$ a basis for $\mathbb{R}^{n}$?
Suppose $\left \{ x_{1},..., x_{n}\right \}$ and $\left \{ y_{1},..., y_{n}\right \}$ are two different bases for $\mathbb{R}^{n}$. Is $\left \{ x_{1}+y_{1},..., x_{n}+y_{n}\right \}$ also a basis for $\mathbb{R}^{n}?$ If yes, prove your claim, if no, show a counterexample.
I think the claim is true. I have picked two different bases in $\mathbb{R}^{2}$ to try to come up with a counterexample, but couldn't. I picked $\left \{ (0,1),(1,0) \right \}$ and $\left \{ (1,1),(-1,1) \right \}$ and added their components to come up with a different set of vectors $\left \{ (1,2),(0,1) \right \}$. However, I have found that this is also a basis for $\mathbb{R}^{2}$ as it spans $\mathbb{R}^{2}$ and also is linearly independent, that is, $Ax=b \Rightarrow x=0$.
If the claim is true, then can anyone help me with the proof?
• What of $y_i = -x_i$? Sep 2, 2015 at 9:28
If you chose $y_i=-x_i$ the claim is false.
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https://www.scaffoldedmath.com/2019/05/voyage-to-the-treasure-collaborative-math-game.html?m=1
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## Pages
### Voyage to the Treasure! collaborative math games
My friend Alex from Middle School Math Man and I have been working together on a new collaborative math game series called Voyage to the Treasure! In this game, students work together to beat the board and get to the treasure first.
Voyage to the Treasure math games are collaborative. Students are on the same team. Similar to a study group with students working together to review their math, students work together in this game to beat the board's Math Monster. This builds teamwork, community and gets students talking with each other in a good way.
### So how do you play Voyage to the Treasure?
The main idea is that students are on the same team, working together to beat the board as they solve and check their math problems.
Setup:
⦁Place 3 Voyager ships and Math Monster on their start spots.
⦁Place MAP on its placement.
Place problem cards face down next to game board.
⦁Cut out 2 Answer Banks to use during game.
Use paperclip and pencil to spin spinner.
student directions card
How to Play:
Students take turns spinning, solving and moving all of the Voyager ships.
Here is a short video that shows an example game:
1: Spin the spinner, move that many spaces. Land on Math Monster? He moves 2 spaces!
2: Take a card, solve its problem.
4: Get problem correct? Check KEY for your bonus move!
5: Pick up the MAP by landing on it and correctly solving a problem.
6: Take turns spinning and moving all Voyagers. Leave no Voyager behind!
Strategize well so that all Voyagers get to the treasure before Math Monster!
students check the answer bank themselves
Once the game is over, students can complete a peer evaluation form to make sure everyone put in their fair share of the work.
Ideas for differentiating:
Students are in charge of moving 3 Voyager ships through the game board to the treasure. If you think this is too much for some of your students, 2 Voyagers can be used instead of 3.
Also, problem cards can be removed from the deck.
The download also includes a peer evaluation sheet to help ensure all students put in an equal team effort.
We're really excited about these games and are adding more topics all of the time! So far we have a variety of games made for middle school and algebra topics.
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https://edurev.in/course/quiz/attempt/5737_JEE-Main-Physics-Test-2/e5cc06cc-4920-4826-b522-c425e76b9c3e
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JEE Main Physics Test- 2 - JEE MCQ
# JEE Main Physics Test- 2 - JEE MCQ
Test Description
## 25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Physics Test- 2
JEE Main Physics Test- 2 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Physics Test- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Physics Test- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Physics Test- 2 below.
Solutions of JEE Main Physics Test- 2 questions in English are available as part of our Mock Tests for JEE Main and Advanced 2024 for JEE & JEE Main Physics Test- 2 solutions in Hindi for Mock Tests for JEE Main and Advanced 2024 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Main Physics Test- 2 | 25 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mock Tests for JEE Main and Advanced 2024 for JEE Exam | Download free PDF with solutions
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JEE Main Physics Test- 2 - Question 1
### A liquid drop tends to have a spherical shape because
Detailed Solution for JEE Main Physics Test- 2 - Question 1
JEE Main Physics Test- 2 - Question 2
### A capillary tube of radius 1 mm is 10 cm long. What maximum height of water it can hold if it is immersed in water upto 5 cm and then taken out of it ? (Surface Tension of water = 98 dynes/cm)
Detailed Solution for JEE Main Physics Test- 2 - Question 2
JEE Main Physics Test- 2 - Question 3
### Ice floats in water contained in a vessel. When whole ice has melted, water temp. falls from . The water level will
Detailed Solution for JEE Main Physics Test- 2 - Question 3
JEE Main Physics Test- 2 - Question 4
Two simple harmonic motions are given by the equations? ? The amplitudes are in the ratio of
Detailed Solution for JEE Main Physics Test- 2 - Question 4
JEE Main Physics Test- 2 - Question 5
Two soap bubbles of radii 4 mm & 6 mm are kept in contact. What is radius of curvature at the junction of bubbles.
JEE Main Physics Test- 2 - Question 6
The time period of a simple pendulum with effective length equal to the radius of Earth (R) will be (g is acceleration due to gravity at the surface of earth)
Detailed Solution for JEE Main Physics Test- 2 - Question 6
JEE Main Physics Test- 2 - Question 7
For the wave represented by the equation y = 3cos cm the wave velocity and the maximum
particle velocity are
Detailed Solution for JEE Main Physics Test- 2 - Question 7
JEE Main Physics Test- 2 - Question 8
Figure shows a capillary tube of radius r dipped into water. If the atmospheric pressure is the pressure at point A is
Detailed Solution for JEE Main Physics Test- 2 - Question 8
JEE Main Physics Test- 2 - Question 9
Ultrasonic waves emitted by dolphins have a frequency of 250 kHz. Their wavelength in water is nearly
Detailed Solution for JEE Main Physics Test- 2 - Question 9
JEE Main Physics Test- 2 - Question 10
An organ pipe closed at one end vibrating in its first harmonic and another pipe open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of to that of is
Detailed Solution for JEE Main Physics Test- 2 - Question 10
JEE Main Physics Test- 2 - Question 11
Match list I & list II & select the correct answer given below
JEE Main Physics Test- 2 - Question 12
Consider the following statements-
IN a stationary wave
1. All the particles perform simple harmonic motion with a frequency which is four times that of the two component waves.
2. Particles on the opposite sides of a node vibrate with a phase difference of
3. The amplitude of vibration of a particle at an antinode is equal to that of either component wave.
4. All the particles between two adjacent nodes vibrate in phase.
Of these statements-
JEE Main Physics Test- 2 - Question 13
The centre of a wheel rolling on a plane surface moves with a speed . A particle on the rim of the wheel at the same level as the centre will be moving at speed
Detailed Solution for JEE Main Physics Test- 2 - Question 13
JEE Main Physics Test- 2 - Question 14
Three identical rods, each of length l are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is
Detailed Solution for JEE Main Physics Test- 2 - Question 14
JEE Main Physics Test- 2 - Question 15
The velocity of a solid sphere after rolling down an inclined plane of height h, from rest, without sliding is
Detailed Solution for JEE Main Physics Test- 2 - Question 15
JEE Main Physics Test- 2 - Question 16
A particle of mass m is moving in a plane along a circular path of radius r. Its angular momentum about the axis of rotation is L. The centripetal force on the particle is
Detailed Solution for JEE Main Physics Test- 2 - Question 16
JEE Main Physics Test- 2 - Question 17
The earth retains its atmosphere because
Detailed Solution for JEE Main Physics Test- 2 - Question 17
JEE Main Physics Test- 2 - Question 18
A solid sphere rolls without slipping and presses a spring of spring constant ‘K’ as in fig. at the moment shown a spring was in relaxed state then the maximum compression in the spring will be
Detailed Solution for JEE Main Physics Test- 2 - Question 18
JEE Main Physics Test- 2 - Question 19
A wheel of M. I.(Moment of inertia) is making 10 revolution per second. It is stopped in 20 Second. Then angular retardation is
Detailed Solution for JEE Main Physics Test- 2 - Question 19
JEE Main Physics Test- 2 - Question 20
We have two sphere one of which is hollow and the other is solid. They have identical masses and moment of inertia about their respective diameter. The ratio of their radii is
Detailed Solution for JEE Main Physics Test- 2 - Question 20
*Answer can only contain numeric values
JEE Main Physics Test- 2 - Question 21
The moment of inertia of a body about a given axis is 1.2 kg-m2. Initially, the body is at rest. In order to produce a rotational KE of 1500 J, an angular acceleration of 25 rad/s2 must be applied about that axis for a duration of :
Detailed Solution for JEE Main Physics Test- 2 - Question 21
*Answer can only contain numeric values
JEE Main Physics Test- 2 - Question 22
One mole of an ideal gas expands at a constant temperature of 300 K from an initial volume of 10 litres to a final volume of 20 litres. The work done (in J) in expanding the gas is
(R = 8.31 J/mole-K) (log 2 = 0.3010)
*Answer can only contain numeric values
JEE Main Physics Test- 2 - Question 23
Time taken (in s) by a 836 W heater to heat one litre of water from 10°C to 40°C is :-
Detailed Solution for JEE Main Physics Test- 2 - Question 23
*Answer can only contain numeric values
JEE Main Physics Test- 2 - Question 24
The escape velocity from a planet is v0. The escape velocity from a planet having twice the radius but same density is nv0 then n is :-
Detailed Solution for JEE Main Physics Test- 2 - Question 24
*Answer can only contain numeric values
JEE Main Physics Test- 2 - Question 25
For an EMwaves, E = E0 sin (12 × 106 [z – 2 × 108t]) In a medium, it's refractive index is P/2 then P is
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http://coding.derkeiler.com/Archive/C_CPP/comp.lang.cpp/2004-03/2401.html
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# Merging mixed data sets
From: Dmitry Epstein (mitia.nospam_at_northwestern.edu.invalid)
Date: 03/14/04
```Date: 14 Mar 2004 22:27:10 GMT
```
I have the following problem (not a "homework" problem - honest :)
I have two data sets consisting of ASCII strings (mostly 10-50
chars). One set is sorted, the other is not. I need to merge the
two into a sorted set while avoiding duplication, i.e. I want to
get a union of the two sets. The choice of containers (list,
vector, etc.) is up to me.
I can think of a few ways of solving the problem but I wonder what
would be the most efficient?
1. Use the STL, Luke! Probably the most straightforward approach
would be to sort the second set first, then use the set_union from
<algorithm> to merge the two. If I take this path, then what
container would be best? Sorting ought to be much more efficient
with lists than with vectors, right?
The downside of this approach is that I can't merge the two sets
in-place, without copying elements. I would've liked to use the
merge member-function of the list class, except it does not do what
I need. On the other hand, set_union will copy elements into a new
container regardless.
Hence,
2. Use STL to sort, then merge manually. With this approach I
would probably use the list container. After sorting the second
set, I would merge it into the first set using splice to avoid
copying elements. What's the best algorithm for such merging?
3. Roll your own. I am wondering if forgoing pre-sorting of the
second set would actually be more efficient? The problem then
would reduce to adding elements to a sorted set. What would be the
best way to do that? A binary tree perhaps? What would be a good
implementation of that?
Thanks,
Dmitry
```--
Dmitry Epstein
Northwestern University, Evanston, IL. USA
mitia(at)northwestern(dot)edu
```
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https://playtaptales.com/numbers/unsexagintatrecentillion/
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## Unsexagintatrecentillion
A Unsexagintatrecentillion (1 Unsexagintatrecentillion) is 10 to the power of 1086 (10^1086). This is an exorbitantly astronomical number!
## How many zeros in a Unsexagintatrecentillion?
There are 1,086 zeros in a Unsexagintatrecentillion.
## What's before Unsexagintatrecentillion?
A Sexagintatrecentillion is smaller than a Unsexagintatrecentillion.
## What's after Unsexagintatrecentillion?
A Duosexagintatrecentillion is larger than a Unsexagintatrecentillion.
## Unsexagintatrecentillionaire
A Unsexagintatrecentillionaire is someone whos assets, net worth or wealth is 1 or more Unsexagintatrecentillion. It is unlikely anyone will ever be a true Unsexagintatrecentillionaire. If you want to be a Unsexagintatrecentillionaire, play Tap Tales!
## Is Unsexagintatrecentillion the largest number?
Unsexagintatrecentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
## Unsexagintatrecentillion written out
Unsexagintatrecentillion is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
## Big Numbers
This is just one of many really big numbers!
| 1,039
| 2,624
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| 2.71875
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http://mathematica.stackexchange.com/questions/45174/2d-plotting-of-triplet-data-set-with-third-axis-as-variation-of-thickness-of-cur
| 1,466,929,205,000,000,000
|
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| 200,285,342
| 17,048
|
# 2D Plotting of triplet data set with third axis as variation of thickness of curve
I have similar question like mentioned at representing triplet data set with color on 2d plot but this time instead of color I need to represent variation of third axis with the thickness of the curve
-
related – Kuba Apr 1 '14 at 16:47
Here Filling can be a big help.
Some sample data:
data = {#, Sin[#], #^2} & /@ Range[0, 10, 1/4];
A function to help create the variable "thickness". The parameter f is like a scaling factor:
addThickness[data_, f_] :=
Transpose[
data /. {x_, y_, z_} :> {{x, y}, {x, y - f z}, {x, y + f z}}]
and now make the FillingStyle match the PlotStyle:
ListLinePlot[addThickness[data, 0.01],
Filling -> {2 -> {1}, 3 -> {1}}, PlotStyle -> Blue,
FillingStyle -> Opacity[1]]
-
How can I apply this method to list data like data = Table[{TL[[jj, 5]], TL[[jj, 6]], TL[[jj, 1]]}, {jj, 1, Length[TL]}]; – physicist Apr 1 '14 at 20:37
I have used the above code but it is not so useful because it is creating negative values of third variable which is not the part of data – physicist Apr 1 '14 at 21:39
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| 2.640625
| 3
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| 0.806225
|
https://goodmancoaching.nl/if-then-statement-and-time-between-antecedent-and-consequent/
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| 309,712,283
| 18,638
|
# If-then statement and time between antecedent and consequent?
Conclusion: that statement which is affirmed on the basis of the other propositions (the premises) of the argument. Conditional statement: an “if p, then q” compound statement (ex. If I throw this ball into the air, it will come down); p is called the antecedent, and q is the consequent.
Contents
## How do you know if its antecedent or consequent?
The Conditional: The Fourth Connective
1. Conditional statement: when two statements are combined by placing the word “if” before the first and “then” before the second.
2. The component statement that follows the “if” is called the antecedent.
3. The component statement that follows the “then” is called the consequent.
## What is conditional statement antecedent and consequent?
Definition1.2.
For propositions P and Q, the conditional sentence P⟹Q P ⟹ Q is the proposition “If P, then Q. ” The proposition P is called the antecedent, Q the consequent. The conditional sentence P⟹Q P ⟹ Q is true if and only if P is false or Q is true.
## What do we call the antecedent or the condition in an IF THEN statement?
In an implication, if implies then is called the antecedent and. is called the consequent. Antecedent and consequent are connected via logical connective to form a proposition.
## When a conditional IF THEN statement has a false antecedent if it is?
When the antecedent is false, the truth value of the consequent does not matter; the conditional will always be true. A conditional is considered false when the antecedent is true and the consequent is false.
Conditional.
P Q P ⇒ Q
F F T
## What is the consequent of a conditional statement?
Conditional statement: an “if p, then q” compound statement (ex. If I throw this ball into the air, it will come down); p is called the antecedent, and q is the consequent.
## What is the difference between antecedent phrase and consequent phrase?
In a period, the phrase ending with the less conclusive cadence is called the “ antecedent ” and the phrase ending with the more conclusive cadence is called the “ consequent .” These can be thought of as being in a “question and answer” relationship.
## What two clauses are in an if-then statement?
Conditional sentences are constructed using two clauses—the if (or unless) clause and the main clause.
## What is IF and THEN statement?
Hypotheses followed by a conclusion is called an If-then statement or a conditional statement. This is noted as. p→q. This is read – if p then q. A conditional statement is false if hypothesis is true and the conclusion is false.
## What are the two parts of conditional statement?
Conditional Statement A conditional statement is a logical statement that has two parts, a hypothesis p and a conclusion q. When a conditional statement is written in if-then form, the “if’ part contains the hypothesis and the “then” part contains the conclusion.
## What does P → Q mean?
The implication p → q (read: p implies q, or if p then q) is the state- ment which asserts that if p is true, then q is also true. We agree that p → q is true when p is false. The statement p is called the hypothesis of the implication, and the statement q is called the conclusion of the implication.
## What does P ↔ Q mean?
P→Q means If P then Q. ~R means Not-R. P ∧ Q means P and Q. P ∨ Q means P or Q. An argument is valid if the following conditional holds: If all the premises are true, the conclusion must be true.
## What is the contrapositive of P → Q?
Contrapositive: The contrapositive of a conditional statement of the form “If p then q” is “If ~q then ~p”. Symbolically, the contrapositive of p q is ~q ~p. A conditional statement is logically equivalent to its contrapositive.
## Which is logically equivalent to P ↔ Q?
P → Q is logically equivalent to ¬ P ∨ Q . Example: “If a number is a multiple of 4, then it is even” is equivalent to, “a number is not a multiple of 4 or (else) it is even.”
## What is a converse inverse and contrapositive?
The converse of the conditional statement is “If Q then P.” The contrapositive of the conditional statement is “If not Q then not P.” The inverse of the conditional statement is “If not P then not Q.”
## What is converse statement?
Definition: The converse of a conditional statement is created when the hypothesis and conclusion are reversed. In Geometry the conditional statement is referred to as p → q. The Converse is referred to as q → p.
## What is converse and inverse?
The converse statement is notated as q→p (if q, then p). The original statements switch positions in the original “if-then” statement. The inverse statement assumes the opposite of each of the original statements and is notated ∼p→∼q (if not p, then not q).
## What is the difference between inverse and converse?
is that converse is familiar discourse; free interchange of thoughts or views; conversation; chat or converse can be the opposite or reverse while inverse is the opposite of a given, due to contrary nature or effect.
## What is the contrapositive statement?
Definition of contrapositive
: a proposition or theorem formed by contradicting both the subject and predicate or both the hypothesis and conclusion of a given proposition or theorem and interchanging them “if not-B then not-A ” is the contrapositive of “if A then B “
## What is the contrapositive of if A then B?
Conditionals: “if A then B” (or “A implies B”) is a conditional statement with antecedent A and consequent B. It’s contrapositive is “if not B then not A” and it’s converse is “if B then A”. Statements with the same truth table are said to be equivalent.
## What is converse statement in logic?
The converse of a statement is formed by switching the hypothesis and the conclusion. The converse of “If two lines don’t intersect, then they are parallel” is “If two lines are parallel, then they don’t intersect.” The converse of “if p, then q” is “if q, then p.”
## What is syllogism law?
In mathematical logic, the Law of Syllogism says that if the following two statements are true: (1) If p , then q . (2) If q , then r . Then we can derive a third true statement: (3) If p , then r .
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# (a) The product of two proper fractions is less than each of the fractions that are multiplied. (b) The product of a proper and an improper fraction is less than the improper fraction and greater than the proper fraction. (c) The product of two imporper fractions is greater than the two fractions.
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A proper fraction is a fraction in which numerator is less than the denominator.<br> An improper fraction is a fraction in which numerator is greater than the denominator.<br> (a) Let `x = 3/5, y = 4/5`<br> Here, `x` and `y` are proper fractions.<br> Then, their product `= xy = (3/5)(4/5) = 12/25`<br> As, `12/25 lt 3/5 and 12/25 lt 4/5.`<br> So, product of two proper fractions is less than each of the fractions.<br> (b) Let `x = 3/5, y = 5/4`<br> Here, `x` is a proper fraction and `y` is an improper fraction.<br> Then, their product `= xy = (3/5)(5/4) = 3/4`<br> As, `3/5 lt 3/4 and 5/4 gt 3/4.`<br> So, the product of a proper and an improper fraction is less than the improper fraction and greater than the proper fraction.<br> (c) Let `x= 5/4, y = 7/5`<br> Here, `x` and `y` are improper fractions.<br> Then, their product `= xy = (5/4)(7/5) = 7/4`<br> As `7/4 gt 5/4 and 7/4 gt 7/5`<br> So, the product of two improper fractions is greater than the two fractions.
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# 6462720
## 6,462,720 is an even composite number composed of five prime numbers multiplied together.
What does the number 6462720 look like?
This visualization shows the relationship between its 5 prime factors (large circles) and 288 divisors.
6462720 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of two hundred eighty-eight divisors.
## Prime factorization of 6462720:
### 28 × 33 × 5 × 11 × 17
(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 11 × 17)
See below for interesting mathematical facts about the number 6462720 from the Numbermatics database.
### Names of 6462720
• Cardinal: 6462720 can be written as Six million, four hundred sixty-two thousand, seven hundred twenty.
### Scientific notation
• Scientific notation: 6.46272 × 106
### Factors of 6462720
• Number of distinct prime factors ω(n): 5
• Total number of prime factors Ω(n): 14
• Sum of prime factors: 38
### Divisors of 6462720
• Number of divisors d(n): 288
• Complete list of divisors:
• Sum of all divisors σ(n): 26490240
• Sum of proper divisors (its aliquot sum) s(n): 20027520
• 6462720 is an abundant number, because the sum of its proper divisors (20027520) is greater than itself. Its abundance is 13564800
### Bases of 6462720
• Binary: 110001010011101000000002
• Base-36: 3UIO0
### Squares and roots of 6462720
• 6462720 squared (64627202) is 41766749798400
• 6462720 cubed (64627203) is 269926809257115648000
• The square root of 6462720 is 2542.1880339581
• The cube root of 6462720 is 186.2680837657
### Scales and comparisons
How big is 6462720?
• 6,462,720 seconds is equal to 10 weeks, 4 days, 19 hours, 12 minutes.
• To count from 1 to 6,462,720 would take you about sixteen weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 6462720 cubic inches would be around 15.5 feet tall.
### Recreational maths with 6462720
• 6462720 backwards is 0272646
• 6462720 is a Harshad number.
• The number of decimal digits it has is: 7
• The sum of 6462720's digits is 27
• More coming soon!
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## Pseudo-Numeric Identifiers
Let’s say you’re a programmer, and your application uses Library of Congress Control Numbers for books, e.g., 2001012345, or ZIP codes, like 90210. What data types would you use to represent them? Or maybe something like the Dewey Decimal System, which uses 320 to classify a book as Political Science, 320.5 for Political Theory, and 320.973 for “Political institutions and public administration (United States)”?
If you said “integer”, “floating point”, or any kind of numeric type, then clearly you weren’t paying attention during the title.
The correct answer was “string” (or some kind of array of tokens), because although these entities consist of digits, they’re not numbers: they’re identifiers, same as “root” or “Jane Smith”. You can assign them, sort them, group them by common features, but you can’t meaningfully add or multiply them together. If you’re old enough, you may remember the TV series The Prisoner or Get Smart, where characters, most of them secret agents, refer to each other by their code numbers all the time; when agents 86 and 99 team up, they don’t become agent 185 all of a sudden.
If you keep in mind this distinction between numbers, which represent quantities, and strings that merely look like numbers because they happen to consist entirely of integers, you can save yourself a lot of grief. For instance, when your manager decides to store the phone number 18003569377 as “1-800-FLOWERS”, dashes and all. Or when you need to store a foreign phone number and have to put a plus sign in front of the country code.
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Curl
(redirected from curls)
Also found in: Dictionary, Thesaurus, Medical, Idioms.
curl
Maths a vector quantity associated with a vector field that is the vector product of the operator ∇ and a vector function A, where ∇ = i∂/∂x + j∂/∂by + k∂/∂z, i, j, and k being unit vectors. Usually written curl A, rot A
Curl
of a vector field A, the vector characteristic of a “rotating component” of field A. The curl is represented by the symbol rot A. It can be interpreted in the following manner: Let A be the velocity field of a fluid flow. At a given point of the flow we place a small wheel with blades and orient its axis in the direction of rot A at that point. Then the angular velocity of the wheel’s rotation from the action of the current will be maximum, and its value will equal (| rot A |)/2. If the field A has the coordinates P(x, y, z), Q(x, y, z), and R(x, y, z), then the curl has the coordinates
E. G. POZNIAK
curl
[kərl]
(forestry)
A block of timber cut from a crotch for cutting into veneers.
(materials)
A defect of paper caused by unequal alteration in the dimensions of the top and underside of the sheet.
(mathematics)
The curl of a vector function is a vector which is formally the cross product of the del operator and the vector. Also known as rotation (rot).
curl
A winding, swirling, or circling in the grain of wood, usually obtained from the crotch or fork of a tree; also see fiddleback.
Curl
(1) A programming environment for developing rich Internet applications (RIAs) from Curl, Inc., Cambridge, MA (www.curl.com). Conceived at MIT, Curl combines HTML markup with an object-oriented programming language. The user's machine requires the Curl runtime engine and browser plug-in for execution. The first Curl implementation was released in 2002. See RIA.
(2) (cURL) A command line utility for executing functions with URL-oriented protocols such as FTP and HTTP. Pronounced "C-URL," there are versions for Unix, Linux, Windows, Mac and other operating systems. For more information, visit http://curl.haxx.se.
References in classic literature ?
However, I took care there should be no further scorning at my good nature: ever since, I've been as stiff as herself; and she has no lover or liker among us: and she does not deserve one; for, let them say the least word to her, and she'll curl back without respect of any one.
Now, I am a fidgety little man, and always love to have something in my fingers; so that, being debarred from my wife's curls, I looked about me for any other plaything.
Even your fair hand could not manage a curl more delicately than mine.
asked one tormentor, tweaking a curl that strayed out from the cushions.
The hair, dark, and still damp from the outdoor air, lay in loose waves about the forehead and curved back over the ears in wonderfully becoming lines, with softening little curls here and there.
And Miss Polly pulled the locks so tightly back that the last curl lay stretched dead at the ends of her fingers.
Then, whimpering again, and at intervals biting the curls, and stopping to look how much was bitten off, 'It's a shame
A gentleman coming in, Miss Bella, with a short and sharp exclamation, scrambled off the hearth-rug and massed the bitten curls together in their right place on her neck.
And never a flake That the vapour can make With the moon-tints of purple and pearl, Can vie with the modest Eulalie's most unregarded curl - Can compare with the bright-eyed Eulalie's most humble and careless curl.
And that one wee curl, always looking as if it were going to drop, but never dropping, is delicious.
He had certainly seen everything and with each curl of his lip, he declared that it amounted to nothing.
And you've got a little curl loose," he said, carefully turning her head round.
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## LCC '18 Contest 2 J1 - Derek and Snow
View as PDF
Points: 3
Time limit: 3.0s
Memory limit: 64M
Author:
Problem type
It's that time of the year again: WINTER! It has just snowed and Derek's driveway is covered in snow. It's Derek's job to shovel the snow but he hates doing it because it delays him from programming. Given , , , and , help Derek calculate how much time is needed to clear his driveway so he can get back to coding the next big game.
, the length of his driveway in metres
, the width of his driveway in metres
, the depth of the snow in metres
, the time (seconds) it takes to shovel 1 cubic metre
#### Input Specification
There will be a 4 lines of input, , , , and .
#### Output Specification
Output one number, rounded to exactly one decimal point, the number of seconds it takes for Derek to clear his driveway
#### Sample Input
3
4
0.5
20
#### Sample Output
120.0
#### Sample Explanation
The dimensions of the snow on his driveway is . It takes Derek 20 seconds per , so it takes him 120 seconds.
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# Rates
### Rates
Who is the fastest runner? Which one is the best buy? Learn how to find the unit rate and unit price to answer these questions. This lesson is very useful to our daily life.
#### Lessons
• 1.
Find the unit rate.
a)
$4.50 for 2 lb of broccoli. b) 5 pizzas for 12 people. c) 2 goals in a 90 minute long soccer game. • 2. Find the unit price and identify the best buy. a) No. of Units Total Price Unit Price 250 ml$3.50 400 ml $4.50 550 ml$6.60
b)
No. of Units Total Price Unit Price 300 g $5.20 550 g$9.20 1 kg \$15.50
• 3.
Runner Distance (km) Time (min) Ann 15 90 Katie 2 120 Dora 3 180
a)
Determine the distance each runner ran per minute.
b)
c)
Who is the fastest runner?
##### Do better in math today
5.
Introduction to Ratios, Rates and Percentages
5.1
Ratios
5.2
Rates
5.3
Proportions
5.4
Representing percents
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## Wein's Displacement Law - MCQ - Basic Level
Physics is an important subject in preparation for various Competitive exams. To keep a track of your preparation we have designed a small quiz of Basic Level Questions on Wein's Displacement Law. You have 10 minutes to solve the quiz, click the start timer button in order to start the timer and then click on submit button at the end to submit the quiz and view your scores and correct answers.
Q1. A black body has maximum wavelength λm at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be
• 3/2 λm
• 2/3 λm
• 4/9 λm
• 9/4 λm
Q2. The wavelength of radiation emitted by a body depends upon
• The nature of its surface
• The area of its surface
• The temperature of its surface
• All the above factors
Q3. On investigation of light from three different stars A, B and C, it was found that in the spectrum of A the intensity of red colour is maximum, in B the intensity of blue colour is maximum and in C the intensity of yellow colour is maximum. From these observations it can be concluded that
• The temperature of A is maximum, B is minimum and C is intermediate
• The temperature of A is maximum, C is minimum and B is intermediate
• The temperature of B is maximum, A is minimum and C is intermediate
• The temperature of C is maximum, B is minimum and A is intermediate
Q4. If black wire of platinum is heated, then its colour first appear red, then yellow and finally white. It can be understood on the basis of
• Wien's displacement law
• Prevost theory of heat exchange
• Newton's law of cooling
• None of the above
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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING: Wein's Displacement Law - MCQ - Basic
Wein's Displacement Law - MCQ - Basic
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Community Profile
# Andriy Kavetsky
Last seen: plus d'un an il y a Actif depuis 2016
I'm student of Applied Mathematics at Ivan Franko National University.
All
#### Content Feed
Afficher par
A résolu
Determine if input is odd or even
Determine if input is odd or even, sample problem
presque 4 ans il y a
A résolu
Calculate the square of a number (Super Easy)
The goal is to calculate the square (y) of a number (x). Good way to start MatLab for beginners.
presque 4 ans il y a
A résolu
Remove all the consonants
Remove all the consonants in the given phrase. Example: Input s1 = 'Jack and Jill went up the hill'; Output s2 is 'a ...
environ 4 ans il y a
A résolu
Pascal's Triangle
Given an integer n >= 0, generate the length n+1 row vector representing the n-th row of <http://en.wikipedia.org/wiki/Pascals_t...
environ 4 ans il y a
A résolu
Get the combinations
Consider p,q = 2 vectors of same or different length. Get a Output Array which has all the possible combinations of Elements o...
environ 4 ans il y a
A résolu
Transpose the Matrix
Transpose the given matrix, e.g. x=[a b;c d] transpose of x = [a c;b d]
environ 4 ans il y a
A résolu
Project Euler: Problem 9, Pythagorean numbers
A Pythagorean triplet is a set of three natural numbers, a b c, for which, a^2 + b^2 = c^2 For example, 3^2 + 4^2 =...
environ 4 ans il y a
A résolu
Project Euler: Problem 2, Sum of even Fibonacci
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 te...
environ 4 ans il y a
A résolu
Find the nearest prime number
Happy 5th birthday, Cody! Since 5 is a prime number, let's have some fun looking for other prime numbers. Given a positive in...
environ 4 ans il y a
A résolu
Extra safe primes
Did you know that the number 5 is the first safe prime? A safe prime is a prime number that can be expressed as 2p+1, where p is...
environ 4 ans il y a
A résolu
Octoberfest festival
A group of students decided to visit Octoberfest festival. First they ordered one beer, then after half-hour they taken one more...
environ 5 ans il y a
Problème
Octoberfest festival
A group of students decided to visit Octoberfest festival. First they ordered one beer, then after half-hour they taken one more...
environ 5 ans il y a | 3 | 458 solveurs
A résolu
Given input in degrees, output to radians
presque 6 ans il y a
A résolu
Calculate area of sector
A=function(r,seta) r is radius of sector, seta is angle of sector, and A is its area. Area of sector A is defined as 0.5*(r^2...
presque 6 ans il y a
A résolu
0, 2, 0, -2, 0, 2, 0, -2, ...
Generate the first n terms of a periodic sequence defined as f(x) = 0, 2, 0, -2, 0, 2, 0, -2, ..., for x = 1, 2, 3, 4, 5, 6...
presque 6 ans il y a
A résolu
true or false
if the matrix has a zero, return true. else, return false
presque 6 ans il y a
A résolu
Rankine to Kelvin Converter
Given input R, degrees Rankine, convert into degrees Kelvin.
presque 6 ans il y a
A résolu
MPS to MPH
Given the speed in miles per second, find the speed in miles per hour.
presque 6 ans il y a
A résolu
Kinetic Energy
Given mass, m, and kinetic energy, KE, find the velocity of the object.
presque 6 ans il y a
A résolu
presque 6 ans il y a
A résolu
presque 6 ans il y a
A résolu
if
if value n is larger than or equal to 100, return true, else false.
presque 6 ans il y a
A résolu
Sum the rows
Sum the rows of the given matrix. Example x = [ 1 2 3 4 ] y = [ 3 7 ]
presque 6 ans il y a
A résolu
Return the length of the longest word in the string.
Example: Input s='A fractal is a natural phenomenon or a mathematical set that exhibits a repeating pattern that display...
environ 6 ans il y a
A résolu
Set the array elements whose value is 13 to 0
Input A either an array or a vector (which can be empty) Output B will be the same size as A . All elements of A equal to 13...
environ 6 ans il y a
A résolu
Solve the following boundary value problem
Return the sum of sum(res.y) y"+|y|=0 y(0)=a y(4)=b Tip: use bvp4c
environ 6 ans il y a
A résolu
Return ! if array element starts with a certain letter. Otherwise return ?
Input x=['try' 'once' 'more']; s='t' Output ans='!' Input x=['try' 'once' 'more']; s='O' Output ...
environ 6 ans il y a
A résolu
Check if a string starts with another string
Input s='He was so tired' q='He' Output ans = true
environ 6 ans il y a
A soumis
CalcSum( x,epselon )
CalculateSumWithPrecision
environ 6 ans il y a | 1 téléchargement |
A résolu
Small aircraft weight limit
Write an if-else statement that evaluates to true if the sum of passengerWeight and cargoWeight is less than or equal maxWeight....
environ 6 ans il y a
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## Problem L. LCA queries ≡
• problems
Author: A. Usmanov Time limit: 1 sec Input / output: interactive Memory limit: 256 Mb
### Statement
This is an interactive problem. You will interact with a server by sending and receiving particular messages.
You are given a complete binary tree of height N having maximum possible number of nodes.
Nodes are numbered from 1 to 2N − 1.
Your task is to determine the number of root. You can perform requests to find least common ancestor (LCA) of any two nodes to achieve this. You are allowed to perform no more than N requests.
### Input format
The first line contains one integer N — height of the tree.
### Interaction protocol
To make a request print a line "? X Y", where X and Y — are integer numbers of nodes. After such a request you receive one integer — the number of least common ancestor of these two nodes.
When your program determines the root of the given tree, it should print "! X" and immediately stop.
If your program makes incorrect query, it will receive "Presentation error" verdict. If your program exceeds allowed number of queries, it will receive "Wrong answer" verdict.
Every query and final output must be a single line ending with single end-of-line (\n). Output buffers must be flushed after every line:
Language C++ Pascal Java Python Code cout.flush() flush(output) System.out.flush() stdout.flush()
1 ≤ N ≤ 10
1 ≤ X, Y < 2N
### Explanation of the samples
The first sample contains the next tree:
### Sample tests
No. Standard input Standard output
1
3
2
5
7
? 3 6
? 1 5
? 2 4
! 7
2
1
! 1
0.077s 0.010s 15
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https://www.accountingformanagement.org/diad-fib-quiz/
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# Depreciation, impairments and depletion Fill-in-the-blanks quiz
### Directions!
Read the whole text carefully and think the correct word that should fill in the blank space(s) and then confirm your answer by pressing the 'see answer' button. Don't press the 'see answer' button before reading the question text.
If you are unable to solve these fill-in-the-blank questions, read "depreciation, impairments and depletion" chapter from explanations section of our website.
Question: 1
is a process of cost allocation, not valuation.
Question: 2
The cost of an asset is initially recorded as a(n) .
Question: 3
Depreciable cost is equal to cost of the asset less .
Question: 4
Under method of depreciation, equal portion of depreciable cost is allocated to each period of the useful life of asset.
Question: 5
The method of depreciation focuses on the actual use of the asset rather than just passage of time.
Question: 6
If the useful life of asset is 20 years, the straight line depreciation rate would be .
Question: 7
method of depreciation assumes that the depreciation is a function of passage of time rather than actual productive use of the asset.
Question: 8
A company uses double declining balance method to compute depreciation on fixed assets. If straight line depreciation rate is 10%, the declining balance rate would be .
Question: 9
The sum of years' digits method assumes that the productivity of asset with the passage of time.
Question: 10
Sum of years’ digits method attempts to charge a depreciation in early years of an asset's useful life.
Question: 11
An asset is purchased on January 1, 2018. The cost of the asset is \$150,000 and the useful life is 5 years with no salvage value. The depreciation charge for the first year under sum of years' digits method would be .
Question: 12
An asset is purchased on January 1, 2018. The cost of the asset is \$150,000 and the useful life is 15 years with \$10,000 salvage value. The estimated productive life of the asset is 20,000 hours. If the asset is used for 2,000 hours during the year 2018, the depreciation charge under activity method for the year 2018 would be .
Question: 13
An asset is purchased on January 1, 2018. The cost of the asset is \$150,000 and the useful life is 20 years with \$10,000 salvage value. The yearly depreciation charge under straight line method would be .
Question: 14
Accelerated depreciation and sum-of-the-years' digits methods are two forms of .
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# 20 Writing Balanced Chemical Equations Worksheet
49 Balancing Chemical Equations Worksheets [with Answers] writing and balancing chemical reactions worksheet answers pdf, writing and balancing chemical equations worksheet answers pdf, writing and balancing chemical equations worksheet 2 answer key, writing balancing chemical equations worksheet, writing and balancing chemical equations from words worksheet answers, via: templatelab.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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# Multiple Regression with Interaction
I've come across somewhat of a confusing topic relating to the syntax of multiple regression with explanatory variables and their interactions. A DataCamp explanation led me to think that:
`lm(formula = y ~ r + r:s , data)`
...is the same as:
`lm(formula = y ~ r + s + r:s , data)`
Which is incorrect. I have found that the latter is in fact the same as the shortened version:
`lm(formula = y ~ r * s , data)`
But the former is certainly different.
What exactly is the difference between these - that is, what does the first model show that the latter two wouldn't?
Thank you.
## Simple Regression:
It is a subtle difference, but there is certainly a difference there. One way you can easily visualize the differences is by using the `summary` command. I will use the `iris` dataset since its already in R. First, a simple linear regression:
``````# Simple regression:
summary(lm(formula = Sepal.Width ~ Sepal.Length,
data = iris))
``````
This will just show the one independent variable, Sepal.Length, on the dependent variable, Sepal.Width:
``````Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.41895 0.25356 13.48 <2e-16 ***
Sepal.Length -0.06188 0.04297 -1.44 0.152
``````
## Interaction and Main Effects
For the next equation with just the `*` input:
``````# Interaction and main effects:
summary(lm(formula = Sepal.Width ~ Sepal.Length*Petal.Length,
data = iris))
``````
It gives us both the main effects of each independent variable/predictor, while also giving us the interaction between the two. You can see them all listed under coefficients now:
``````Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.51011 0.64336 2.347 0.020257 *
Sepal.Length 0.46940 0.12954 3.624 0.000400 ***
Petal.Length -0.42907 0.11832 -3.626 0.000397 ***
Sepal.Length:Petal.Length 0.01795 0.02186 0.821 0.413063
``````
## Only Interaction
For the `:` input, it gives us only the interaction and nothing else:
``````# Only interaction:
summary(lm(formula = Sepal.Width ~ Sepal.Length:Petal.Length,
data = iris))
``````
Which you can see below:
``````Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.31473 0.06852 48.375 < 2e-16 ***
Sepal.Length:Petal.Length -0.01108 0.00257 -4.312 2.93e-05 ***
``````
## Manually Adding Both Interactions and Effects
Finally, if you are entering interactions AND manually adding main effects, you would simply use the `:` input again, but then use `+` to add a main effect:
``````# Only interaction and one main effect:
summary(lm(formula = Sepal.Width ~ Sepal.Length + Sepal.Length:Petal.Length,
data = iris))
``````
As seen below:
``````Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.299034 0.422673 -0.707 0.48
Sepal.Length 0.807410 0.093603 8.626 9.44e-15 ***
Sepal.Length:Petal.Length -0.058626 0.005899 -9.939 < 2e-16 ***
``````
Notice when I do the same call of using `+` and `*` now, it still just gives both the interaction and main effects without specifying.
``````summary(lm(formula = Sepal.Width ~ Sepal.Length + Sepal.Length*Petal.Length,
data = iris))
``````
In a sense it actually ignores the plus sign:
``````Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.51011 0.64336 2.347 0.020257 *
Sepal.Length 0.46940 0.12954 3.624 0.000400 ***
Petal.Length -0.42907 0.11832 -3.626 0.000397 ***
Sepal.Length:Petal.Length 0.01795 0.02186 0.821 0.413063
``````
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# Physics
posted by .
A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance.
I still get 0.315 m as the answer, but it's incorrect. What am I doing wrong?
• Physics -
If you say the final speed is zero, where it stops, then .315 meters is correct
However the final speed is .95 m/s, not zero. That gives me .238 meters.
I will type it up in a minute.
• Physics -
F = m g (.35cos 15 + sin15)
F = m (9.8) (.5969)
a = F/m = 5.85 m/s^2 down the ramp
Vo = 1/92 m/s
Vf = .95 m/s
v = Vo - 5.85 t
.95 = 1.92 - 5.85 t
t = .1658
x = 1.92 t - .5 (5.85) t^2
x = .318 - .0804
x = .238
• Physics -
Ahhh ok that was what was wrong.
Thanks a ton. :)
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# Lang's Algebra: Herbrand quotient
I've looked around a lot and couldn't find much help (at least that I could understand) on this question - it is 1.45 in Lang's Algebra book:
Let $G$ be a cyclic group of order $n$, generated by $\sigma$. Assume $G$ acts on an abelian group $A$ as groups s.t. $\sigma(x+y) = \sigma(x)+\sigma(y)$ for $x,y \in A$, and let $f,g: A \to A$ be the homomorphisms defined as:
$$f(x) = \sigma\,x - x$$ and $$g(x) = x + \sigma\,x + \cdots + \sigma^{n-1}\,x$$
Herbrand quotient is given as $q(A) = (A_f: A^g)/(A_g:A^f)$, provided both indices are finite. And, $A_f$ and $A^f$ are the kernel and image of map $f$. Assume $B$ is a subgroup of $A$ s.t. $GB \subset B$. Then
a.) Define in a natural way an operation of $G$ on $A/B$
b.) Prove that $q(A) = q(B)\,q(A/B)$ Hint: consider complex: $E: 0 \to A_g \to A \overset{g}{\to} A \overset{f}{\to} A \overset{g}{\to} A^g \to 0$
Hint: $K(A): \cdots A_i \overset{d_i}{\to} A_{i+1} \overset{d_{i+1}}{\to} \cdots$ where $A_i = A$ for all $i$ and $d^i = f$ if $i$ is even and $d^i = g$ if $i$ is odd. Similarily consider $K(B)$ and $K(A/B)$. Examine long exact sequence on cohomology associated to the exact sequence of complexes $0 \to K(B) \to K(A) \to K(A/B) \to 0$. Keep in mind complexes $K$ are periodic so the long exact sequence will also be periodic, of the form $H^0(K(B)) \to H^0(K(A)) \to H^0(K(A/B)) \to H^1(K(B)) \to H^1(K(A)) \to H^1(K(A/B))$
c.)If $A$ is finite, then $q(A) = 1$.
So I fiddled around with $C_n$ groups and their subgroups to see the homomorphism work, and I can think of the isomorphism theorems - making me think of the quotient group - but I am stuck on the first part of the question.
I would appreciate any help with this!!
-
a) is trivial.
b) Let $K(A)$ be the following complex, where $A_i = A$ for all $i$ and $d^i = f$ if $i$ is even and $d^i = g$ if $i$ is odd.
$$\cdots\rightarrow A_i\rightarrow A_{i+1} \rightarrow\cdots$$
Similarly we define $K(B)$ and $K(A/B)$. Then there exits the following exact sequence of complexes.
$$0\rightarrow K(B) \rightarrow K(A) \rightarrow K(A/B) \rightarrow 0$$
Let $H_i(A)$(resp. $H_i(B)$, $H_i(A/B)$) be the $i$-th homology group of $K(A)$(resp. $K(B)$, $K(A/B)$).
Then we get the following exact sequence.
$\cdots \rightarrow H_1(A/B) \rightarrow H_0(B) \rightarrow H_0(A) \rightarrow H_0(A/B) \rightarrow H_1(B) \rightarrow H_1(A) \rightarrow H_1(A/B) \rightarrow H_0(B) \rightarrow\cdots$
We denote $|H_0(A)|$ by $h_0(A)$. Similarly we define $h_1(A)$, $h_0(B)$, etc..
We denote by $m_0(A)$ the order of image of $H_0(B) \rightarrow H_0(A)$. Similarly we define $m_1(A)$, $m_0(B)$, etc..
Then
$h_0(B)/m_0(B) = m_0(A)$
$h_0(A)/m_0(A) = m_0(A/B)$
$h_0(A/B)/m_0(A/B) = m_1(B)$
$h_1(B)/m_1(B) = m_1(A)$
$h_1(A)/m_1(A) = m_1(A/B)$
$h_1(A/B)/m_1(A/B) = m_0(B)$
Hence
$h_0(A) = m_0(A)m_0(A/B)$
$h_1(A) = m_1(A)m_1(A/B)$
$h_0(B) = m_0(B)m_0(A)$
$h_1(B) = m_1(B)m_1(A)$
$h_0(A/B) = m_0(A/B)m_1(B)$
$h_1(A/B) = m_1(A/B)m_0(B)$
Hence
$$q(A) = \frac{h_0(A)}{h_1(A)} = \frac{m_0(A)m_0(A/B)}{m_1(A)m_1(A/B)}$$
$$q(B) = \frac{h_0(B)}{h_1(B)} = \frac{m_0(B)m_0(A)}{m_1(B)m_1(A)}$$
$$q(A/B) = \frac{h_0(A/B)}{h_1(A/B)} = \frac{m_0(A/B)m_1(B)}{m_1(A/B)m_0(B)}$$
Hence
$$q(A) = q(B)q(A/B)$$
c)
$$0 \subset A^g \subset A_f \subset A$$
$$0 \subset A^f \subset A_g \subset A$$
Hence
$|A| = [A:A_f][A_f : A^g]|A^g|$
$|A| = [A:A_g][A_g : A^f]|A^f|$
Since $[A : A_f] = |A^f|$ and $[A : A_g] = |A^g|$, $[A_f : A^g] = [A_g : A^f]$. Hence $q(A) = 1$.
-
Firstly, thank you! I was hoping that maybe you could explain a little more in the steps. I am just learning about complexes and chains and, while I have yet to digest all of this, I am afraid I may not be able to understand it... For instance, could you explain a little bit more about the use of the complex E in the hint, especially with respect to the kernels and images of the maps? Thank you again! – nate Sep 14 '12 at 3:50
@nate Do you understand that there exits the following exact sequence of complexes? $0\rightarrow K(B) \rightarrow K(A) \rightarrow K(A/B) \rightarrow 0$ – Makoto Kato Sep 14 '12 at 4:00
Hi, I read about complex chains and found a link onsite, http://math.stackexchange.com/questions/26328/homology-of-a-simple-chain-complex?rq=1, and even though I'm learning dealing with cochains, I was able to follow it and do it myself. Now to try your question.... – nate Sep 14 '12 at 5:27
@nate Chapter 20 of Lang's algebra treats chain complexes and their homologies. I think it gives you enough knowledge to understand my proof. – Makoto Kato Sep 14 '12 at 5:31
Okay, well thanks a lot for your help - I'll give the Lang book another try (it is sort of terse for me!) :) – nate Sep 14 '12 at 6:29
This should be very standard: To define an action of $G$ on $A/B$, you need to specify $g(aB)$ for $g\in G$ and cosets $a+B\in A/B$. The obvious way is to set $g(a+B)=g(a)+B$, but you need to check that this is well-defined: If $a+B=a'+B$, then $g(a)-g(a')=g(a-a')\in g(B)\subseteq B$, hence $g(a)+B=g(a')+B$ as desired.
-
Okay, I understand the action definition - was pretty standard looking. Would anyone happen to have class notes / links of something to study? Not a tome or in-depth book at first ;) Thanks! – nate Sep 14 '12 at 1:10
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[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
There are infinitely many functions which have those characteristics. However, here's how to sketch a function like that.
g(0)=1 tells us that the function passes through the point (0,1).
g'(0)=4 tells us that the function is increasing at 0 at a rate of about 4:1.
g'(1) = 0 tells us that the function has a critical point at 1 (the tangent line has slope 0 (horizontal)).
g'(2) = -2 tells us that the function is decreasing at a rate of -2:1.
Really, that's all we have to go on. You can graph any old function that does those things and you're good.
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# Contour integration
Homework Helper
Gold Member
[SOLVED] contour integration
## Homework Statement
I'm really rusty with this. I need to calculate
$$2\pi i Res\left(\frac{1}{1+z^4},e^{i\pi/4}\right)$$
## The Attempt at a Solution
Well,
$$2\pi iRes\left(\frac{1}{1+z^4},e^{i\pi/4}\right)=\int_{C_{\rho}}\frac{1}{1+z^4}dz$$
where $$C_{\rho}$$ is a little circle of radius rho centered on $$e^{i\pi/4}$$, on which there are no singularities. Fine, so let's take $$\rho=1/\sqrt{2}$$.
Now I need to parametrize C_rho. Take
$$\gamma(t)=e^{i\pi/4}+\frac{e^{i2\pi t}}{\sqrt{2}} \ , \ \ \ \ 0\leq t < 1$$
We have
$$\frac{d\gamma}{dt}=i\sqrt{2}\pi e^{i2\pi t}$$
So that
$$\int_{C_{\rho}}\frac{1}{1+z^4}dz=\int_0^1\frac{i\sqrt{2}\pi e^{i2\pi t}}{1+(e^{i\pi/4}+\frac{1}{\sqrt{2}}e^{i2\pi t})^4}dt$$
Now what?? :grumpy:
I believe the answer is supposed to be $$-e^{i\pi/4}/4$$
Last edited:
Why don't you just calculate the residue directly?
Homework Helper
Gold Member
The singularity at $$e^{\frac{{i\pi }}{4}}$$ is a simple pole and the numerator of the integrand is non zero and holomorphic at the singularity. So you can calculate the residue by using the formula
$${\mathop{\rm Re}\nolimits} s\left( {\frac{1}{{z^{^4 } + 1}}} \right) = \frac{{1_{z = z_0 } }}{{\frac{d}{{dx}}\left( {z^4 + 1} \right)}}_{z = z_0 } ,z_0 = e^{\frac{{i\pi }}{4}}$$
In the calculation you can simplify the algebra a little by using $$z_0 ^4 = - 1$$.
The result should coincide with the answer that you expected to get as indicated in your original post.
Last edited:
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A differential equation is a relation between a mathematical function and its derivates. If the function depend on more than one variable, the differential equation is said to be partial. These are a lot harder to solve analytically and numerically than the ordinary differential equations. The theories for PDEs has been developed mainly for the field of theoretical physics, where they play a very important part, especially in quantum mechanics. Below I list the most important PDEs in physics and how you would solve them.
(An excuse goes out to all Netscape users who may not be able to see the equations below.)
One of the most important PDEs in physics is Poisson's partial differential equation, which is what mechanical problems usually reduces to:
``` n ∂2u
Δu = ∑ ----- = f (1)
j=1 ∂x2j
```
Where u is the sought function, f is a known function, and x is a coordinate in space. In physics, n usually equals 3, for our 3 dimensional world. When f=0 the equation is called Laplace's equation. The Newton potential gives solutions to this as:
``` 1 f(y)
u(x) = ---- ∫∫∫------- dy1dy2dy3
4π |x-y| ```
In physical applications of this, you usually search for solutions within a domain Ω where you have a Dirichlet boundary condition u = φ on the boundary ∂Ω. There are many methods to find this solution analytically, and usually they are focused on finding upper and lower values of the function . This PDE can also be solved for other boundary conditions such as Neumann boundary condition. Equation (1) is and example of a elliptical differential equation. The term "elliptical" refers to the characteristic polynomial of the equation, which is a way to classify PDEs.
Another common PDE is the heat equation, or diffusion equation:
```∂u
--- - Δu = f (2)
∂t ```
where u is a function both in the room (x1, x2, x3... xn) and time (t). For n=3 this equation describes the distribution of heat in an homogenous material or some other process of diffusion in the physical world. If we have the initial value of u, u(t=0) = u0, the solutions for t > 0 can be derived from
```
u(t,x) = (4πt)-π/2∫e-|x-y|2/4t·u0(y)dy ```
For solutions within a domain Ω you can also apply boundary conditions which makes the PDE solvable. Equation (2) is a hypo elliptical differential equation. In quantum mechanics, the Schrödinger equation is an important PDE, and it is fairly similar to the heat equation. The difference is that it has a imaginary factor:
``` ∂u
i--- - Δu + Vu = 0 (3)
∂t ```
This equation is not easily solved, but great progress has been made during the last decades. In physics, where V is the same potential function as in the N-body problem, and u = u(x,y,z), this is the equation for the quantum mechanical, non-relativistic N-body problem. For each φ where |φ(x)|2dx < there is a solution u to (3) which is equal to φ for t = 0 and |u(t,x)|2dx is independent of t.
The wave equation, which describes the movement of light and electromagnetic waves, is also a PDE:
```∂2u
----- - Δu = f (4)
∂t2```
Maxwell's equations reduce to the above. For u = u(x,y,z) a solution is given by
``` f(t-|y|,x-y)
u(t.x)= ∫∫∫------------ dy1dy2dy3
4π|y|```
The wave equation (4) is an example of a hyperbolical differential equation. Cauchy's problem, to find a solution for u=φ and ∂u/∂t=Ψ, can easily be derived from this last equation. Other problems lead to more complex calculations. For instance, the Dirichlet problem leads to a complicated study of singularities, which is related to the study of refraction and diffraction in optics
All of the above are examples of linear partial differential equations. However, many of the fundamental equations in physics are non-linear, such as Navier-Stokes equations in fluid dynamics. These are a bit harder to solve analytically and are usually the subject to numerical methods. Usually one has to study the linear solution to the non-linear problem, and then based upon this adjust the solution in the surrounding you are interested in.
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# Gartley Pattern and Trading the Patterns
### INTRODUCTION TO GARTLEY PATTERN
The Gartley pattern was first introduced by H.M. Gartley in his book “Profits in the Stock Market”, which was published in 1935. The pattern was named “The Gartley,” but in fact, many variations of the Gartley pattern have become common ever since the release of that book. Gartley patterns are chart patterns used in technical analysis and are known for their relationship using Fibonacci numbers and ratios. The Gartley pattern is a reversal pattern with clear rules and provides an excellent reward to risk.
“Classical” chart patterns are considered to be: flag, pennant, wedge pattern, triangle, range, rectangle, flat, head and shoulders, inverted head and shoulders, double top and bottom, trip top and bottom, gap, cup and handle, broadening top. Some of these patterns are reversal signals, others are continuation patterns. Most of the classical charts patterns use Fibonacci levels as well. A flag will typically find support levels at the various Fibonacci points (such as 23.6% and 38.2%) of various swing highs and lows but they are not so prominently used as in Gartley. Read more here.
5 LETTERS
The Gartley is using 5 letters to distinguish the 5 separate moves/waves/impulses. Here is an introduction:
The letter X is the start of the trend;
The letter A is the end of the trend;
The letter B is the first pullback of the trend;
The letter C is the pullback of the pullback (not breaking point A);
The letter D is the target of the letter C.
The various Fibonacci relationships between XA and AB have a value when calculating targets for B, C and D. Depending on the type of Fibonacci level the pattern is commonly named differently. The pattern is valid for both a down and an uptrend. In general, though, there is an also a close link to the Elliott Wave Theory. The AB, BC, and CD legs are also known in EW as an ABC correction of XA and a continuation of the XA direction can be expected at point D.
We will now go into the specific Gartley patterns which are usually called Bat, Crab, Gartley, Butterfly, etc.
GARTLEY
This pattern is valid when price respects and bounces off of the XA swing high swing low to form point B at the 61.8% Fibonacci retracement level. The target of point D is, in fact, using the same XA swing high swing low and is aiming for the 78.6% Fibonacci retracement level of XA. The CD leg is therefore often equal to the AB leg.
Other modern variations that have become popular are listed here below.
BAT PATTERN
This pattern is valid when price respects and bounces off of the XA swing high swing low to form point B at the 38.2% or 50% Fibonacci retracement levels (but not more than 61.8%). The target of point D is, in fact, using the same XA swing high swing low and is aiming for the 88.6% Fibonacci retracement level of XA. The CD leg is, therefore, longer than the AB leg.
ALTERNATE BAT PATTERN
This pattern is valid when price respects and bounces off of the XA swing high swing low to form point B at the 38.2% Fibonacci retracement level. The target of point D is beyond the origin of XA and is 1.13 of XA.
BUTTERFLY
This pattern is valid when price respects and bounces off of the XA swing high swing low to form point B at the 78.6% Fibonacci retracement level. The target of point D is beyond the origin of XA and is 1.27 – 1.618 of XA.
CRAB
This pattern is valid when price respects and bounces off of the XA swing high swing low to form point B at the 38.2%-61.8% Fibonacci retracement levels. The target of point D is beyond the origin of XA and is 1.618 of XA.
DEEP CRAB
This pattern is valid when price respects and bounces off of the XA swing high swing low to form point B at the 88.6% Fibonacci retracement level. The target of point D is beyond the origin of XA and is 1.618 of XA.
AB=CD, 5-0, DRIVERS
Here are the remainder of the popularized patterns. An example of 5-0:
An example of AB=CD:
An example of drivers:
SIMPLIFY: SPLIT OF LEVELS
When analyzing the patterns, it becomes obvious that different patterns play out depending on where letter B stops in relationship with XA. This is my attempt to make the patterns easier to interpret (drivers and 5-0 excluded).
PART I: Let us break it down into Fibonacci levels.
1) 38.2% – Bat / Alternate Bat / Crab
2) 50% – Bat / Crab
3) 61.8% – (Bat) / Crab / Gartley / ab=cd
4) 78.6% – Butterfly
5) 88.6% – Deep Crab
That means that if the currency bounces up at the 38.2% for instance, then there could 3 Gartley patterns in play.
PART II: Let us continue with this breakdown and analyze the likely Fibs where letter C can stop when Fibbing AB and the answer is simple: C can stop at any Fib of AB, which is 38.2%, 50%, 61.8%, 78.6%, 88.6%.
PART III: The last but not least, the target D.
1) Bat – 88.6% Fib of XA OR 2.618 of BC
2) Alternate Bat – 113% Fib of XA (below X) OR 2.0 of BC
3) Crab – 161.8% Fib (below X) OR 3.14 of BC
4) Gartley – 78.6% Fib of XA OR 1.27 of BC
5) Butterfly – 161.8% Fib (below X) OR 1.618 of BC
6) Deep Crab – 161.8% Fib (below X) OR 2.618 of BC
Trading the patterns forex is as always a matter of entry methodology. We discussed entry techniques in a previous article: please read the entire article here. In general, it boils down to either entering upon a direct level, a confirmation or a momentum break.
For the Gartley patterns mentioned here, a direct level entry means a pending entry order at a specific Fibonacci level. A confirmation would be to wait for a candlestick reversal pattern at the Fib. And the break out would occur when price bounces off the Fib and breaks a trend line in the anticipated direction.
Please note that trading letter B is a with the trend setup but with a limited target (target is letter C). Trading letter C is a reversal trade but with good reward to risk (target is letter D). Trading letter D could be seen as with the trend trade (very close to support and resistance in any case) and good reward to risk as well (target can be the top in up trend example, bottom in down trend example OR any Fib from C to D).
What is your opinion on Gartley Pattern? Do like trading the patterns? Do you want to use them? Do you like it? Do you already use Fibs? Has this helped your trade management? We gave a practical example of Gartley this week in the GBPAUD reversal article. Here is part 1 and 2 (part 2 has the pattern).
Update 1: the bat screenshot has been changed on 13th of October.
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• Zaid Sarwar
Great Great break down of patterns. You guys helped me alot . I would say best article for peoplle trying to know pattern trading. I hope you guys keep on going
• Nathan
Hi Chris: One more question pls: While I have found it easier to identify points, B,C and the XA leg, typically there are several other swings before point D is reached. Hence I get confused as to which point to pick for B, since there are multiple swing points which look like candidates for B. Is this common? Or am I reading the price action incorrectly?
• Nathan
Excellent Chris! Quick question: Is there a time relationship between the XA, AB, CD etc legs as well?
• Chris Svorcik
Oops – great catch! Thanks for that heads-up!
The bat screenshot has been modified in the meantime. Thanks!
• Chris Svorcik
Thank you Rohan! 🙂 IT is definately good pattern to keep in mind when trading 🙂 thanks for the feedback – much appreciated!
• Chris Svorcik
Thanks Russell, good tip for our readers. Much appreciated.
• Chris Svorcik
That’s great! 🙂
• Chris Svorcik
Hi Adam, that is great! These patterns and Fibs are indeed very interesting. Glad you agree. Good Trading!
• Chris Svorcik
Hi Selemon, happy to hear that you found the article interesting and useful! 🙂
Your feedback is much appreciated 🙂 wish you Good Trading this week!
• smayer97
BTW, your diagrams for the Bat and Alternate Bat are IDENTICAL!. One of them is wrong. You may want to correct that.
• Rohan De Villiers
EXCELLENT article !! going to look for them in last weeks charts and see where i missed them
• Russell Daily
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Physics
posted by .
There is a small amount of water at the bottom of a sealed container of volume 5.2 liters which is otherwise full of an ideal gas. A thin tube open to the atmosphere extends down into the water. The system is initially at atmospheric pressure and temperature 113 Celsius.
If we increase the temperature of the gas until water rises in the tube to a height of 120 cm, then what is the temperature at that instant?
• Physics -
If the system is initially at 1 atm, then there can be no liquid water at the bottom when the temperature is 113 C. It would all have evaporated.
Maybe they are expecting you to assume that the water has not yet heated up to the gas temperature. This would be a nonequilibrium state. You could use the water column height to get the new water pressure, and use that to compute the new temperature.
This is a poorly thought out question, in my opinion.
• Physics -
how would i do that using the water coumn height?
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# Wavelets and Multiwavelets
Wavelets and Multiwavelets Fritz Keinert Studies in advanced mathematics, vol. 42 Chapman & Hall/CRC Press, Boca Raton, FL ISBN 1-58488-304-9 QA403.3 K45 2003 515'.2433—dc22 CRC Press web site for this book
## Back Cover
(with description of contents)
## Errata
Known errors are listed in file errata.pdf. If you find any others, please report to the author.
## Software
A toolbox of Matlab subroutines is made available with the book. Documentation (in both .pdf and .ps format) is included in the archive.
mw.tar Unix tar format, 960 KB mw.tar.Z compressed Unix tar format, 604 KB mw.zip PC zip archive of Matlab subroutines, 460 KB
Preface
Contents
Part I - Scalar Wavelets
1 Basic Theory
1.1 Refinable Functions
1.2 Orthogonal MRAs and Wavelets
1.3 Wavelet Decomposition
1.4 Biorthogonal MRAs and Wavelets
1.5 Moments
1.6 Approximation Order
1.7 Symmetry
1.8 Point Values and Normalization
2 Practical Computation
2.1 Discrete Wavelet Transform
2.2 Pre- and Postprocessing
2.3 Handling Boundaries
2.3.1 Data Extension Approach
2.3.2 Matrix Completion Approach
2.3.3 Boundary Function Approach
2.4 Putting It All Together
2.5 Modulation Formulation
2.6 Polyphase Formulation
2.7 Lifting
2.8 Calculating Integrals
2.8.1 Integrals with Other Refinable Functions
2.8.2 Integrals with Polynomials
2.8.3 Integrals with General Functions
3 Creating Wavelets
3.1 Completion Problem
3.1.1 Finding Wavelet Functions
3.1.2 Finding Dual Scaling Functions
3.2 Projection Factors
3.3 Techniques for Modifying Wavelets
3.4 Techniques for Building Wavelets
3.5 Bezout Equation
3.6 Daubechies Wavelets
3.6.1 Bezout Approach
3.6.2 Projection Factor Approach
3.7 Coiflets
3.7.1 Bezout Approach
3.7.2 Projection Factor Approach
3.7.3 Generalized Coiflets
3.8 Cohen Wavelets
3.9 Other Constructions
4 Applications
4.1 Signal Processing
4.1.1 Detection of Frequencies and Discontinuities
4.1.2 Signal Compression
4.1.3 Denoising
4.2 Numerical Analysis
4.2.1 Fast Matrix--Vector Multiplication
4.2.2 Fast Operator Evaluation
4.2.3 Differential and Integral Equations
5 Existence and Regularity
5.1 Distribution Theory
5.2 L^1-Theory
5.3 L^2-Theory
5.3.1 Transition Operator
5.3.2 Sobolev Space Estimates
5.4 Pointwise Theory
5.5 Smoothness and Approximation Order
5.6 Stability
Part II - Multiwavelets
6 Basic Theory
6.1 Refinable Function Vectors
6.2 MRAs and Multiwavelets
6.2.1 Orthogonal MRAs and Multiwavelets
6.2.2 Biorthogonal MRAs and Multiwavelets
6.3 Moments
6.4 Approximation Order
6.5 Point Values and Normalization
7 Practical Computation
7.1 Discrete Multiwavelet Transform
7.2 Pre- and Postprocessing
7.2.1 Interpolating Prefilters
7.2.3 Hardin--Roach Prefilters
7.2.4 Other Prefilters
7.3 Balanced Multiwavelets
7.4 Handling Boundaries
7.4.1 Data Extension Approach
7.4.2 Matrix Completion Approach
7.4.3 Boundary Function Approach
7.5 Putting It All Together
7.6 Modulation Formulation
7.7 Polyphase Formulation
7.8 Calculating Integrals
7.8.1 Integrals with Other Refinable Functions
7.8.2 Integrals with Polynomials
7.8.3 Integrals with General Functions
7.9 Applications
7.9.1 Signal Processing
7.9.2 Numerical Analysis
8 Two-Scale Similarity Transforms
8.1 Regular TSTs
8.2 Singular TSTs
8.3 Multiwavelet TSTs
8.4 TSTs and Approximation Order
8.5 Symmetry
9 Factorizations of Polyphase Matrices
9.1 Projection Factors
9.1.1 Orthogonal Case
9.1.2 Biorthogonal Case
9.2 Lifting Steps
9.3 Raising Approximation Order by Lifting
10 Creating Multiwavelets
10.1 Orthogonal Completion
10.1.1 Using Projection Factors
10.1.2 Householder-Type Approach
10.2 Biorthogonal Completion
10.3 Other Approaches
10.4 Techniques for Modifying Multiwavelets
10.5 Techniques for Building Multiwavelets
11 Existence and Regularity
11.1 Distribution Theory
11.2 L^1-Theory
11.3 L^2-Theory
11.3.1 Transition Operator
11.3.2 Sobolev Space Estimates
11.4 Pointwise Theory
11.5 Smoothness and Approximation Order
11.6 Stability
Appendix A: Standard Wavelets
A.1 Scalar Orthogonal Wavelets
A.2 Scalar Biorthogonal Wavelets
A.3 Orthogonal Multiwavelets
A.4 Biorthogonal Multiwavelets
Appendix B: Mathematical Background
B.1 Notational Conventions
B.2 Derivatives
B.3 Functions and Sequences
B.4 Fourier Transform
B.5 Laurent Polynomials
B.6 Trigonometric Polynomials
B.7 Linear Algebra
Appendix C: Computer Resources
C.1 Wavelet Internet Resources
C.2 Wavelet Software
C.3 Multiwavelet Software
References
Index
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Breaking
# Crime Analysis
10 months ago
## Milepost GIS Location Estimation in R
I was recently forced at gunpoint to estimate the positions of mileposts along the highways in Maricopa County. This was to estimate stops and arrest positions for law enforcement. Like …
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## Comparison of R and Python Parallel Computing in Address Geocoding
I recently needed to process a bunch of addresses that our proprietary geocoding software was having a problem with. I won’t say the name of the geocoder, but it rhymes …
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## R Function to Calculate Bearing Between Two Positions (lat-lon)
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## Build your own discrete event simulation framework in Python Part III
In Part II, we added a queue and a PatrolCar class to our Python simulation to answer call events in our police patrol simulation. In this installment, we’ll add calls to …
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## Build your own discrete event simulation framework in Python Part II
In Part I, we learned about creating a simple discrete event simulation controller in Python, using a clock that iterated over a minute counter and translated this minute counter to …
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## Build your own discrete event simulation framework in Python Part I
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I wouldn’t say a million. …
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## Excel to R Transhipment Problem Optimization Tutorial Part II
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In this tutorial I’m going to go through creating an optimization model in Excel. The next part of the tutorial will be scaling a small optimization program into a larger …
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# 10-10-02
General Physics
9. Assignment (Subjective Problems)
LEVEL-I
1. 10 rotations of the cap of a screw gauge is equivalent to 5 mm. The cap has 100
dimensions. Find the least count. A reading taken for the diameter of wire with
the screw gauge shows 4 complete rotations and 35 on the circular scale. Find
the diameter of the wire.
2. A certain pendulum clock with a 12hr dial happens to gain 1 min/day. After setting
the clock to the correct time how long it will take to indicate correct time again?
3. The mass of a block is 87.2g and its volume is 25cm3. What are its density upto
correct significant figures?
4. The radius of a sphere is (5.3 0.1) cm. Find the percentage error in its volume.
## 5. The Van-der-Waals interaction between two molecules separated by a distance r
is given by the energy
A B
E = 6 + 12 . Find the dimensions of A and B.
r r
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Part-I-PH-GP-2
LEVEL-II
## 1. In an experiment for finding the specific heat of alcohol, as copper calorimeter of
mass 190 gm is filled with alcohol, and the total mass is found to be 390 gm.
When it is heated by using a 50 W heater for 9 minute the following reading were
recorded with a thermometer and stop watch.
Time (minute) 0 1.5 3 4.5 6 7.5 9
Temperature ( C) 25 38 49 58 65 70 75
Find the specific heat capacity of alcohol. It is not required to take the radiation
correction into account.
MI
2. Suppose, the torque acting on a body, is given by = KL +
Where L = angular momentum, I = moment of inertia & = angular speed
What is the dimensional formula for KM?
3. When a current of (2.5 0.1)A flows through a wire it develops a potential
difference of (20 1)V. What is the resistance of wire?
## 4. What is the fractional error in g calculated from T = 2 . Given fractional errors
g
in T and are x and y respectively?
## 5. A planet of mass m rotates around a star of mass M. The time period of
revolution is T, while the average distance of the planet from the star is a. It is
known that there exists a relationship between them: find it. Assume the
dimensional expression for G.
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Part-I-PH-GP-3
## 10. Assignment (Objective Problems)
LEVEL- I
1. ML 1T 2
is the dimensional formula of
(A) force (B) coefficient of friction
(C) modulus of elasticity (D) energy
2. The dimensional formula of coefficient of viscosity is
(A) MLT 1 (B) M 1L2 T 2
(C) ML 1T 1
(D) none of these
3. On the basis of dimensional equation, the maximum number of unknown that can
be found, is
(A) one (B) two
(C) three (D) four
4. If v stands for velocity of sound, E is elasticity and d the density, then find x in the
x
d
equation v .
E
(A) 1 (B)
(C) 2 (D) 1/2
5. The multiplication of 10.610 with 0.210 upto correct number of significant figure is
(A) 2.2281 (B) 2.228
(C) 2.22 (D) 2.2
## 6. The measurement of radius of a circle has error of 1%. The error in
measurement of its area is
(A) 1% (B) 2%
(C) 3% (D) none of these
7. Dimensional formula of latent heat
(A) M0L2T-2 (B) MLT-2
(C) ML 2T-2 (D) ML2T-2
8. In case of measurement of g, if error in measurement of length of pendulum is
2%, the percentage error in time period is1 %. The maximum error in
measurement of g is
(A) 1 % (B) 2 %
(C) 4 % (D) no error.
9. If length of pendulum is increased by 2%. The time period will
(A) increases by 1% (B) decreases by 1%
(C) increases by 2% (D) decreases by 2%
10. If radian correction is not considered in specific heat measurement. The
measured value of specific heat will be
(A) more than its actual value. (B) less than its actual value.
(C) remains same as actual value. (D) none of these.
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Part-I-PH-GP-4
## 11. The S.I. unit of universal gas constant is
(A) Watt K-1mol-1 (B) N K-1mol-1
(C) JK-1mol-1 (D) erg K-1mol-1
## 12. The dimensional formula of couple
(A) ML2T-2 (B)MLT-1
(C) ML -1T-1 (D) M1L1T-2
## 13. An experiment measures quantities a, b, c and x is calculated from x = ab2/c3. If
the maximum percentage error in a, b and c are 1%, 3% and 2% respectively,
the maximum percentage error in x will be
(A) 13% (B) 17%
(C) 14% (D) 11%
## 14. Dimensional formula of thermal conductivity is
(A) ML2T-3 -1 (B) ML2T-2 -4
(C) ML 2T-2 -1 (D) MLT-3 -1
15. Three measurements 7.1J, 7.2J and 6.7J are made as experiment the result with
correct number of significant figures is
(A) 7.1 J (B) 7.06 J
(C) 7.0 J (D) 7J
## 16. If P represents radiation pressure, c represents speed of light and Q represents
radiation energy striking a unit area per second, then non-zero integers x, y and
z, such that PxQycz is dimensionless, may be
(A) x = 1, y =1, z = 1. (B) x = 1, y = 1, z = 1.
(C) x = 1, y =1, z = 1. (D) x = 1, y = 1, z = 1
17. A spherical ball of mass m and radius r is allowed to fall in a medium of viscosity
. The time in which the velocity of the body increases from zero to 0.63 times
the terminal velocity is called time constant ( ). Dimensionally can be
represented by
mr 2 6 mr
(A) (B)
6 g2
m
(C) (D) none of these.
6 r
## 18. Which of the following is a possible dimensionless quantity?
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Part-I-PH-GP-5
19. In specific resistance measurement of a wire using a meter bridge, the key k in
the main circuit is kept open when we are not taking readings. The reason is
(A) the emf of cell will decrease.
(B) the value of resistance will change due to joule heating effect.
(C) the galvanometer will stop working.
(D) none of these.
20. In the experiment of verification of Ohms law the error in the current
measurement is 1%, while that in the voltage measurement is 2%. The error in
the resistance has a maximum value of
(A)1% (B) 2%
(C) 3% (D) none of these.
LEVEL - II
## 1. Which of the following are not the dimensions of calorie?
(A) [ML2T-2] (B) [MLT-2] (C) [ML-2T-1] (D) [ML2T-1]
2. Which of the following are not the dimensional formula for kinetic energy?
(A) [M2L2T] (B) [ML2T-2] (C) [M0L-1] (D) [ML2T]
## 3. Which of the following are dimensionally wrong?
(A) Pressure = Energy per unit area
(B) Pressure = Energy per unit volume
(C) Pressure = Force per unit volume
(D) Pressure = Momentum per unit volume per unit time
dv
4. Given that: F A ,
dx
dv
where F is force, A is area and velocity gradient, then which of the following are not
dx
the dimensional formula of ?
(A) [ML-1T-1] (B) [ML-1T] (C) [ML-2T-2] (D) [ML2T-1]
5. A particle moving along a straight line with uniform acceleration has velocities 7 m/s at
P and 17 m/s at Q,R is the mid point of PQ. Then:
(A) the average velocity between R and Q is 15 m/s
(B) the ratio of time to go from P to R and that from R to Q is 3:2
(C) the velocity at R is 10 m/s
(D) the average velocity between P and R is 10 m/s
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LEVEL - III
## MATCH THE FOLLOWING
1. Column I Column II
(A) Amount of substance (P) Second
(B) Time (Q) Kelvin
(C) Temperature (R) Mole
(D) Electric current (S) Ampere
(T) Kilogram
2. Column I Column II
(A) Speed (P) MLT1 1 2
## (D) Work (S) M2L2 T 2
(T) M0L1T 1
3. The unit of force and length are doubled, the unit of energy will be ____ times.
## 4. The heat dissipated in a resistance can be obtained by the measurement of resistance,
the current and time. If the maximum error in the measurement of these quantities is 1%,
2% and 1% respectively, the maximum error in the determination of the dissipated heat
is ___%.
5. The relative density of a material is found by weighing the body first in air and then in
water. If the weight in air is (10.0 0.1) gm and weight in water is (5.0 0.1) gm then the
maximum permissible percentage error in relative density is ____%
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LEVEL - I
## 1. 2.175 mm 2. 720 days
3. 3.5g/cc 4. 5.7%
## 5. [A] = ML8T 2, [B] = ML4T 2
LEVEL-II
1. 2300 J/kg-k 2. T4
3. (8 0.7) 4. (y+2x)
1/2
5. T= (GM) a 3/2
## 12. Answers to the Objective Assignment
LEVEL -I
1. C 2. C
3. C 4. D
5 B 6. B
7. A 8 C
9. A 10. A
11. C 12. A
13. A 14. D
15. C 16 B
17. C 18. C
19. B 20. C
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LEVEL-II
## 5. (A), (B), (D)
LEVEL-III
1. A-R, B-P, C-Q, D-S
2. A-T, B-P, C-Q, D-R
3. 4
4. 6
5. 5
***
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KINEMATICS
10. Assignment (Subjective Problems)
LEVEL I
1. A cricketer hits a ball from the ground level with a velocity v 0 = (20 i + 10 j )
m/sec. Find the velocity of the ball at t = 1 sec, from the instant of projection
(g = 10 m/sec2).
2. A body is projected vertically up with a speed V0. Find the magnitude of time
average velocity of the body during its ascent.
## 3. A bomb is released from an aeroplane flying with a horizontal velocity of
magnitude 100 m/sec at an altitude of 1 km. What is the displacement during the
time of its flight?
4. A football player kicks the football so that it will have a hang time (time of flight)
of 5s and lands 50 m away. If the ball leaves the players foot 1.5m above the
ground, what is its initial velocity (magnitude and direction)? (g = 10 m/sec2)
5. A rocket is fired vertically up from the ground with a resultant vertical acceleration
of 10 m/s2. The fuel is finished in 1 minute and it continues to move up
(a) what is the maximum height reached ?
(b) After how much time from then will the maximum height be reached?
(Take g = 10 m/s2 )
6. A ball is falling from the top of a cliff of height h with an initial speed V. Another
ball is simultaneously projected vertically up with the same speed. When do they
meet ?
7. If an object travels one-half its total path in the last second of its fall from rest,
find (a) the time and (b) the height of its fall. Explain the physically unacceptable
solution of the quadratic time equation.
8. A particle starts moving due east with a velocity v1 = 5 m/sec. for
10 sec. and turns to north with a velocity v2 =10 m/sec. for 5 sec. Find the
average velocity of the particle during 15 sec. from starting.
9. A particle is moving with a speed v0 in a circular path of radius R. Find the ratio of
average velocity to its instantaneous velocity when the particle describes an
angle (< /2).
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10. To a man moving due east with a speed v in a rain, the rain appears to fall
vertically. If he changes his speed by a factor n, the rain appears to fall at an
angle to vertical. Find the speed of the rain.
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LEVEL II
1. Two particles are projected horizontally in opposite directions with v1 & v2 from
the top of a pole. If the particles move perpendicular to each other just before
striking the ground, find the height of the pole.
2. A cannon fires successively two shells with velocity vo = 250 m/s; the first at the
angle 1 = 60o and the second at the angle 2 = 45o to the horizontal, the
azimuth being the same. Neglecting the air drag, find the time interval between
firings leading to the collision of the shells.
3. An aeroplane flies in still air at a speed of 400 km/hr. Air is blowing from the
south at a speed of 50 km/hr. The pilot wants to travel from point A to point B
north-east of A and then to return. Calculate the direction he must steer (a) on
his onward journey (b) on his return journey. If the distance AB is 1000 km then
calculate the time taken in two journeys.
## 4. Two particles move in a uniform gravitational field with an acceleration g. At the
initial moment the particles were located at one point in space and moved with
velocities v1 = 3.0 m/s and v2 = 4.0 m/s horizontally in opposite directions. Find
the distance between the particles at the moment when their velocity vectors
become mutually perpendicular.
## 5. A point moves rectilinearly with deceleration whose modulus depends on the
velocity v of the particle as w = a v, where a is a positive constant. At the initial
moment the velocity of the point is equal to vo. What distance will it traverse
before it stops? What time will it take to cover that distance?
6. Find the ratio between the normal and tangential acceleration of a point on the
rim of a rotating wheel when at the moment when the vector of the total
acceleration of this point forms an angle of 30 with the vector of the linear
velocity.
## 7. A fan rotates with a velocity corresponding to a frequency of 900 rev/min. When
its motor is switched off, the fan uniformly slows down and performs 75
revolutions before it comes to a stop. How much time elapsed from the moment
the fan was switched off to the moment it stopped?
8. A motor cyclist, going due east with a velocity of 10 m/s, finds that the wind is
blowing directly from the north. When he doubles his speed, he finds that the
wind is blowing from north east. In what direction and with what velocity is the
wind blowin?.
## 9. The acceleration vector of a particle having initial speed V0 changes with
distance as a x . Find the distance covered by the particle when its speed
becomes twice the initial speed.
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10. An observer in a train moving with a uniform velocity finds that a car moving
parallel to the train has a speed of 10 km/h in the direction of motion of the train.
An object falls from the car and the observer in the train notices that the car has
moved on for one minute, turned back, and moved with a speed of 10 km/h and
picked up the object two minutes after turning. Find
(a) the velocity of the train relative to the ground and
(b) the velocity of the car during its forward and reverse journeys.
Assume that the object comes to rest immediately on fall from the point of
view of the observer on the ground.
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## 11. Assignment (Objective Problems)
LEVEL- I
1. A stone is released from an elevator going up with acceleration 5 m/s2. The
acceleration of the stone after the release is:
(A) 5 ms-2 (B) 4.8 ms-2 upward
(C) 4.8 down ward (D) 9.8 ms-2 down ward
## 2. The locus of a projectile relative to another projectile is a
(A) straight line (B) circle
(C ) ellipse (D) parabola
3. A car accelerates from rest at constant rate of 2 m/s2 for some time. Then its
retards at a constant rate of 4 m/s2 and comes to rest. What is the maximum
speed attained by the car if it remains in motion for 3 seconds
(A) 2 m/s (B) 3 m/s
(C) 4 m/s (D) 6 m/s
4. The co-ordinates of a moving particle at any time t are given by x = ct2 and
y = bt2. The speed of the particle is given by:
(A) 2t (c + b) (B) 2t (c 2 b2 )
(C) t (c 2 b2 ) (D) 2t (c 2 b2 )
## 5. A particle parallel to x-axis as shown in the figure y
such that at all instant the y axis component of its
v
position vector is constant and is equal to b. The
angular velocity of the particle about the origin is
v v
(A) (B) sin
b b O x
v
(C) sin2 (D) vb
b
## 6. A particle is projected vertically upwards and it attains maximum height H. If the
ratio of times to attain height h(h < H) is 1/3, then h equals
(A) 2/3. H (B) 3/ 4. H
(C) 4/3 . H (D) 3/2. H
## 7. A boy B drags a wedge A by an in
extensible string passing over the 1
pulleys 1,2,3 & 4 as shown in the 2
figure If all the pulleys are smooth and A
3 4
the boy walks with constant velocity of
magnitude v, the magnitude of relative V B
velocity between the boy and the
wedge is equal to
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(A) v (B) 2v
(C) 1.5v (D) 1.25v
8. A swimmer wishes to reach directly opposite bank of a river, flowing with velocity
8 m/s. The swimmer can swim 10 m/s in still water. The width of the river is 480
m. Time taken by him to do so:
(A) 60 sec (B) 48 sec
(C) 80 sec (D) None of these
9. A man can swim at a speed of 5 km/h w.r.t. water. He wants to cross a 1.5 km
wide river flowing at 3 km/h. He keeps himself always at an angle of 60 o with the
flow direction while swimming. The time taken by him to cross the river will be
(A) 0.25 hr. (B) 0.35 hr.
(C) 0.45 hr. (D) 0.55 hr.
10. A disc of radius R is rotating inside a room A boy standing near the rim of the
disc, finds the water droplets falling from the ceiling is always hitting on his head.
As one drop hits his head the next one starts from the ceiling. If height of the
roof above his head is H. Angular velocity of disc is:
2gR 2gH
(A) 2
(B)
H R2
2g 2g
(C) (D) 2
H H
## 11. The acceleration-time graph of a particle moving a (m/s2)
along a straight line is as shown in figure. At what 10
time the particle acquires its initial velocity?
(A) 12 sec.
(B) 5 sec.
(C) 8 sec.
(D) 16 sec. 4 t(sec.)
12. What are the speeds of two objects if they move uniformly towards each other,
they get 4m closer in each second and if they move uniformly in the same
direction with the original speeds they get 4m closer in each 10 sec ?
(A) 2.8 m/s and 1.2 m/s (B) 5.2 m/s and 4.6 m/s
(C) 3.2 m/s and 2.1 m/s (D) 2.2 m/s and 1.8 m/s
13. From the top of a tower, a stone is thrown up. It reaches the ground in 5 sec. A
second stone is thrown down with the same speed and reaches the ground in
1sec. A third stone is released from rest and reaches the ground in
(A) 3 sec. (B) 2 sec.
(C) 5 sec. (D) 2.5 sec.
## 14. A particle has an initial velocity of (3 i 4j )m/s and a constant acceleration of
(4 i 3j ) m/s2. Its speed after one second will be equal to
(A) 0 (B) 10 m/sec
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## (C) 5 2 m/sec (D) 25 m/sec
15. A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8
seconds, a stone is released from the balloon. After releasing, the stone will
(A) cover a distance of 40 m till it strikes the ground.
(B) have a displacement of 50 m till it reaches the ground
(C) reach the ground in 4 seconds.
(D) begin to move down instantaneously
16. Two balls are dropped from the top of a high tower with a time interval of t0
second, where t0 is smaller than the time taken by the first ball to reach the floor
which is perfectly inelastic. The distance S between the two balls, plotted against
the time lapse t from the instant of dropping the second ball is best represented
by
(A) (B)
S S
t t
O O
(C) (D)
S S
t t
O O
17. The K.E. (K) of a particle moving along a circle of radius R depends on the
distance covered s as K as 2 . The force acting on particle is
2as2 2as
(A) (B) 1/2
R s2
1
R
1/2
s2
(C) 2as 1 (D) none of these.
R2
18. Two particles P and Q start from rest and move for equal time on a straight line.
Particle P has an acceleration of X m/s2 for the first half of the total time and 2x
m/s2 for the second half. Particle Q has an acceleration of 2X m/s2 for the first
half of the total time and X m/s2 for the second half. Which particle has covered
larger distance?
(A) Both have covered the same distance
(B) P has covered the larger distance
(C) Q has covered the larger distance
(D) data insufficient
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19. A particle is moving in a circle of radius R in such a way that at any instant the
normal and tangential components of its acceleration are equal. If its speed at
t = 0 is v0 the time taken to complete the first revolution is
(A) R/v0 (B) v0 /R
(C) R/v0 (1 e-2 ) (D) R/v0 (e-2 )
20. A motor boat of mass m moves along a lake with velocity v0. At t = 0, the engine
of the boat is shut down. Resistance offered to the boat is equal to v. What is
the total distance covered till it stops completely?
mv 0 3mv 0
(A) (B)
2
mv 0 2mv 0
(C) (D)
2
LEVEL- II
x t3
1. A particle moves along positive branch of the curve y where x , x and y
2 3
are measured in metres and t in seconds, then
1
(A) the velocity of particle at t = 1 s is i j
2
1
(B) the velocity of particle at t = 1 s is i j
2
(C) the acceleration of particle at t = 2 s is 2i j
(D) the acceleration of particle at t = 2 s is i 2j
## 2. Figure shows two blocks, each of mass m. The system
is released from rest at the position, shown in fig. If v2,a2
initial acceleration of blocks A and B are a1 and a2 B
respectively and during the motion velocities of A and
B are v1 and v 2 respectively, then :
A v1
(A) a1 a2 cos (B) a2 a1 cos 1
## 3. The string shown in the figure is passing over
small smooth pulley rigidly attached to trolley
A. If speed of trolley is constant and equal to
VA. Speed and magnitude of acceleration of A
## block B at the instant shown in figure is
(A) vB = vA, aB = 0 (B) aB = 0 x=3cm
3 16v A B
(C) vB vA (D) aB
5 125
## 4. A particle is projected from the ground in earths gravitational field at an angle
with the horizontal then
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(A) center of curvature of projectiles trajectory at the highest point is below the
ground level if tan 1 2
(B) centre of curvature of projectiles trajectory at the highest point is above the
ground level if tan 1 2
(C) center of curvature of projectiles trajectory at the highest point is above
ground level if tan 1 2
(D) center of curvature of projectiles trajectory at the highest point is below the
ground level is tan 1 2
## 5. Two different balls of masses m1 and m2 are m1 m2
allowed to slide down from rest and from
same height h along two smooth inclined h
planes having different angles of inclination
and . Then
(A) the final speed velocity acquired by them
will be the same
(B) the final speed acquired by them will be
different
(C) the times taken by them to reach the bottom will be the same
(D) the times taken by them to reach the bottom will be in the ratio (sin /sin )
## 6. A particle is moving along a curve. Then
(A) if its speed is constant it has no acceleration
(B) if its speed is increasing the acceleration of the particle is along its direction
of motion
(C) if its speed is constant the magnitude of its acceleration is proportional to its
curvature.
(D) the direction of its acceleration cannot be along the tangent.
## 7. A particle of mass m moves along x-axis as follows : it starts from rest at t = 0
from the point x = 0, and comes to rest at t = 1 sec at the point x = 1. No other
information is available about its motion at intermediate times (0 < t < 1). If
denotes the acceleration of the particle then,
(A) cannot remain positive for all t in the interval 0 t 1.
(B) | | can not exceed 2at any point in its path.
(C) | | must be 4 at some point or points in its path.
(D) must change sign during the motion, but no other assertion can be made
with the information given :
## 8. Mark the correct statements for a particle going on a straight line
(A) if the velocity is zero at any instant, the acceleration should also be zero at
that instant
(B) if the velocity is zero for a time interval, the acceleration is zero at any instant
within the time interval
(C) if the velocity and acceleration have opposite sign, the object is slowing down
(D) if the position and velocity have opposite sign, the particle is moving towards
the origin
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9. A particle initially at rest is subjected to two forces. One is constant, the other
is a retarding force proportional to the particle velocity. In the subsequent motion
of the particle :
(A) the acceleration will increase from zero to a constant value
(B) the acceleration will decrease from its initial value to zero
(C) the velocity will increase from zero to maximum & then decrease
(D) the velocity will increase from zero to a constant value.
10. Which of the following statements are true for a moving body?
(A) If its speed changes, its velocity must change and it must have some
acceleration
(B) If its velocity changes, its speed must change and it must have some
acceleration
(C) If its velocity changes, its speed may or may not change, and it must have
some acceleration
(D) If its speed changes but direction of motion does not change, its velocity may
remain constant
COMPREHENSIONS
Comprehension I:
Using the concept of relative motion, answer the following Y
question. Velocity of the river with respect to ground is given by
V0. Width of the river is d. A swimmer swims (with respect to V0
d
water) perpendicular to the current with acceleration a = 2t
O X
(where t is time) starting from rest form the origin O at t = 0.
## 1. The time of crossing the river is
(A) (D)1/3 (B) (2d)1/3
(C) (3d)1/3 (D) Information is insufficient
## 2. The drift of the swimmer is
(A) V0(D)1/3 (B) V0(2d)1/3
(C) V0(3d)1/3 (D) None of these
## 3. The equation of trajectory of the path followed by the swimmer
x3 x2
(A) y (B) y
3V03 2V02
x x
(C) y (D) y
V0 V0
Comprehension II :
When a particle is projected at some angle with the horizontal, the path of the
particle is parabolic in nature. In the process the horizontal velocity remains
constant but the magnitude of vertical velocity changes. At any instant during
flight the acceleration of particle remains g in vertically downward direction.
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During flight at any point the path of particle can be considered as a part of circle
and radius of that circle is called the radius of curvature of the path of particle.
Consider that a particle is projected with velocity u = 10 m/s at an angle = 60
with the horizontal then :
4. The radius of curvature of path of particle at the instant when the velocity vector of the
particle becomes perpendicular to initial velocity vector is
20 10
(A) m (B) m
3 3 3 3
40 80
(C) m (D) m
3 3 3 3
## 5. The magnitude of acceleration of particle at that instant is
(A) 10 m/s 2 (B) 5 3 m / s2
(C) 5 m/s2 (D) 10 3 m / s2
## 6. Tangential acceleration of particle at that instant will be
(A) 10 m/s 2 (B) 20 m/s2
(C) 5 m/s2 (D) 5 3 m / s2
## 1. A body is projected with a velocity of 60 m/s at 30 to horizontal (g = 10 m/s2) :
Column A Column B
(A) Initial velocity vector (p) 60 3 i 40 j
(B) Velocity just after 3 sec (q) 30 3 i 10 j
(C) Displacement just after 2 sec (r) 30 3 i 30 j
(D) Velocity just after 2 sec (s) 30 3 i
## 2. Match the following :
Column II
Column I (velocitytime graph)
(accelerationtime graph)
(A) +a (p)
V
t
a
t
(B) (q)
a V
t t
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(C) (r)
a
V
t
t
(D) (s)
a
V
t t
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LEVEL - I
## 1. Apply vy = vo sin gt; vx = vocos
2. Find the maximum height reached.
3. Vertical displacement of bomb = 1km
Horizontal displacement of bomb = velocity of plane time of flight.
4. Vertical displacement of the projectile is 1.5 meter and horizontal displacement
is 50 meter.
5. Find velocity of the rocket after 1 minute.
6. They meet when their separation is zero.
7. Use kinematical equations.
8. Use vector method to find out displacement vector.
9. First find out time to move through an angle . Then find the displacement during
this time and average velocity.
10. Draw the vector diagram clearly.
LEVEL- II
## 1. Use the concept of dot product.
2. Write equations for their positions at a time t. Solve these equations.
3. Draw the vector diagram clearly in two cases. Note that velocity of plane in still
air + velocity of air = velocity of plane in blowing air.
## 4. Use the concept of dot product.
dv
5. Deceleration equals to v = a v.
dx
6. Normal acceleration is (v2/R) while tangential acceleration is R .
7. Use simple kinematical equations.
8. Draw the vector diagram clearly.
dv
9. Acceleration; a = v = x.
dx
10. Velocity of car as observed from train
= velocity of car relative to ground velocity of train relative to ground.
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## 13. Answers to Subjective Assignments
LEVEL - I
v0
1. 20 i 2.
2
3. 1732 m, tan-1(0.707) 4. 26.64 m/s, 67.96
h
5. (a) 36 km (b) 1 minute 6.
2v
5 2
7. (a) 3.4 sec (b) 57m 8. due north of east
3
2 sin 2
9. 10. v 1 n2 cot 2
LEVEL II
v1v 2
1.
2g
2v 0 sin 1 2
2. t , 10.7 sec.
g cos 1 cos 2
3. (a) 390 56 north of east, 2.3 hour (b) 390 56 west of south, 2.75 hour
4. 2.5 m
2 2 vo
5. v 3/2
0 ,
3a a
ar 1
6.
at 3
7. 10 sec
8. 10 2 m/s, from north west
4
3v 0 3
9.
2
## 10. (a) 3.33 km/hr (b) 13 km/hr, 6.67 km/hr
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LEVEL - I
1. (D) 2. (A)
3. (C) 4. (D)
5. (C) 6. (C)
7. (D) 8. (C)
9. (B) 10. (C)
11. (C) 12. (D)
13. (C) 14. (C)
15. (C) 16. (D)
17. (C) 18. (C)
19. (C) 20. (A)
LEVEL II
## 9. (B), (D) 10. (A), (C)
COMPREHENSIONS
1. (C) 2. (C)
3. (A) 4. (A)
5. (A) 6. (C)
MATCH THE FOLLOWING
## 1. (A) (r); (B) (s); (C) (p); (D) (a)
2. (A) (q); (B) (r); (C) (s) ; (D) (p)
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LAWS OF MOTION
9. Assignments (Subjective Problems)
LEVEL I
A
1. Two blocks A and B of masses M1 and M2 B
respectively kept in contact with each other on a F
M1
M2
smooth horizontal surface. A constant horizontal
force (F) is applied on A as shown in figure. Find
the acceleration of each block and the contact
force between the blocks
## 2. A bob of mass m = 50 gm is suspended from the
ceiling of a trolley by a light inextensible string. If the a
trolley accelerates horizontally, the string makes an
m
angle = 30 o with the vertical. Find the acceleration of
the trolley.
## 3. Two small bodies connected by a light
inextensible string passing over a smooth m1 m2
pulley are in equilibrium on a fixed smooth
wedge as shown in the figure. Find the ratio of
the masses. Given that = 600 and = 300.
k
4. Both the springs shown in Figure are k
## unstretched. If the block is displaced by a m
distance x and released, what will be the
initial acceleration?
F
5. A block of mass m = 1 kg is at rest on a rough horizontal m
surface having coefficient of static friction 0.2 and kinetic
force 0.15. Find the frictional forces if a horizontal force (a)
F = 1 N, (b) F = 1.96 N and (c) F = 2.5 N are applied on a
block which is at rest on the surface.
## 6. Two masses m1 = 5 kg, m2 = 2 kg placed on a smooth horizontal surface are
connected by a light inextensible string. A horizontal force F = 1 N is applied on
m1. Find the acceleration of either block. Describe the motion of m1 and m2 if the
string breaks but F continues to act.
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## 7. The coefficient of static friction between a block of mass m and an incline is s =
0.3. (a) What can be the maximum angle of the incline with the horizontal so
that the block does not slip on the plane? (b) If the incline makes an angle /2
with the horizontal, find the frictional force on the block.
8. A 20 kg box is dragged across a rough level floor having a coefficient of kinetic
friction of 0.3 by a rope which is pulled upward at angle of 30 to the horizontal
with a force of magnitude 80 N.
(a) What is the normal force?
(b) What is the frictional force?
(c) What is the acceleration of the box?
(d) If the force is reduced until the acceleration becomes zero, what is the tension
in the rope?
## top of a wedge (fig.) whose base is equal to
= 2.10 m. The coefficient of friction between
the body and the wedge surface is k = 0.140.
For what value of the anlge will the time of
sliding be the least? What will it be equal to?
10. A chain of length is placed on a smooth spherical surface of radius R with one
of its ends fixed at the top of the sphere. What will be the acceleration a of each
element of the chain when its upper end is released? It is assumed that the
length of the chain < ( R/2).
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LEVEL II
## 1. A smooth wedge with elevation is fixed in an elevator moving up with uniform
acceleration a0 = g/2. The base of the wedge has a length L. Find the time
taken by a particle sliding down the incline to reach the base.
2. A body of mass 2 kg is lying on a rough inclined plane of inclination 30. Find the
magnitude of the force parallel to the incline needed to make the block move (a)
up the incline (b) down the incline. Coefficient of static friction = 0.2.
3. A spring has its end fixed to the ceiling of the elevator rigidly. It has spring
constant = 2000 N/m. A man of mass 50 kg climbs along the other end of the
spring vertically up with an acceleration of 2 m/s2 relative to the elevtater. The
elevater is going up with retardation 3 m/s2. Find extension in the spring.
4. A bar of mass m resting on a smooth horizontal plane starts moving due to the
force F = mg/3 of constant magnitude. In the process of its rectilinear motion the
angle between the direction of this force and the horizontal varies as = as,
where a is a constant, and s is the distance traversed by the bar from its initial
position. Find the velocity of the bar as a function of the angle .
## 5. Two blocks in contact of masses 2 kg and 4 kg in succession from down to up
are sliding down an inclined surface of inclination 30. The friction coefficient
between the block of mass 2.0 kg and the inclines is 1, and that between the
block of mass 4.0 kg and the incline is 2. Calculate the acceleration of the 2.0 kg
block if (a) 1 = 0.20 and 2 = 0.30, (b) 1 = 0.30 and 2 = 0.20. Take g = 10 m/s2.
## 6. A balloon is descending with a constant acceleration a, less than the acceleration
due to gravity g. The weight of the balloon, with its basket and contents, is w.
What weight, w, should be released so that the balloon will begin to accelerate
upward with constant acceleration a? Neglect air resistance.
## 7. In the figure shown co-efficient of friction
between the block B and C is 0.4. There is no B
## on which it is placed. The block A is released
from rest, find the distance moved by the block A
C when block A descends through a height 2m. Given masses of the blocks are
mA = 3 kg, mB = 5 kg and mC = 10 kg.
F
8. Two masses m1 and m2 are connected by
means of a light string, that passes over a light
pulley as shown in the figure. If m1 = 2kg and
m2 = 5 kg and a vertical force F is applied on
the pulley then find the acceleration of the m2
m1
masses and that of the pulley when
(a) F = 35 N (b) F = 70 N (c) F = 140 N
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## 9. In the given figure the co-efficient of friction between the m
M F
walls of block of mass m and the plank of mass M is . The
same co-efficient of friction is there between the plank and
the horizontal floor. The force F is of 100 N and the masses
m and M are of 1 kg and 3 kg respectively. Find the value
of , if the block does not slip along the wall of the plank.
## 10. In figure, a bar of mass m is placed on the smooth
surface of a wedge of mass M. The bar is connected to
an inextensible string passing over a light smooth m
pulley fitted with the wedge. The string is connected to M
the vertical wall. The angle of inclination of the slant
surface of the wedge is . If all contacting surfaces are
smooth, find the acceleration of the wedge.
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LEVEL I
## 1. When a body is stationary:
(A) There is no force acting on it
(B) The forces acting on it are not in a contact with it
(C) The combination of forces acting on it balance each other
(D) The body is in vacuum
## 2. A toy train consists of three identical compartments X, Y and Z. It is pulled by a
constant horizontal force F applied on Z horizontally. Assuming there is negligible
friction, the ratio of tension in string connecting XY and YZ is:
(A) 2:1 (B) 3:2
(C) 1:2 (D) 2:3
## 3. Two blocks of masses 2 kg and 1 kg are in contact with each other on a
frictionless table, when a horizontal force of 3.0 N is applied to the block of mass
2 kg the value of the force of contact between the two blocks is:
(A) 4 N (B) 3 N
(C) 2 N (D) 1 N
## 4. A block of metal weighing 2 Kg is resting on a frictionless plane. It is struck by a
jet releasing water at a rate of 1 Kg/sec and at a speed of 5m/sec. The initial
acceleration of the block will be:
(A) 2.5 m/sec2 (B) 5.0 m/sec2
(C) 10 m/sec2 (D) none of above
## 5. When a force of constant magnitude always act perpendicular to the motion of a
particle then:
(A) Velocity is constant (B) Acceleration is constant
(C) KE is constant (D) None of these
6. Two masses M1 and M2 are attached to the ends of string which passes over the
pulley attached to the top of a double inclined plane. The angles of inclination of
the inclined planes are and . Take g = 10 ms-2. If M1 = M2 and = , what is
the acceleration of the system?
(A) zero (B) 2.5 ms-2
(C) 5 ms-2 (D) 10 ms-2
7. Starting from rest, a body slides down a 45 inclined plane in twice the time it
takes to slide down the same distance in the absence of friction. The coefficient
of friction between the body and the inclined plane is:
(A) 0.33 (B) 0.25
(C) 0.75 (D) 0.80
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## 8. A trolley car slides down a smooth inclined plane of angle of inclination . If a
body is suspended from the roof of the trolley car by an inextensible string of
length l, the corresponding tension in the string will be
(A) mg (B) mg cos
(C) mg sin (D) None of these
## 9. A block of mass m is held at the top of an inclined rough plane of angle of
inclination . The coefficients of static and kinetic friction are 1 and 2
respectively. If the block is pushed down at the verge of slipping, assuming <
tan-1 1, Its acceleration down the plane is :
(A) g[Sin - 1 Cos ] (B) g[Sin - 2 Cos ]
(C) g( Sin - 1 Cos ] (D) g
## 10. A satellite in force-free space sweeps stationary interplanetary dust at a rate
(dM/dt) = v. The acceleration of satellite is:
(A) -2v2/M (B) - v2/M
(C) - v2/2M (D) - v2
11. Three blocks are connected on a horizontal frictionless table by two light strings,
one between m1 and m2, another between m2 and m3 . The tensions are T1
between m1 and m2 and T2 between m2 and m3. if m1 = 1 kg, m2 = 8 kg, m3 = 27
kg and F = 36 N applied on m3, then T2 will be
(A) 18 N (B) 9 N
(C) 3.375 N (D) 1.75 N
## 12. A force-time graph for the motion of a body is
1
shown in figure. Change in linear momentum F
between 0 and 8 s is:
(A) Zero (B) 4 N-s
0 8
(C) 8 Ns (D) None t
## 13. A block of mass M is pulled along a horizontal frictionless surface by a rope of
mass m. If a force F is applied at one end of the rope, the force which the rope
exerts on the block is:
(A) F/(M+m) (B) F
(C) FM/(m+M) (D) 0
14. A chain of length L and mass M is hanging by fixing its upper end to a rigid
support. The tension in the chain at a distance x from the rigid support is:
(A) Zero (B) F
(L x ) (L x )
(C) Mg (D) Mg
L M
15. Two masses m and m are tied with a thread passing over a pulley, m is on a
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## frictionless horizontal surface and m is hanging freely. If acceleration due to
gravity is g, the acceleration of m in this arrangement will be
(A) g (B) g/(m+ m )
(C) g/m (D) g/(m- m )
16. A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N
on the block. If the coefficient of friction the block and the wall is 0.5, the
magnitude of the frictional force acting on the block is:
(A) 2.5 N (B) 0.98 N
(C) 4.9 N (D) 0.49 N
## 17. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a
horizontal floor. The coefficient of friction between A and B is 0.2 while that
between B and floor is 0.5. When a horizontal force of 25 N is applied on the
block B. The force of friction between A and B is:
(A) Zero (B) 3.9 N
(C) 5.0 N (D) 49 N
18. A ball weighing 10 gm hits a hard vertical surface with a speed of 5m/s and
rebounds with the same speed. The ball remains in contact with the surface for
(0.01) sec. The average force exerted by the surface on the ball is:
(A) 100 N (B) 10 N
(C) 1 N (D) 0.1 N
## 19. Two masses A and B each of mass M are A F
fixed together by a massless spring. A
force F acts on the mass B as shown in figure. At the instant shown the mass A
has acceleration a. What is the acceleration of mass B?
(A) (F/M)-a (B) a
(C) -a (D) (F/M)
## 20. An object is placed on the surface of a smooth inclined plane of inclination . It
takes time t to reach the bottom. If the same objective is allowed to slide down a
rough inclined plane of same inclination , it takes nt to reach the bottom where n
is a number greater than 1. The coefficient of friction is given by
(A) = tan (1 1/n2) (B) = cot (1 1/n2)
(C) = tan (1 1/n2)1/2 (D) = cot (1 1/n2)1/2
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LEVEL II
## 1. In the figure small block is kept on m then =0
=0 m
F
(A) the acceleration of m w.r.t. ground is M F
m
(B) the acceleration of m w.r.t. ground is zero A B
2 m
(C) the time taken by m to separate from M is
F
2 M
(D) the time taken by m to separate from M is
F
## 2. In the figure, a man of true mass M is standing on a weighing m
machine placed in a cabin. The cabin is joined by a string with a
body of mass m. Assuming no friction, and negligible mass of cabin
and weighing machine, the measured mass of man is (normal force
between the man and the machine is proportional to the mass)
Mm mg
(A) measured mass of man is (B) acceleration of man is
(M m) (M m)
Mg
(C) acceleration of man is (D) measured mass of man is M.
(M m)
## 3. The figure shows a block of mass m placed on a smooth m
wedge of mass M. Calculate the value of M and tension in
the string, so that the block of mass m will move vertically
downward with acceleration 10 m/s2 Smooth
(Take g = 10 m/s2)
Mcot
(A) the value of M is
1 cot
M tan
(B) the value of M
1 tan
Mg
(C) the value of tension in the string is
tan
g
(D) the value of tension is
cot
## 4. Two blocks of masses m1 and m2 are connected g=10m/s2
through a massless inextensible string. Block of
mass m1 is placed at the fixed rigid inclined surface m1=4kg m2=2kg
while the block of mass m2 hanging at the other end Fixed
30
of the string, which is passing through a fixed
massless frictionless pulley shown in figure. The
coefficient of static friction between the block and the
inclined plane is 0.8. The system of masses m1 and
m2 is released from rest.
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## (A) the tension in the string is 20 N after releasing the system
(B) the contact force by the inclined surface on the block is along normal to the inclined
surface
(C) the magnitude of contact force by the inclined surface on the block m1 is 20 3N
(D) none of these
## 5. A force F is applied vertically upward to the pulley and it
is observed that the pulley in the figure moves upward F
with a uniform velocity of 2 m/s. The possible value(s) light pulley
of F is/are (in newtons)
(A) 150 (B) 120 g=10m/s2
(C) 75 (D) 400.
10kg 6kg
Ground
6. The coefficient of friction of all the surfaces is .
The string and the pulley are light. The blocks T1 T1
A 2m
are moving with constant speed. Choose the
correct statement T1 T1
F 3m
B
(A) F = 9 mg (B) T1 = 2 mg
(C) T1= 3 mg (D) T1 = 4 mg
## 7. Two masses of 10 kg and 20 kg are connected
10 kg 20 kg F
by a light spring as shown. A force of 200 N acts
on a 20 kg mass as shown. At a certain instant
F = 200 N
the acceleration of 10 kg mass is 12 ms-2.
(a) At that instatnt the 20 kg mass has an acceleration of 2 ms-2
(b) At that instant the 20 kg mass has an acceleration of 4ms-2
(c) The stretching force in the spring is 120 N
(d) Spring force have different magnitudes for both blocks.
## 8. A particle stays at rest as seen in a frame. We can conclude that
(a) the frame is inertial
(b) resultant force on the particle is zero
(c) the frame may be inertial but the resultant force on the particle is zero
(d) the frame may be non-inertial but there is a nonzero resultant force
9. A particle is observed from two frames S1 and S2. The frame S2 moves with respect to
S1 with an acceleration . Let F1 and F2 be the pseudo forces on the particle when seen
from S1 and S2 respectively. Which of the following are not possible?
(a) F1 = 0, F2 0 (b) F1 0, F2 = 0
(c) F1 0, F2 0 (d) F1 = 0, F2 = 0
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## 10. A block of mass m slides down on a wedge of mass M
as shown in figure. Let a1 be the acceleration of the
M m
wedge and a2 the acceleration of block. N1 is the
normal relation between block and wedge and N2 the
normal reaction between wedge and ground. Friction is
absent everywhere. Select the correct alternative (s)
## (a) N2 < (M + m) g (b) N1 = m (g cos q - | a1 | sin q)
(c) N1 sin q = M | a1 | (d) ma2 Ma1
COMPREHENSIONS
Comprehension I:
A car engine is so constructed as to exert a torque onto the wheels causing them to
rotate, and the wheels move forward on the road due to static friction. The force of static
friction, acting between the wheels and the road, is responsible for the forward
acceleration of the car. The force of static friction has an upper limit, known as the
limiting force of static friction proportional to the normal reaction between the wheels
and the road; this limits the maximum forward acceleration of the car.
In recent years, there has been a tendency to design lighter and more fuel efficient cars,
which drive faster than conventional cars. At very high speeds, it is observed that
conventional cars lose out on manoevrability, as friction is no longer sufficient. This is
caused by airflow around the body of the car, which produces pressure differentials that
increase the tendency of the car to get airborne. Modern designers have tried to
manipulate this airflow so as to reduce lift, decrease drag, and in some cases even
cause a downward force resulting in better traction.
1. A motorist, driving a car on a level road, desires to take a tight circular turn of radius r at
a constant speed v. The coefficient of static friction between the wheels of the car is S
and that of kinetic friction is K. He will be able to take this turn without skidding if
(choose the most appropriate option)
gr gr
(A) K (B) S
v2 v2
gr v2
(C) S , K (D) S
v2 rg
2. The car drives onto a bridge, which is convex upward, maintaining a constant speed v.
The driver, when he is on the bridge,
(A) can take a tighter turn, than when he is on a level road.
(B) cannot take a tighter turn, than when he is on a level road.
(C) can take a turn of the same radius as he could on a level road.
(D) can take tighter turns when he is getting upon the bridge, and looser turns when he
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## is getting off the bridge.
Assume that the driver always takes a full circular turn without skidding.
3. The car is driven at a very high constant speed v, on a straight level road. This causes a
lift L to act on the car due to airflow. The force of friction (f) acting between the driving
wheels and the road (choose the most appropriate option)
(A) is zero, since the car is moving with constant velocity.
(B) Satisfies f = S mg, where m is the mass of the car.
(C) Satisfies f = S (mg L), assuming that the wheels do not lose contact with the road.
(D) Satisfies f = D, where D is the resultant backward force on the car due to air drag
and other contact forces.
Comprehension II :
If a man is measuring his actual weight by
weighing machine as shown in the figure.
The mass of man is 60 kg, mass of
weighing machine is 20 kg and mass of lift
is 30 kg (pulley is smooth and string is
massless).
(Take g = 10 ms2)
## 4. The tension exerted by the man on the string
(A) 6600 N (B) 1800 N
(C) 1110 N (D) 800 N
## 5. Acceleration of the lift is
(A) 30 ms2 (B) 18.5 ms2
(C) 110 ms2 (D) 13.3 ms2
## 6. Normal reaction exerted by man on weighing machine
(A) 600 N (B) 900 N
(C) 800 N (D) 1110 N
## MATCH THE FOLLOWING
1. On a rough surface
Column A Column B
(A) Body is stationary it is possible that (p) Frictional force acting on it is zero
(B) Body is just about to move (q) Frictional force acting on it is static
(C) Body is moving with uniform (r) Frictional force acting on it is limiting
acceleration then it is possible that frictional force
(D) Body is moving with uniform (s) Frictional force acting on it is kinetic
velocity
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## 2. System of blocks are placed on a smooth String2
String1
horizontal system, as shown. For a particular
value of F, 3 kg block is just about to leave 5kg 37 3kg 1kg F
3
ground. (tan 37 = )
4
Column II
Column I
(A) Tension in string 1 (p) 80 N
(B) Tension in string 2 (q) 64 N
(C) Net force by ground on 5 kg block (r) 50 N
(D) Net force on 3 kg block (s) 24 N
5kg
3. In the above arrangement all pulleys are light =0.5
A
and frictionless, threads are ideal, = 0.5 at all 4kg
surfaces. 37 B =0.5
C 2.5kg
Column I Column II
(A) Net force on block A (p) 0
(B) Net force on block B (q) 5
(C) Net force on block C (r) 10
(D) Tension in string (s) 20
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## 11. Hints to Subjective Assignments
LEVEL-I
1. Draw the F.B.D. and find the forces acting on A & B. Vertical forces will balance
each other and apply Newton's second law in horizontal
2. Draw the F.B.D. and resolve the forces in vertical and horizontal direction, write
down eqs. and find a.
3. Draw the F.B.D. and resolve all forces along and perpendicular to plane, apply
Newton's second law.
4. F=ma
a=2kx/m 2 k x
F
6. a
m1 m2
## 7. Calculate the angle of repose.
8. a = g(sin - cos )
## 10. Constant velocity means a = 0
T mg= 0
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LEVEL-II
1. Pseudo force (ma0) and weight mg act vertically downward. Resolve along the
incline.
## 2. Take a small element of chain at an angular position
3. a = m2g/(m1+m2)
## 5. a relative = a1 - a2, where a1 and a2 be the retardation of m and M respectively.
6. Find acceleration of mass relative to earth. Then apply Newtons second law.
7. First consider relative motion between the blocks. Find common acceleration
and sec whether Fe = Mc .a flim
8. Draw the F.B.D. of pulley block system .Find the maximum tension
corresponding to the force F and check whether motion is possible or not.
If possible then apply equations of motion and constraint relation for the
accelerations of masses and pulley.
9. mg ma
F g(M m)
g .
M m
(M + m) g F - 2 g (M + m)
dv
10. Write the acceleration as v and integrate.
ds
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## 12. Answers to Subjective Assignments
LEVEL I
F M2F
1. a= ,N 2. 5.7 m/s2
M1 M2 M1 M2
m1 1 2kx
3. 4.
m2 3 m
5. (a) 1N (b) 1.96 N (c) 1.5 N
F 1
6. m / s2 , m2 will move with constant velocity and m will accelerate
m1 m2 7
## with 1/5 m/s2
7. (a) tan-1 s = tan-1 0.3 = 16.70 (b) 0.145 mg
8. (a) 160 N, (b) 48 N (c) 1.06 m/s2 (d) 55.42 N
1
9. tan 2 = - , = 490 , tmin= 1.0 s
k
10. (Rg/l) / {1- cos(l/R) }
LEVEL II
16L
1. 2. (a) 13.46 N (b) 0
3 3g
## 3. 0.225 m 4. v= (2g / 3a) sin
2 wa
5. 3.7 m/s2, 2.97 m/s2 6.
g a
7. 2m.
15 15
8. (a) a1 = a 2 = 0 = ap (b) a1 = m/s2, a2 = 0 , ap = m/s2
2 4
29
(c) a1 = 25 m/s2, a2 = 4 m/s2, ap = m/s2
2
mg sin
9. = 0.5 10.
M 2m(1 cos
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LEVEL I
1. (C) 2. (C)
3. (D) 4. (A)
5. (C) 6. (A)
7. (C) 8. (B)
9. (B) 10. (B)
11. (B) 12. (B)
13. (C) 14. (C)
15. (B) 16. (B)
17. (B) 18. (B)
19. (A) 20. (A)
LEVEL II
## 1. (B), (D) 2. (A), (C)
3. (A), (D) 4. (A), (B), (C)
5. (A), (B) 6. (A), (B)
7. (B), (C) 8. (C), (D)
9. (D) 10. (A), (B), (C)
COMPREHENSION
## 10. (D) 11. (B)
12. (D) 13. (A)
14. (C) 15. (A)
MATCH THE FOLLOWING
1. (A) (p), (q), (r); (B) (q), (r); (C) (s); (D) (s)
## 3. (A) (s); (B) (p); (C) (q); (D) (r)
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## WORK, ENERGY & POWER
7. Assignments (Subjective Problems)
LEVEL I
## 1. An object of mass 5 kg falls from rest through a vertical distance of 20 m and
attains a velocity of 10 m/s. How much work is done by the resistance of the air
on the object? (g = 10m/s2).
## 2. A train of mass 100 metric tons is drawn up an incline of 1 in 49 at the rate of 36
km per hour by an engine. If the resistance due to friction be 10 N per metric ton,
calculate the power of the engine. If the steam is shut off, how far will the train
move before it comes to rest?
A
3. Shown in the figure is a smooth
vertical frame of wire along which B
H
a small bead moves from the point h
A. Find its speed at the point B.
4. A rubber ball after falling through a height h penetrates into the water through a
distance x. Find the average force imparted by water on the rubber ball in ideal
conditions.
## 5. A bus of mass 1000 kg has an engine which produces a constant power of 50
kW. If the resistance to motion, assumed constant is 1000 N, find the maximum
speed at which the bus can travel on level road and the acceleration when it's
travelling at 25 m/s.
## 6. A particle of mass m is projected up the
v0
smooth inclined plane of inclination with a
speed v0. ( 2gl sin ) as shown in the l
figure. Find the maximum height travelled m
by the particle.
## 7. A block of mass m collides with a horizontal weightless spring of force constant k.
The block compresses the spring by x. Calculate the maximum momentum of the
block.
## 8. Prove that K. E. of two identical trains with respect to heliocentric frame of
reference moving in opposite directions on equatorial line with same speed (w.r.t
earth) are not equal.
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9. A small mass m starts from rest and slides down the smooth spherical surface of
R. Assume zero potential energy at the top. Find (a) the change in potential
energy (b) the kinetic energy (c) the speed of the mass as a function of the angle
10. An ideal massless spring can be compressed
by 1 m by a force of 100 N. This same spring is
placed at the bottom of a frictionless inclined
plane which makes an angle = 30o with the
horizontal. A 10 kg mass is released from rest
o
at the top of the incline and is brought to rest 30
momentarily after compressing the spring 2
meters.
(a) Through what distance does the mass slide before coming to rest?
(b) What is the speed of the mass just before it reaches the spring?
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LEVEL - II
## 1. A spring of mass m and stiffness k is fitted
v
to a block of mass M. The system is moving m
with a constant velocity v on a smooth M
## horizontal surface. If the system collides
with a wall, find the maximum compression
of the spring before it recoils, assuming that
the total energy is conserved.
## 2. A body of mass m is pushed with the initial velocity v0 up an inclined plane of
angle of inclination . The co- efficient of friction between the body and the plane
is . What is the net work done by friction during the ascent of the body, comes
to stop.
3. A block of mass m slides from the top of an inclined plane of angle of inclination
& length l. The coefficient of friction between the plane and the block is . Then it
is observed cover a distance d along the horizontal surface having the same
coefficient of friction , before it comes to a stop. Find the value of d.
## 4. A particle of mass m moves along a straight line on smooth horizontal plane,
acted upon by a force delivering a constant power P. If the initial velocity of the
particle is zero, then find its displacement as a function of time t.
## 5. The figure shows a ball A of mass m A
connected to a light spring of stiffness k. R
Another identical ball B is connected with
the ball A by a light inextensible string as B /4
shown in the figure. Other end of the
spring is fixed. Initially the spring is in
relaxed position. A vertical force F acts m1
## on B such that the balls move slowly.
What is the work done by the force in pulling the ball B till the ball A reaches at
the top of the cylindrical surface. The ball A remains in contact with the surface
and co-
## 6. A body of mass m was slowly hauled up the hill. F
(Fig.) by a force F which at each point was directed
along a tangent to the trajectory. Find the work m
h
performed by this force, if the height of the hill is h,
the length of its base l, and coefficient of kinetic l
friction k.
## 7. A particle of mass m moves along a circle of radius R with a normal acceleration
varying with time as wn = at2, where a is a constant. Find the time dependence of
the power developed by all the forces acting on the particle, and the mean value
of this power averaged over the first t seconds after the beginning of motion.
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## 8. A bob hangs from a rigid support by an inextensible string of length . If it is
displaced through a distance keeping the string straight & released, find the
speed of the bob at the lowest position.
9. A body is projected from the top of a smooth fixed frictionless semi circular
vertical tube of radius R with a speed gR . Find the speed of the body when it
descends through a vertical distance R/2.
## 10. Two blocks of masses m1, and m2 connected by a light m1
spring of stiffness k, are kept on a smooth horizontal
surface as shown in the figure. What should be the
initial compression of the spring so that the system will
be about to break off the surface, after releasing the
m2
block m1?
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## 8. Assignment (Objective Problems)
LEVEL I
1. Two bodies of masses m1 and m2 have equal momenta. Their kinetic energies E1
and E2 are in the ratio:
(A) m1 : m2 (B) m1 : m2
(C) m2 : m1 (D) m12 : m22
2. A chain of mass M, length hangs from a pulley. If it is wound such that half of
the chain remains overhung, the work done by the external agent is equal to
Mg 3
(A) (B) Mg
2 4
3
(C ) Mg (D) None of these
8
## 3. A block of mass m is suspended by a light thread T
from an elevator. The elevator is accelerating
upward with uniform acceleration a. The work
done during t secs by the tension in the thread is: m
m m
(A) ( g a )at 2 (B) ( g a)at 2
2 2
m
(C) gat2 (D) 0.
2
## 4. It is easier to draw up a wooden block along an inclined plane than to haul it
vertically, principally because:
(A) the friction is reduced
(B) the mass becomes smaller
(C) only a part of the weight has to be overcome
(D) g becomes smaller
5. A motor boat is travelling with a speed of 3.0 m/sec. If the force on it due to water
flow is 500 N, the power of the boat is:
(A) 150 KW (B) 15 KW
(C) 1.5 KW (D) 150 W
6. Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio
of the magnitudes of their linear momenta is:
(A) 4 : 1 (B) 2 : 1
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(C) 1 : 2 (D) 1 : 16
7. A system of two bodies of masses m
and M being interconnected by a k
spring of stiffness k moves towards a m
rigid wall with a K.E. E . If the body M M
sticks to the wall after the collision, the
maximum compression of the spring
will be
m.E 2m.E
(A) (B)
k (M m)k
2m.E 2M.E
(C ) (D)
Mk (M m)k
## 8. Two identical blocks each of mass m being interconnected by a light spring of
stiffness k is pushed by a force F as shown in the figure. The maximum potential
energy stored in the spring is equal to:
F2 F2
(A) (B) F
2k 4k
F2
(C) (D) None of these
8k
9. A stone tied to a string of length is whirled in a vertical circle with the other end
of the string at the centre. At a certain instant of time, the stone is at its lowest
position and has a speed u. The magnitude in its velocity as it reaches a
position, where the string is horizontal, is
(A) u2 2g (B) 2g
(C) u2 g (D) 2(u2 g )
## 10. A cord is used to raise a block of mass m vertically through a distance d at a
constant downward acceleration g/4. The work done by the chord is
(A) mgd/4 (B) 3Mgd/4
(C) -3Mgd/4 (D) Mgd
11. A uniform chain of length L and mass M is lying on a smooth table and one-third
of its length is hanging vertically down over the edge of the table. If g is
acceleration due to gravity, the work required to pull the hanging part onto the
table is:
(A) MgL (B) MgL/3
(C) MgL/9 (D) MgL/18
## 12. An engine pumps a liquid of density d continuously through a pipe of area of
cross-section A. If the speed with which the liquid passes through a pipe is v,
then the rate of liquid flow is
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2
13. A body of mass m accelerates uniformly from rest to v1 in time t1. As a function of
t, the instantaneous power delivered to the body is:
(A) mv1/t1 (B) mv12/t1
(C) mv1t2/ t1 (D) mv12t/t12
## 14. A man M1 of mass 80 Kg runs up a staircase in 15 s. Another man M2 also of
mass 80 Kg runs up the stair case in 20 s. The ratio of the power developed by
them will be:
(A) 1 (B) 4/3
(C) 16/9 (D) None of the above
15. How much work is done in raising a stone of mass 5 Kg and relative density 3
lying at the bed of a lake through height of 3 meter? (Take g = 10 ms-2):
(A) 25 J (B) 100 J
(C) 75 J (D) None of the above
## 16. A person is pulling a mass m from
ground on a rough hemispherical surface
upto the top of the hemisphere with the R
help of a light inextensible string as m
shown in the figure. The radius of the
hemisphere is R. The work done by the
tension in the string is:
(A) mgR(1+ ) (B) mgR
(C) mgR(1- ) (D) mg(R/2).
17. A small mass m is sliding down on a smooth m
curved incline from a height h and finally
k
moves through a horizontal smooth surface. A h
light spring of force constant k is fixed with a
vertical rigid stand on the horizontal surface,
as shown in the figure. The maximum
compression in the spring if the mass m
released from rest from the height h and hits
the spring on the horizontal surface is:
2mgh mgh
(A) (B)
k k
mgh
(C) (D) None of these.
2k
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## 18. A block of mass m moves towards a light
spring of stiffness k on a smooth horizontal v k
plane. If it compresses the spring through a m
## distance x0, the magnitude of total change in
momentum of the block is:
(A) 2 km x 0 (B) 0
(C) km x 0 (D) - 2 km x 0 .
19. If v, P and E denote the velocity, momentum and kinetic energy of the particle,
then:
(A) P = dE/dv (B) P = dE/dt
(C) P = dv/dt (D) none of these
20. Energy required to accelerate a car from 10 to 20 m/s compared with that
required to accelerate from 0 to 10 m/s is
(A) twice (B) four times
(C) three times (D) same
LEVEL II
## Other end of the string is fixed point O. Bob is rotating in a
circular path of radius l in horizontal plane about O with constant v
O
speed v, as shown in the figure. The average force exerted by l
string on the bob during its :
m
A
mv 2 2mv 2
(A) half revolution will be (B) half revolution will be
l l
2mv 2
(C) one fourth revolution will be (D) one revolution will be zero
l
2. A ball of mass m attached to the lower end of a light vertical spring of force constant k.
The upper end of the spring is fixed. The ball is released from rest with the spring at its
normal (unstretched) length, and comes to rest again after descending through a
distance x
(A) x = mg/k
(B) x = 2mg/k
(C) the ball will have no acceleration at the position where it has descended through x/2
(D) the ball will have an upward acceleration equal to g at its lowermost position
## 3. A long block A is at rest on a smooth horizontal surface. A small v
B
block B, whose mass is half of A, is placed on A at one end and A
projected along A with some velocity u. The coefficient of friction
between the blocks is
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u
(A) the blocks will reach a final common velocity
3
(B) the work done against friction is two-thirds of the initial kinetic energy of B
(C) before the blocks reach a common velocity, the acceleration of A relative to B is
2
g
3
(D) before the blocks reach a common velocity, the acceleration of A relative to B is
3
g
2
4. The ring shown in the figure is given a constant m smooth horizontal rail
## horizontal acceleration (a0 = g / 3 ). Maximum
deflection of the string from the vertical is 0, l
then
(A) 0 = 30 particle
m
(B) 0 = 60
(C) at maximum deflection, tension in string is equal to mg
2mg
(D) at maximum deflection, tension in string is equal to
3
## 5. Figure shows a block P of mass m resting on a horizontal P
smooth floor at a distance l from a rigid wall. Block is k
m
pushed toward right by a distant 3l and released. When
2 l
block passes from its mean position another block of
mass m1 is placed on it which sticks to it due to friction so
that the combined block just collides with the left wall.
5m
(A) m1 .
8
5m
(B) m1 .
4
3 K
(C) Velocity of block m at mean position is
2 m
(D) Velocity of block m at mean position is
6. A heavy stone is thrown from a cliff of height h in a given direction. The speed with which
it hits the ground
(A) must depend on the speed of projection
(B) must be larger then the speed of projection
(C) must be independent of the speed of projection
(D) may be smaller than the speed of projection
7. You lift a suitcase from the floor and keep it on a table. The work done by you on the
suitcase does not depend on
(A) the path taken by the suitcase (B) the tame taken by you in doing so
(C) the weight of the suitcase (D) your weight
## 8. No work is done by a force on an object if
(A) the force is always perpendicular to its velocity
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## (B) the force is always perpendicular to its acceleration
(C) the object is stationary but the point of application of the force moves on the object
(D) the object moves in such a way that the point of application of the force remains
fixed
9. A particle of mass m is attached to a light string of length l, the other end of which is
fixed. Initially the string is kept horizontal and the particle is given an upward velocity .
The particle is just able to complete a circle.
(A) the string becomes slack when the particle reaches its highest point
(B) the velocity of the particle becomes zero at the highest point
1
(C) the kinetic energy of the ball in initial position was m 2 mg l
2
(D) the particle again passes through the initial position
## 10. The kinetic energy of a particle continuously increases with time.
(A) the resultant force on the particle must be parallel to the velocity at all instants
(B) the resultant force on the particle must be at an angle less than 90 all the time
(C) its height above the ground level must continuously decrease
(D) the magnitude of its linear momentum is increasing continuously
COMPREHENSION
Comprehension I :
## According to the principle of conservation of linear strut 3 string
strut
momentum if the external force acting on the
system is zero, the linear momentum of the M
system will remain conserved.
It means if the centre of mass of a system is initially at rest, it will remain at rest in the
absence of external force, that is the displacement of centre of mass will be zero. A
plank of mass M is placed on a smooth horizontal surface. Two light identical springs
each of stiffness K are rigidly connected to struts at the end of the plank as shown in the
figure. When the springs are in their unextended position the distance between their free
ends is 3 . A block of mass m is placed on the plank and pressed against one of the
springs so that it is compressed to l. To keep the block at rest it is connected to the strut
by means of a light string. Initially the system is at rest. Now the string is burnt.
5m 5m
(A) (B)
M M m
3m 4m
(C) (D)
M m M m
## 2. Maximum velocity of plank is
Km m
(A) (B)
(M m) (M m)
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Km Km
(C) (D)
M(M m) (M m)
## 3. Maximum kinetic energy of block m is
Km2 Km2
(A) (B)
2M(M m) M(M m)
KM 2
(C) (D) none of these
2(M m)
Comprehension II :
v0
A small particle of mass m is given an initial velocity v 0 tangent to A
r0
the horizontal rim of a smooth cone at a radius r0 from the vertical
centerline as shown at point A. As the particle slides to point B, a
r B
vertical distance h below A and a distance r from the vertical
centerline, its velocity v makes an angle with the horizontal v
tangent to the cone through B.
4. The value of is
1 v 0r0 1 v 0r0
(A) cos (B) cos
v 20 2gh(r0 h tan ) v 02 2gh(r0 h tan )
1 v 0r0 1 v 0r0
(C) cos (D) cos
v 20 2gh(r0 h tan ) r0 v 02 2gh
## 5. The speed of particle at point B
(A) v 02 2gh (B) v 02 2gh
(C) v 02 gh (D) 2v 02 2gh
6. The minimum value of v0 for which particle will be moving in a horizontal circle of radius
r0.
2gr0 gr0
(A) (B)
tan 2 tan
gr0 4gr0
(C) (D)
tan tan
## MATCH THE FOLLOWING
1. A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has length
4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under
a tension of 80 N. A sinusoidal wave-pulse of amplitude 3.5 cm is sent along the wire PQ
from the end P. No power is dissipated during the propagation of the wave-pulse.
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Column I Column II
(A) The time taken by the wave pulse to reach the (p) 0.14
other end R of the wire (in sec)
(B) The amplitude of the reflected wave pulses (q) 2.0
after the incident wave pulse cross the joint Q.
(in cm)
(C) The amplitude of the transmitted wave pulses (r) 32
after the incident wave pulse cross the joint Q.
(in cm)
(D) Velocity of wave in QR (in m/s) (s) 1.5
2. A particle of 500 gm mass moves along a horizontal circle of radius 16 m such that
normal acceleration of particle varies with time as an = 9t2.
Column II
Column I (given values are in proper unit)
(A) Tangential force on particle at t = 1 second (in (p) 72
Newton)
(B) Total force on particle at t = 1 second (in Newton) (q) 36
(C) Power delivered by total force at t = 1 second (in (r) 7.5
Newton)
(D) Average power developed by total force over first (s) 6
one second (in watt)
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LEVEL I
## 1. Use work energy theorem
2. Work done by engine + work done by gravity + work done by frictional force = 0, as
there is no change in kinetic energy.
3. Use conservative work concept
4. Work done by resistive force of water = change in K.E.
5. Use relation for power
6. Use work energy theorem
7. Consider F.B.D. of both the components
8. Resultant velocity of a train w.r.t. sun (heliocentric frame) = vector sum of velocity of
train w.r.t. earth velocity of earth.
9. (a) Decrease in P.E. = increase in K.E.
(b) Net force along radius and towards the centre provides the necessary centripetal
force.
10. Consider equilibrium of body and apply work energy.
LEVEL II
## 1. Think about the reference frame and apply energy concept
2. Use work energy theorem
3. Use conservation of energy to find out the velocity at the bottom of incline plane.
Then use work-energy theorem.
4. Use the concept of work.
5. Use work energy theorem.
6. Use work energy theorem.
7. Normal acceleration is given as time dependent, hence v is time dependent now find
dv /dt
8. Use work energy theorem (Note: work done by tension is zero)
9. Consider F.B.D. of the particle and apply work energy theorem
10. Think the condition of breaking off
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LEVEL I
## 1. -750 J 2. 210 kW, 238 m
h
3. 2g(H h) 4. mg 1
x
2
(v 2
0 2 g sin ) sin 2
5. 50 m./sec., 2 m/sec . 6. sin
2g
7. ( km ) x
## 10. (a) S = 4 m, (b) v = 2 5 m/s
LEVEL II
(3M m) mv 02
1. v 2.
3k 2(tan )
1/ 2
8Pt3
3. [sin - cos ]/ 4.
9m
1
5. mgR [ 1 ]
2 4 2
mRat
6. mg (h + kl) 7. P = mRat, <P> =
2
8. gl 9. 2gR
( 2m1 m2 )g
10.
k
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## 11. Answers to the Objective Assignment
LEVEL I
1. (C) 2. (C)
3. (A) 4. (C)
5. (C) 6. (C)
7. (B) 8. (A)
9. (D) 10. (B)
11. (D) 12. (D)
13. (D) 14. (B)
15. (B) 16. (A)
17. (A) 18. (C)
19. (A) 20. (C)
LEVEL II
1. (B), (D) 2. (B), (C), (D)
3. (A), (B), (D) 4. (A), (D)
5. (B), (C) 6. (A), (B)
7. (A), (B), (D) 8. (A), (C), (D)
9. (A), (D) 10. (B), (D)
COMPREHENSION
1. (B) 2. (C)
3. (C) 4. (A)
5. (A) 6. (C)
MATCH THE FOLLOWING
## 2. (A) (s); (B) (r) ; (C) (p); (D) (q)
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COLLISION
9. Assignments (Subjective Problems)
LEVEL I
1. Two particles A and B of which lighter particle has mass m, are released from
infinity. They move towards each other under their mutual force of attraction. If
their speeds are v and 2 v respectively find the K.E. of the system.
2. A bullet of mass 0.01 kg travelling at a speed of 500 m/s strikes a block of mass
2 kg which is suspended by a string of length 5 m. The centre of gravity of the
block is found to rise a vertical distance of 0.2 m. What is the speed of the bullet
after it emerges from the block?
3. A body of mass 1 kg initially at rest, explodes and breaks into three fragments of
masses in the ratio 1 : 1 : 3. The two pieces of equal mass fly off perpendicular to
each other with a speed of 15 m/s each. What is the velocity of the heavier
fragment?
4. Steel ball of mass 0.5 kg is fastened to a cord 20 cm long and fixed at the far end
and is released when the cord is horizontal. At the bottom of its path the ball
strikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is
elastic. Find the speed of the block, just after the collision.
5. A particle loses 25% of its energy during collision with another identical particle
at rest. Find the coefficient of restitution.
6. A body of mass 3 kg collides elastically with another body at rest and then
continues to move in the original direction with one half of its original speed.
What is the mass of the target body?
7. An automatic gun fires 600 bullets a minute. The mass of each bullet is 4 gm and
its initial velocity is 500 m/s. Find the mean impact force experienced by the gun.
## 8. A skater of mass m standing on ice throws a stone of mass M with a velocity of v
m/s in a horizontal direction. Find the distance over which the skater will move
back if the coefficient of friction between the skaters and the ice is .
## 9. A steel ball with a mass of m = 20 g falls from a height of h1 = 1 m onto a steel
plate and rebounds to a height of h2 = 81 cm. Find: the impulse of the force
received by the plate during the impact.
10. A ball collides with an inclined plane of inclination after falling through a
distance h. If it moves horizontally just after the impact, find the coefficient of
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restitution.
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LEVEL II
## 1. A block of mass m moving with a velocity v hits a light spring of stiffness K
attached rigidly to a stationary sledge of mass M. Neglecting friction between all
contacting surface, find the maximum compression of the spring.
## 2. A small empty bucket of mass M is attached to a long inextensible cord of length
. The bucket is released from rest when the cord is in a horizontal position. In its
lowest position the bucket scoops up a mass m of water, what is the height of the
swing above the lowest position?
3. A uniform thin rod of mass M and length L is standing vertically along the y-axis
on a smooth horizontal surface, with its lower end at the origin (0,0). A slight
disturbance at t = 0 causes the lower end to slip on the smooth surface along the
positive x-axis, and the rod starts falling. (a) What is the path followed by the
centre of mass of the rod during its fall. (b) Find the equation of trajectory of a
point on the rod located at a distance r from the lower end. What is the shape of
the path of this point?
## 4. A block of mass M with a semicircular track of radius R rest on a horizontal
frictionless surface. A uniform cylinder of radius r and mass m is released from
rest at the top point A. The cylinder slips in the semicircular frictionless track.
How far the block moved when the cylinder reaches the bottom of the track?
## 5. A block of mass M is hanging from a rigid support by an in-extensible light string.
A ball of mass m hits it with a vertical velocity v at its bottom. Find the change in
momentum of the ball assuming inelastic collision.
6. Two identical blocks A and B of mass M each are kept on each other on a
smooth horizontal plane. There exists friction between A and B. If a bullet of
mass m hits the lower block with a horizontal velocity v and gets embedded into
it. Find the work done by friction between A and B.
7. A cannon and a supply of cannon balls are inside a sealed railroad car. The
cannon fires to the right, the car recoils to the left. The cannon balls remain in the
car after hitting the far wall. Show that no matter how the cannon balls are fired,
the railroad car cannot travel more than , assuming it starts from rest
8. Two wooden blocks of mass M1 = 1 kg, M2 = 2.98 kg lie separately side by side
smooth surface. A bullet of mass m = 20 gm strikes the block M1 and pierces
through it, then strikes the second block and sticks to it. Consequently both the
blocks move with equal velocities. Find the percentage change in speed of the
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## bullet when it escapes from the first block.
9. A steel ball is suspended by a light inextensible string of O
length from a fixed point O. When the ball is in equilibrium
it just touches a vertical wall as shown in the figure. The ball
is first taken aside such that string becomes horizontal and
then released from rest. If co-efficient of restitution is e, then
find the maximum deflection of the string after nth collision.
## 10. A body of mass M with a small disc of mass m
m v
placed on it rests on a smooth horizontal plane.
M
The disc is set in motion in the horizontal direction
with velocity v. To what height (relative to the
initial level) will the disc rise after breaking off the
body M? All surfaces are frictionless.
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## 10. Assignments (Objective Problems)
LEVEL I
1. If the KE of a body becomes four times of its initial value, then the new
momentum will be more than its initial momentum by;
(A) 50% (B) 100%
(C) 125% (D) 150%
2. A bullet in motion hits and gets embedded in a solid block resting on a frictionless
table. What is conserved?
(A) Momentum and KE (B) Kinetic energy alone
(C) Neither KE nor momentum (D) Momentum alone
## 3. The centre of mass of a body:
(A) Lies always at the geometrical center
(B) Lies always inside the body
(C) Lies always outside the body
(D) Lies within or outside the body
4. The spacecraft of mass M moves with velocity V in free space at first, then it
explodes breaking into two pieces. If after explosion a piece of mass m comes to
rest, the other piece of space craft will have a velocity:
(A) MV/(M m) (B) MV/(M + m)
(C) mV/(M m) (D) mV/(M + m)
5. A bomb travelling in a parabolic path under the effect of gravity, explodes in mid
air. The centre of mass of fragments will:
(A) Move vertically upwards and then downwards
(B) Move vertically downwards
(C) Move in irregular path
(D) Move in the parabolic path the unexploded bomb would have traveled
## 6. If a ball is thrown upwards from the surface of earth:
(A) The earth remains stationary while the ball moves upwards
(B) The ball remains stationary while the earth moves downwards
(C) The ball and earth both moves towards each other
(D) The ball and earth both move away from each other
## 7. Fast neutrons can easily be slow down by
(A) The use of lead shield. (B) Passing them through heavy water.
(C) Elastic collision with heavy nucleus (D) Applying a strong electric field.
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## 8. A body is hanging from a rigid support by an inextensible string of length 'l' . It is
struck inelastically by an identical body of mass m with horizontal velocity
v = 2gl , the tension in the string increases just after the striking by
(A) mg (B) 3 mg
(C) 2 mg (D) None of these
9. A bullet weighing 10 gm and moving at 300 m/s strikes a 5 kg ice and drops
dead. The ice block is sitting on a frictionless level surface. The speed of the
block after the collision, is
(A) 6 cm/sec (B) 6 m/sec
(C) 60 mc/sec (D) 60 m/sec.
10. A massive ball moving with speed v collides with a tiny ball having a mass very
much smaller than the mass of the first ball. If the collision is elastic, then
immediately after the impact, the second ball will move with a speed
approximately equal to:
(A) v (B) 2v
(C) v/2 (D)
11. A bag of mass M hangs by a long thread and a bullet (mass m) comes
horizontally with velocity v and gets caught in the bag. Then for the combined
system (bag + bullet):
(A) Momentum is mMv/(M + m) (B) KE is (1/2) Mv2
(C) Momentum is mv (D) KE is m2v2/2(M + m)
12. A surface is hit elastically and normally by n balls per unit time, all the balls
having the same mass m and moving with the same velocity u. The force on the
surface is:
(A) mnu2 (B) 2mnu
2
(C) (1/2)mnu (D) 2mnu2
13. A bullet of mass m hits a target of mass M hanging by a string and gets
embedded in it. If the block rises to a height h as a result of this collision, the
velocity of the bullet before collision is:
(A) v 2gh (B) v 2gh [1 (m / M)]
(C) v 2gh [(1 M / m)] (D) v 2gh [1 (m / M)]
14. Two particles of mass M and 2 M are at a distance D apart. Under their mutual
force they start moving towards each other. The acceleration of their centre of
mass when they are D/2 apart is:
(A) 2 GM/D2 (B) 4 GM/D2
(C) 8 GM/D2 (D) Zero
15. A body of mass 1 kg initially at rest, explodes and breaks into three fragments of
mass in the ratio 1 : 1 :3. The two pieces of equal mass fly off perpendicular to
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each other with speed of 15 m/sec. each. The speed of havier fragment is:
(A) 5 2 m/s (B) 45 m/se
(C) 5 m/s (D) 156 m/s
16. A sphere of mass m1 = 2kg collides with a sphere of mass m2 = 3kg which is at
rest. Mass m1 will move at right angle to the line, joining centres at the time of
collision, if the coefficient of restitution is
(A) 4/9 (B) 1/2
(C) 2/3 (D) 23
17. In a free space, a rifle of mass M shoot a bullet of mass m at a stationary block of
mass M distance D away from it. When the bullet has moved through a distance
d towards the block, the centre of mass of the bullet-block system is at a distance
from rifle of
mD d dM m
(A) (B) from a rifle
M m M
md MD
(C) . (D) None of these.
M m
18. Two identical billiard balls are in contact on a table. A third identical ball strikes
them symmetrically and come to rest after impact. The restitution is
2 1
(A) (B)
3 3
1 3
(C) (D)
6 2
19. A sphere of mass m moving with a constant velocity hits another stationary
sphere of the same mass. If e is the coefficient of restitution, then ratio of
velocities of the two spheres after collision will be:
1 e 1 e
(A) (B)
1 e 1 e
e 1 e 1
(C) (D)
e 1 e 1
20. A neutron travelling with a velocity v and kinetic energy E collides elastically head
on with the nucleus of an atom of mass number A at rest. The fraction of total
energy retained by the neutron is:
2 2
A 1 A 1
(A) (B)
A 1 A 1
2 2
A 1 A 1
(C) (D)
A A
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LEVEL II
## 1. Velocity of a particle of mass 2 kg changes from v1 2i 2j m / s to v 2 ( i j)m / s
after colliding with a plane surface
1
(A) the angle made by the plane surface with the positive xaxis is 90 + tan1
3
1 1
(B) the angle made by the plane surface with the positive xaxis is tan
3
1 1
(C) the direction of change in momentum makes an angle tan with the +ve xaxis
3
1 1
(D) the direction of change in momentum makes an angle 90 tan with the
3
plane surface
## 2. Choose the correct statement (s) of the following :
(A) force acting on a particle for equal time intervals can produce the same change in
momentum but different change in kinetic energy
(B) force acting on a particle for equal displacements can produce same change in
kinetic energy but different change in momentum
(C) force acting on a particle for equal time intervals can produce different change in
momentum but same change in kinetic energy
(D) force acting on a particle for equal displacements can produce different change in
kinetic energy but same change in momentum
## 3. In an elastic collision between two particles
(A) the total kinetic energy of the system is always conserved
(B) the kinetic energy of the system before collision is equal to the kinetic energy of the
system after collision
(C) the linear momentum of the system is conserved
(D) the mechanical energy of the system before collision is equal to the mechanical
energy of system after collision
## 4. A ball of mass 1 kg is thrown up with an initial speed of 4 2 kg
m/s. A second ball of mass 2 kg is released from rest from u=0
## some height as shown in the figure.
(A) The centre of mass of the two balls comes down with 4 m/s
acceleration g /3.
1 kg
(B) The centre of mass first moves up and then comes
down
(C) The acceleration of the centre of mass is g downwards
(D) The centre of mass of the two balls remains stationary.
## 5. Two bodies of masses m1 and m2 are placed on a m2 m1 F
horizontal table with coefficient of friction and are
joined by a spring. Initially, the spring has its natural
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## length. If F is minimum force which, when applied on
m1, will make the other block m2 just move. (k is the
spring constant) and x is elongation in spring at that
instant.
m1g m2g
(A) F m2 g (B) x
2 k
m2g 2 m2g
(C) F m1g (D) x
2 k
6. A ball A of mass m moving with velocity n collides head on with a stationary ball B of
mass m. If e be the coefficient of restitution, then which of the following is correct?
1 e
(A) The ratio of velocities of balls A and B after the collision is
1 e
1 e
(B) The ratio of the final and initial velocities of ball A is
2
1 e
(C) The ratio of velocities of balls A and B after collision is
1 e
1 e
(D) The ratio of the final and initial velocities of the ball B is
2
7. When a bullet is fired from a gun
(A) Kinetic energy of bullet is more than that of gun
(B) Acceleration of bullet is more than that of gun
(C) Momentum of bullet is more than that of gun
## (D) Velocity of bullet is more than that of gun
8. A ball of mass m moving horizontally at a speed n collides with the bob of a simple
pendulum at rest. The mass of the bob is also m.
(A) If the collision is inelastic, the height to which the two balls rise after the collision is
v2
8g
(B) If the collision is inelastic, the kinetic energy of the system immediately after the
collision becomes half of that before collision
v2
(C) If the collision is perfectly elastic, the bob of the pendulum will rise to a height of
2g
(D) If the collision is perfectly elastic, the kinetic energy of the system immediately after
the collision is equal to that before collision
9. In which of the following cases the center of mass of a rod is certainly not at its centre?
(A) the density continuously increases from left to right
(B) the density continuously decreases from left to right
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(C) the density decreases from left to right upto the centre and then increases
(D) the density increases from left to right upto the centre and then decreases
10. A nonzero external force acts on a system of particles. The velocity and the acceleration
of the center of mass are found to be 0 and a0 at an instant t. It is possible that
(A) 0 = 0, a0 = 0 (B) 0 = 0, a0 0
(C) 0 0, a0 = 0 (D) 0 0, a0 0
COMPREHENSION
## Two blocks A and B of mass m and 2 m respectively are
connected together by a light spring of stiffness K. The system
m 2m
is lying on a smooth horizontal surface with the block A in
A B
contact with a fixed vertical wall as shown in the figure. The
block B is pressed towards the wall by a distance x0 and then
released. There is no friction anywhere.
1. If spring takes time t to acquire its natural length then average force on the block A by
the wall is
2mK x 0
(A) 0 (B)
t
mK 0x Km x 0
(C) 3 (D)
2 t 3 t
## 2. The maximum speed acquired by the block A is
2 K 4 K
(A) x0 (B) x0
3 2m 3 2m
K 1 K
(C) x0 (D) x0
2m 3 2m
3. If natural length of the spring is l, then maximum separation between the blocks is
x0
(A) (B) x0
3
x0
(C) (D) x0
3
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## If external forces on a system is zero. Linear momentum
remains conserved. Coefficient of restitution e is
defined as ratio of velocity of separation and velocity of M
approach along line of impact. A disc of mass M and
radius of curvature R is lying on the ground (frictionless). V R
A small ball of mass M moving on the ground with a
velocity V strikes the disc and collides with it. The
coefficient of restitution being e.
## 4. Velocity of approach for the system (ball + disc) is
(A) V cos (B) V sin
(C) V (D) zero
## 5. Velocity of the disc after collision is
V cos eV cos
(A) (B)
2 2
V cos (1 e) V cos (1 e)
(C) (D)
2 2
## 6. The speed of the ball after the collision.
cos2 (1 e)2
(A) eV cos (B) V sin2
4
sin2 cos2 (1 e)2
(C) V cos2 (1 e)2 (D) V sin2
4 4
## MATCH THE FOLLOWING
1. A ball A of mass m1, initially moving with velocity u 1, collides with another ball B of mass
m2, moving with velocity u2 in a straight line in the same direction. After collision
velocities of the balls A and B are 1 and 2 respectively then (e is the coefficient of
restitution between the balls).
Column A Column B
(A) If u2 = 0, e = 1/2 and m1 >> m2 then (p)
u1
1
(B) If u2 = 0 , e = 1/3 and m2 >> m1 then (q)
3u1/2
1
(C) If u2 = 0, e = 1 and m1 >> m2 then 1 (r) u1/3
(D) If m1 = m2, u2 = 0, e = 1 then 2 1 = (s) Zero
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## 2. Initially spring connecting A and B are 3kg 3kg 6kg
elongated by a distance of 3 cm and placed on C A B
smooth horizontal surface. When spring is in its
N
natural length (block A moving right and block B Spring constant K 200
is moving left) block C moving towards A with m
speed 0.4 m/s (towards right) collides and get
stuck with it.
Column I Column II
(in proper unit)
(A) Velocity of B before collision (m/s) (p) 0.1
(B) Velocity of center of mass of whole (q) 0.2
system after colliding (m/s)
(C) Amplitude of oscillation of combined (r) 0.03
body (m) (rounded off to one significant
figure)
(D) Loss of energy during collision (J) (s) 0.05
3. A body is initially moving towards right explodes into two pieces 1 and 2. Direction of
motion of the pieces is shown in Column I and possible mass ratio are shown in Column
II.
Column I Column II
(A) V1 (p) m1>m2
V2
(B) V1 (q) m1=m2
1
2
V2
(C) V1 (r) m1<m2
V2
(D) V1 (s) Impossible for any masses
1
2
V2
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## 11. Hints to Subjective Assignments
LEVEL I
1. Apply COM.
2. Apply COM then COE.
3. Apply COM.
4. Apply COE then COM.
5. From KE find final velocity of first particle. Then apply conservation of linear
momentum
6. Apply COM .
7. Apply Newtons Second Law for variable mass.
8. Apply COM then relate work- energy.
9. Impulse equals to change of momentum
10. Remember momentum conservation along horizontal
LEVEL II
## 1. Remember that during maximum compression of spring, relative velocity of the
bodies is zero.
2. This is completely inelastic collision.
3. x coordinate of cm remains constant
4. x coordinate of cm remains constant
5. Apply COM
6. Apply COM
7. System cm has same x coordinate
8. Apply COM
9. Apply definition of e and apply COE
10. Apply COM in horizontal direction then apply COE.
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LEVEL I
## 1. 3mv2 2. 100 m/sec.
2
3. 5 2 m/sec. 4. m/sec.
3
5. 1/ 2 6. 1 kg
M2v 2
7. 20 N 8.
2m2 g
9. 0.171 kgm/sec. 10. e = tan2
LEVEL II
Mm
1. v 2. {M/(M+m)}2L
k(M m)
m
3. Straight line 4. (R r )
M m
Mmv 1 Mm 2 v 2
5. 6.
(M m) 2 (m 2M)((m M)
8. 25 %
v2 M
9. cos-1 (1e2n) 10.
2g M m
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LEVEL - I
1. (B) 2. (D)
3. (D) 4. (A)
5. (D) 6. (D)
7. (B) 8. (C)
LEVEL - II
COMPREHENSION
1. (B) 2. (B)
3. (A) 4. (A)
5. (C) 6. (B)
## MATCH THE FOLLOWING
8. (A) (p); (B) (r); (C) (p); (D) (p)
9. (A) (p); (B) (p); (C) (r); (D) (s)
10. (A) (s); (B) (s); (C) (p), (q), (r); (D) (p), (q), (r)
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ROTATIONAL MECHANICS
7. Assignment (Subjective Problems)
LEVEL I
## 1. A ring having mass M, radius R is kept on F
horizontal surface as shown in the figure. Find
the minimum value of co-efficient of friction so
that ring will not slip. Also find
(a) the direction and the magnitude friction acting on the ring.
(b) acceleration of ring.
(c) angular acceleration of ring.
2. A disc is rotating about one of its diameters with a kinetic energy E. If the mass
and the radius of the disc are m and r respectively, find its angular momentum.
3. A solid uniform disk of mass m and radius R is pivoted about a horizontal axis
tangential to the rim of disc. A particle of mass m is attached to a point on the
rim of disk, diametrically opposite to the pivot. The combination is now released
from rest, with the plane of disc initially horizontal. Find the angular velocity
when the small particle reaches its lowest point.
## 4. The flywheel of a gasoline engine is required to give up 300 J of kinetic energy
while its angular velocity decreases from 600 rev min-1 to 540 rev. min-1. What
is the moment of inertia of the flywheel ?
5. A cord, with one end fixed to a horizontal ceiling, is wrapped over a flywheel of
radius 'r'. The wheel is allowed to fall. Find the angular acceleration of the
wheel and the tension in the cord.
6. A uniform disc of radius r, and mass `M kg can rotate without friction about a
fixed vertical axis passing through its center and perpendicular to its plane. A
cord is wound at the rim of the disc and a uniform force of F Newton is applied
on the cord. Find the tangential acceleration of a point on the rim of the disc.
7. A ball is thrown in such a way that it slides with a speed v0 initially without rolling
on a rough horizontal plane. Prove that it will roll without sliding when its speed
5
falls to v 0 .
7
8. A disc of mass m, radius r being wrapped over by a light and inextensible string
is pulled by force F at the free end of the string. If it moves on a smooth
horizontal surface, find (a) linear (b) angular acceleration of the disc.
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## 9. Show that a cylinder will slip on an inclined plane of inclination if the
1
coefficient of static friction between plane and cylinder is less than tan .
3
## 10. A uniform rod of mass M and length a lies M
on a smooth horizontal plane. A particle of
A
mass m moving at a speed v A a V
perpendicular to the length of the rod a
4
strikes it at a distance a/4 from the centre m v m
and stops after the collision. Find
Before Collision After Collision
(a) the velocity of the centre of the rod
and
(b) the angular velocity of the rod about its
centre just after the collision.
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LEVEL - II
1. A boy stands on a freely rotating platform with his arms stretched. His rotation
speed is 0.25 rev./s. But when he draws them in, his speed is 0.80 rev./s. Find
(a) the ratio of his moment of inertia in the first case to that in the second.
(b) the ratio of K.E. in the first case to that in the second.
## 2. A sphere of mass m and radius R rolls without sliding on a horizontal surface. It
collides with a light spring of stiffness K with a kinetic energy E. If the surface
(AB) under the spring is smooth, find the maximum compression of the spring.
v
A B
## 3. A uniform rod of mass m & length 0 is rotating
with a constant angular speed about a vertical
axis passing through its point of suspension. Find
the moment of inertia of the rod about the axis of
rotation if it makes an angle to the vertical (axis
of rotation) .
## 4. Where should a spherical shell placed
h m
on a smooth horizontal surface (shown R
in the figure) be hit by a cue that it will R
roll without sliding ?
## 5. In the figure shown two particles m & M are
interconnected by an inextensible and light
l m
string. M is in equilibrium due to revolution
of particle m. Now M is pulled down slowly
through a distance l/2. Find the change in
angular speed of particle m.
M
## 6. A solid sphere is projected up along an inclined plane of inclination =300 with a
speed v = 2m/sec. If it rolls without slipping, find the maximum distance
traversed by it (g = 10 m/sec -2).
v
m
7. A bullet of mass m collides inelastically at the
R
periphery of a disc of mass M and radius R, with
a speed v. The disc rotates about a fixed O
M
horizontal axis. Find the angular velocity of the
disc bullet system just after the impact.
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## 8. Two heavy metallic plates are joined together at A
900 to each other. A laminar sheet of mass Q
30Kg is hinged at the line AB joining the two
heavy metallic plates. The hinges are P B
frictionless. The moment of inertia of the
laminar sheet about an axis parallel to AB and
passing through its centre of mass is 1.2 Kg-m2.
Two rubber obstacles P and Q are fixed, one on each metallic plates at a
distance 0.5 m from the line AB. This distance is chosen so that the reaction
due to the hinges on the laminar sheet is zero during the impact. Initially the
laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and
turns back. If the impulse on the sheet due to each obstacle is 6N-s,
(a) Find the location of the centre of mass of the laminar sheet from AB.
(b) At what angular velocity does the laminar sheet come back after the first
impact?
(c) After how may impacts, does the laminar sheets come to rest..
9. A boy rolls a hoop over a horizontal path with a speed of 7.2 km/h. Over what
distance can the hoop roll uphill at the expense of its kinetic energy? The slope
of the hill is 1 in 10.
## 10. A solid sphere of radius R is moving on a rough horizontal plane. At certain
instant, it has translational velocity v0 in right direction and an angular velocity
v0
in clockwise sense. When its translational velocity is 0.75 v0. It has a
4R
perfectly elastic collision with a smooth vertical wall which is normal to its path.
Find the speed of the sphere when the sphere starts rolling.
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## 8. Assignment (Objective Problems)
LEVEL I
1. The ratio of the radii of gyration of a circular disc and a circular ring of the same
masses and radii about a tangential axis parallel to the their planes is
(A) 6 : 5 (B) 1 : 2
(C) 5 : 6 (D) none of these
2. A wheel of mass 2 kg having practically all the mass concentrated along the
circumference of a circle of radius 20 cm, is rotating on its axis with an angular
velocity of 100 rad/s. The rotational kinetic energy of the wheel is
(A) 4J (B) 70J
(C) 400 J (D) 800 J
## 3. A rod of length L is hinged from one end. It is brought to a horizontal position
and released. The angular velocity of the rod when it is in vertical position is
2g 3g
(A) (B)
L L
g g
(C) (D)
2L L
4. If a solid sphere, disc and cylinder are allowed to roll down an inclined plane
from the same height
(A) cylinder will reach the bottom first
(B) disc will reach the bottom first
(C) sphere will reach the bottom first
(D) all will reach the bottom at the same time
## angular velocity 0 as shown in the figure. The speed of cylinder
when it starts rolling
3
(A) 5/2 0R (B) 0R
2
5 2
(C) 0R (D) 0R
3 3
6. When there is no external torque acting on a body moving in elliptical path, which of
the following quantities remain constant
(A) kinetic energy (B) potential energy
(C) linear momentum (D) angular momentum
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## 7. A solid homogeneous sphere is moving on a rough horizontal surface, partly rolling
and partly sliding. During this kind of motion of this sphere
(A) total kinetic energy is conserved
(B) angular momentum of the sphere about the point of contact with the plane is
conserved
(C) only the rotational kinetic energy about the centre of mass is conserved.
(D) angular momentum about the centre of mass is conserved.
8. A thin circular ring of mass M and radius R is rotating about its axis with a constant
angular velocity . Two objects, each of mass m are attached gently to the opposite
ends of the diameter of the ring. The wheel now rotates with an angular velocity.
(A) M/(M + m) (B) {(M - 2m)/(M +2m)}
(C) {M/(M + 2m)} (D) {(M + 2m)/M}
9. A sphere moving at some instant with horizontal velocity vc in right and angular
velocity in anti clockwise sense. If v c = R . The instantaneous centre of
rotation is
(A) at the bottom of the sphere (B) at the top of the sphere
(C) at the centre of the sphere (D) any where inside the sphere
10. A thin bar of mass M and length L is free to rotate about a fixed horizontal axis
through a point at its end. The bar is brought to a horizontal position and then
released. The angular velocity when it reaches the lowest point is
(A) directly proportional to its length and inversely proportional to its mass.
(B) independent of mass and inversely proportional to the square root of its length
(C) dependent only upon the acceleration due to gravity.
(D) directly proportional to its length and inversely proportional to the acceleration
due to gravity.
## 11 A triangular plate ABC is free to rotate about two points C
A and B on smooth horizontal floor. A force F is applied
perpendicular to AB so as to rotate the plate about A
A B
and B separately. If 1 & 2 are the corresponding r
accelerations for the cases then 1/ 2 will be
(A) 1
(B) <1
(C) > 1
(D) dependent on the force and the dimensions of the plate.
12. A uniform rod AB of mass m and length at rest on a smooth horizontal surface.
An impulse P is applied to the end B. The time taken by the rod to turn through
a right angle is
(A) m /12P (B) m /6P
(C) m / 6P (D) none of these
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## 13. A thin hollow sphere of mass m is completely filled m
with an ideal liquid of mass m. When the sphere
m
rolls with a velocity v kinetic energy of the system
is equal to
1
(A) mv2 (B) mv2
2
4 4
(C) mv2 (D) mv2
3 5
14. A string is wrapped several times round a solid cylinder and then the end of the
string is held stationary while the cylinder is released from rest with an initial
motion. The acceleration of the cylinder and tension in the string will be
2g mg mg
(A) and (B) g and
3 3 2
g mg g mg
(C) and (D) and
3 2 2 3
## shaped frame at its mid point as shown in the
figure. The moment of inertia of the system about
an axis passing through O perpendicular to the
plane of the frame is equal to
m 2 m 2
(A) (B)
12 3
2
m m 2
(C) sin2 (D) sin
12 3
16. A cubical block of mass M and edge a slides down a rough inclined plane of
inclination with a uniform velocity. The torque of the normal force on the block
about its centre has a magnitude
(A) zero (B) Mga
(C) Mg(a/2)sin (D) Mga cos
17. A string of negligible thickness is wrapped several times around a cylinder kept
on a rough horizontal surface. A man standing at a distance from the cylinder
holds one end of the string and pulls the cylinder towards him. There is no
slipping anywhere. The length of the string passed through the hand of the man
while the cylinder reaches his hands is
(A) (B) 2
(C) 3 (D)
## 18. A uniform circular disc of radius r is placed on a
v0
rough horizontal surface and given a linear
velocity v0 and angular velocity 0 as shown. The 0
## disc comes to rest after moving some distance to
the right. It follows that
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(A) 3 v0 = 2 0 r (B) 2 v0 = 0 r
(C) v0 = 0 r (D) 2 v0 = 3 0 r
19. Two uniform solid spheres having unequal masses and unequal radii are
released from rest from the same height on a rough incline. If the spheres roll
without slipping,
(A) the heavier sphere reaches the bottom first
(B) the bigger sphere reaches the bottom first
(C) the two spheres reach the bottom together
(D) the information given is not sufficient to tell which sphere will reach the
bottom first.
20. A rod of mass m is released on smooth horizontal surface making angle with
horizontal. Then which of the following statement is incorrect.
(A) Acceleration of rod along vertical is less than g.
(B) Acceleration of centre of mass along horizontal is zero.
(C) Angular acceleration of rod is not constant.
(D) Momentum of the rod along vertical will remain constant.
LEVEL II
1. The mathematical statement v vc v ' , where v c is the velocity of centre of mass, v '
is the velocity of the point with respect to the centre of mass and v is the total velocity
of the point with respect to ground
(A) is true for a rolling sphere
(B) is true for a block moving on frictionless horizontal surface
(C) is true for a rolling cylinder
(D) none of these
F3 F2
2. A spool of wire rests on a horizontal surface as shown in
figure. As the wire is pulled, the spool does not slip at
contact point P. On separate trails, each one of the forces F4
F1, F2, F3 and F4 is applied to the spool. For each of these
F1
forces the spool
(A) will rotate anticlockwise if F1 is applied
P
(B) will not rotate if F2 is applied
(C) will rotate anticlockwise if F3 is applied
(D) will rotate clockwise is F4 is applied
## 3. A wheel is rolling on a horizontal plane. At a certain C
instant, it has velocity v and acceleration a of c.m.
as shown in the figure. Acceleration of a
B
(A) A is vertically upwards v
(B) B may be vertically downwards A
(C) C cannot be horizontal
(D) Some point on the rim may be horizontal leftwards.
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## 4. A thin rigid uniform circular disc rolls without slipping Y Translatory
on a horizontal rigid surface (or the ground). At a B C.M. frame
certain instant, its position w.r.t. ground frame is as Y
shown in the figure. v
A X
(A) sector ABC has greater kinetic energy than sector O C
X
(B) sector BO C has greater kinetic energy than D
Ground frame
sector CO D w.r.t ground frame
(C) sector BO C has the same kinetic energy as sector AO B w.r.t. ground frame
(D) all the sectors AO B, BO C, CO D and AO D have same kinetic energy w.r.t. the
centre of mass frame
## 5. A particle moving on a horizontal frictionless plane collides elastically with a uniform
smooth rod lying on the same plane and which is hinged to rotate about a vertical axis
passing through one of its ends. Then,
(A) angular momentum of the system about any point is conserved
(B) linear momentum of the particle is conserved perpendicular to the rod
(C) kinetic energy of the system (rod + particle) remains unchanged after the collision
(D) linear momentum of the system (rod + particle) is conserved
6. In the figure shown, the plank is being pulled to the right with a constant speed v. If the
cylinder does not slip then:
## (A) the speed of the centre of mass of the cylinder is 2v.
(B) the speed of the centre of mass of the cylinder is zero.
(C) the angular velocity of the cylinder is v/R.
(D) the angular velocity of the cylinder is zero.
## 7. If a cylinder is rolling down the incline with sliding.
(A) after some time it may start pure rolling
(B) after sometime it will start pure rolling
(C) it may be possible that it will never start pure rolling
(D) none of these
## 8. Which of the following statements are correct.
(A) friction acting on a cylinder without sliding on an inclined surface is always upward
along the incline irrespective of any external force acting on it.
(B) friction acting on a cylinder without sliding on an inclined surface is may be upward
may be downwards depending on the external force acting on it.
(C) friction acting on a cylinder rolling without sliding may be zero depending on the
external force acting on it.
(D) nothing can be said exactly about it as it depends on the friction coefficient on
inclined plane.
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9. A paritcle falls freely near the surface of the earth. Consider a fixed point O (not
vertically below the particle) on the ground.
(A) Angular momentum of the particle about O is increasing .
(B) Torque of the gravitational force on the particle about O is decreasing.
(C) The moment of inertia of the particle about O is decreasing .
(D) The angular velocity of the particle about O is increasing.
10. A body is in equilibrium under the influence of a number of forces. Each force has a
different line of action. The minimum number of forces required is
(A) 2, if their lines of action pass through the centre of mass of the body.
(B) 3, if their lines of action are not parallel.
(C) 3, if their lines of action are parallel.
(D) 4, if their lines of action are parallel and all the forces have the same magnitude.
COMPREHENSION
## A circular ring having mass M = 4 m and radius y
A
R is attached to a small smooth ring of mass m
B
kept at point A. The ring is threaded onto a x
horizontal fixed frictionless wire.
Initially the bigger ring is held horizontally alongside the wire in such a manner that
initially the planes of both the rings are at 90 to each other and the planes of smaller
and bigger rings are perpendicular to the plane of the paper (x-y plane). Now the bigger
ring is released from rest. After release the planes of smaller and bigger ranges remain
perpendicular to the plane of paper (x-y plane).
5 g g
(A) (B) 2
6 3R 3R
g g
(C) 10 (D)
85R 3R
## 2. The speed of small ring at this instant is
5 gR
(A) gR (B) 8
3 85
20 4 gR
(C) gR (D)
15 3 3
3. Speed of point B is
gR gR
(A) (B) 2
3 3
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gR gR
(C) 4 (D) 12
3 85
## The figure shows a uniform smooth solid A
4m
cylinder A of radius 4 m rolling without
slipping on the 8 kg plank which in turn is
8 kg
supported by a fixed smooth surface. Block
B, is known to accelerate down with 6 m/s 2.
(Take g = 10 m/s2) B
2 kg
2
6 m/s
## 4. What is the angular acceleration of the cylinder ?
4 2 6 2
5 5
5. What is the ratio of the mass of the cylinder to the mass of block B ?
(A) 1 (B) 2
(C) 3 (D) 4.
6. If unwrapped length of the thread between the cylinder and block B is 20 m at the
beginning, when the system was released from rest, what would it be 2 s later ?
(A) 28 m (B) 30 m
(C) 22 m (D) 32.5 m
## MATCH THE FOLLOWING
C
1. A rigid body is rolling without slipping on the horizontal
surface :
B V
60
D
A
Column A Column B
(A) Velocity at point A i.e., V A (p) V 2
(B) Velocity at point B i.e., V B (q) Zero
(C) Velocity at point C i.e., VC (r) V
(D) Velocity at point D i.e., VD (s) 2V
2. A uniform disc is acted upon by some forces and it rolls on a horizontal plank without
slipping from north to south. The plank, in turn lies on a smooth horizontal surface.
Match the following regarding this situation :
Column II
Column I
(A) Frictional force on the disc by the (p) May be directed towards north
surface
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(B) Velocity of the lowermost point of the (q) May be directed towards south
disc
(C) Acceleration of centre of mass of the (r) May be zero
disc
(D) Vertical component of the acceleration (s) Must be zero
of centre of mass
3. In each case, there is sufficient friction for regular rigid uniform body to undergo pure
rolling on a rigid horizontal surface. Now Match the Column I and II
Column II
Column I
(A) (p) The direction of static friction may be
h F forward or may be backward or static
R friction may be zero
disc
(B) (q) The direction of static friction is
F towards backward
R
disc
(C) (r) The angular acceleration will be
R
clockwise
h
F
disc
(D) F
(s) Acceleration of the centre mass will
h be along direction F
disc
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9. HINTS (Subjective)
LEVEL- I
2. L=I
5. Draw FBD.
## 7. Draw a FBD, assume friction, pure rolling starts when v = R
8. Draw FBD, apply Newton 2nd law, find torque and relate it to angular
acceleration.
9. Draw FBD on the inclined plane, assume backward friction apply 2nd law.
LEVEL- II
## 1. Conserve angular momentum.
2. KEsphere = + PEspring = 0
md
3. I = r2 dm where r = sin , dm = .
0
6. KE + PE = 0
## 8. Moment of impulse = charge in angular momentum
9. KE + PE = 0
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## 10. Answers to the Subjective Assignment
LEVEL- I
F F
1. opposite to the applied force , clockwise sense.
2M 2MR
mE
2. .R
2
12g
3. 4. 0.81 kg-m2
11r
g mg 2F
5. , 6.
2r 2 M
F 2F mv 3mv
8. (a) (b) 10. (a) V = (b) =
m mR M Ma
LEVEL II
10E
1. (a) 16/5 (b) 5/16 2.
7k
m 20 sin2 2
3. 4. R
3 3
Mg
5. 3 6. 0.56m
m
v 12 mv
7. 8. 0,
M (M 6m)
R 1
2m
9. 4m
v0
10.
28
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LEVEL - I
1. (C) 2. (C)
3. (B) 4. (C)
5. (C) 6. (D)
7. (B) 8. (C)
LEVEL - II
## 1. (A), (B), (C) 2. (B), (C)
3. (A), (B), (C), (D) 4. (A), (B), (C), (D)
5. (C) 6. (B), (C)
7. (A), (C) 8. (B), (C)
9. (A), (C), (D) 10. (B), (C), (D)
COMPREHENSION
1. (C) 2. (B)
3. (D) 4. (D)
5. (B) 6. (A)
## 1. (A) (q); (B) (p); (C) (s); (D) (r);
2. (A) (p), (q), (r); (B) (p), (q), (r); (C) (p), (q), (r); (D) (s)
3. (A) (p), (s), (r); (B) (q), (r), (s); (C) (q), (r), (s); (D) (p), (r), (s)
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GRAVITATION
10. Assignment (Subjective Problems)
LEVEL I
1. Three identical bodies, each of mass m, are separated by a distance a and are
found to start moving towards one another under mutual force of gravitational
attraction. If at t = t0, the separation between any two is a/2, find the speed of
each body at time t0.
2. If the radius and density of a planet are two times and half respectively of those
of earth, find the intensity of gravitational field at planet surface and escape
velocity from planet.
3. A mass M is split into two parts m and (M m), which are the separated by a
m
certain distance. What ratio of maximizes the gravitational force between
M
the parts?
4. Four massive particles, each of mass m, are kept at the vertices of square of side
. With what speed should the system rotate in its plane about its centre so as to
remain stable?
5. Two concentric spherical shells have masses M1, M2 and radii R1, R2 (R1< R2).
What is the force exerted by this system on a particle of mass m if it is placed at
a distance (R1 + R2)/2 from the centre?
## 6. A body stretches a spring by a particular length at the earth's surface at the
equator. At what height above the South Pole will it stretch the same spring to
the same length? Assume the earth to be spherical.
7. Find the radius of the circular orbit of a satellite moving with an angular speed
equal to the angular speed of earth's rotation.
8. What is the true weight of an object, that weighed exactly 10.0 N at the north
pole, at the position of a geostationary satellite?
9. What should be the period of rotation of the earth so that every object on the
equator is weightless?
10. A particle is fired vertically upward with a speed 15 km/s. (a) Show that it will
escape from the earth and (b) With what speed will it move in interstellar space?
Assume the presence of the earths gravitational field only.
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LEVEL II
1. With what speed should a satellite be projected from earths surface so that it
starts resolving around earth at a height of 2600 km in circular orbit? (Radius of
earth = 6400 km, g at surface = 9.8 m/sec2)
2. A thin spherical shell of radius 3R and mass M and a hollow sphere of mass 3M
with R and 2R as internal and external radii are placed concentrically at O. Find
5R
the gravitational field & the gravitational potential at Q where OP = and
2
OQ = 4R.
3. Two small dense stars rotate about their common centre of mass, as a binary
system with the period of 1 year for each. One star is of double the mass of the
other and the mass of the lighter one is 1/3 the mass of the sun. The distance
between the earth and the sun is R. If the distance between two stars is r, then
obtain the relation between r and R.
## 4. What is the magnitude of the m M
d
gravitational force on the particle of L
mass m due to the rod?
5. A system consists of a thin ring of radius r and of mass M and a straight wire of
linear mass density of infinite length placed along the axis of the ring with one
of its ends at the centre of the ring. Find the force of interaction between the wire
and the ring.
6. Two massive particles of mass m1 and m2 are released from rest from a very
large distance. Find the speeds of the particles when their distance of separation
is r.
7. A particle of mass m is kept on the axis of a fixed circular ring of mass M and
radius R at a distance x from the centre of the ring. Find the maximum
gravitational force between the ring and the particle.
8. A double-star, with two stars masses m1 and m2, rotates with constant angular
speed. If the maximum distance of separation is R, then find the minimum value
of angular speed.
9. A projectile is fired vertically upward from the surface of earth with a velocity Kve
where ve is escape velocity and K<1. Neglecting air resistance, show that the
maximum height to which it will rise, measured from the centre of earth, is
R
where R is the radius of earth.
1 K2
10. A planet of mass m moves along an elliptical orbit around the sun so that its
maximum and minimum distances from the sun (mass = M) are equal to r1 and r2
respectively. Find the angular momentum of this planet relative to the sun.
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LEVEL I
## 1. The ratio of the inertial mass to gravitational mass is equal to
(A) 0.5 (B) 1
(C) 2 (D) no fixed number.
2. If the radius of earth were to shrink by one percent, its mass remaining the same,
the acceleration due to gravity on the earths surface would
(A) decrease (B) remains unchanged
(C) increase. (D) none of these
## 3. Keplers third law of planetary motion provides information about
(A) areal velocity of a planet (B) nature of motion of a planet
(C) ratio of time periods of two planets (D) all the above
## 4. If a particle is slowly brought from reference point to another point P in a
gravitational field, then work done per unit mass by the external agent is (at that
point)
(A) gravitational force (B) gravitational field intensity
(C) gravitation potential (D) none of the above
## 5. The atmosphere is held to the earth by
(A) winds (B) gravity
(C) clouds (D) none of the above.
## 6. The time period of simple pendulum of infinite length is
R
(A) infinite (B) 2
g
g 1 R
(C) 2 (D)
R 2 g
## 7. When a satellite has an elliptical orbit, the plane of the orbit
(A) sometimes passes through the centre of earth
(B) does not pass through the centre of earth
(C) passes through the centre of earth always
(D) none of the above.
8. The earth revolves round the sun in an elliptical orbit. Its speed is
(A) going on decreasing continuously
(B) greatest when it is closest to the sun
(C) greatest when it is farthest from the sun
(D) constant at all the points on the orbit.
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9. Two satellites of masses m1 and m2 (m1 > m2) are revolving round the earth in
circular orbits of radii r1 and r2 (r1 > r2) respectively. Which of the following
statements is true regarding their speeds v1 and v2 ?
(A) v1 = v2 (B) v1 < v2
(C) v1 > v2 (D) (v1/r1) = (v2/r2).
10. Two satellites are orbiting around the earth in circular orbits of same radius. One
of them is 10 times greater in mass than the other. Their period of revolutions are
in the ratio
(A) 100:1 (B) 1:100
(C) 10:1 (D) 1:1
11. A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at
rest but it moves at a speed of 2 m/s as it reaches A. The work done by a person
on the mass is 3J. The potential at A is:
(A) -3 J/kg (B) -2 J/kg
(C) -5 J/kg (D) none of these.
12. Let V and E be the gravitational potential and gravitational field at a distance r
from the centre of a uniform spherical shell. Consider the following two
statements, (A) The plot of V against r is discontinuous and (B) The plot of E
against r is discontinuous.
(A) Both A and B are correct (B) A is correct but B is wrong
(C) B is correct but A is wrong (D) both A and B are wrong.
13. Two satellites A and B move round the earth in the same orbit. The mass of B is
twice the mass of A. Which of the following is correct?
(A) Speeds of A and B are equal
(B) The potential energy of earth + A is same as that of earth + B
(C) The kinetic energy of A and B are equal
(D) The total energy of earth + A is same as that of earth + B.
14. The minimum speed of a particle projected from earths surface so that it will
never return is/are
GM
(A) (B) 22.1 km/sec
R
(C) 4g0R (D) none of above
## 15. A body of mass m is approaching towards the centre of a hypothetical hollow
planet of mass M and radius R. The speed of the body when it passes the centre
of the planet through a diametrical tunnel is
GM 2GM
(A) (B)
R R
(C) Zero (D) none of these.
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16. The energy required to remove a body of mass m from earths surface is/are
equal t
GMm
(A) (B) mgR
R
(C) mgR (D) none of these.
17. A small mass m is moved slowly from the surface of earth to a height h from
the surface. The work done (by external agent) in doing this is
(A) mgh, for all values of h. (B) mgh, for h << R.
1 1
(C) mgR for h = R (D) mgR , for h = R
2 2
18. The escape velocity of a particle of mass m varies as:
(A) m2 (B) m
(C) m0 (D) m-1.
19. Two particles of masses m1 & m2 are infinitely separated and their gravitational
potential energy is chosen zero. Their gravitational energy, when they are
separated by r, is
Gm1m2 Gm1m2
(A) 2
(B)
r r
Gm1m2 Gm1m2
(C) 2
(D)
r r
## 20. The gravitational field at a point on the axis of a uniform disc,
(mass = M, radius = a) forming at the point, is
GM Gm
(A) cos (B) 1 cos
a a
2GM GM
(C) 2
1 cos (D) 2 1 sin
a a
LEVEL II
## 1. A particle of mass m and charge q is projected vertically upwards. A uniform electric
field E is applied vertically downwards. If the total potential energy is U (gravitational
plus electrostatic) and height is h (<< radius of earth) (Assume U to be zero on
earths surface) then :
(A) work done by the gravity mgh
(B) work done by electrostatic force qEh
(C) graph potential energy versus height (h) is straight line
(D) graph of potential energy versus height (h) is parabola.
2. A body of mass m is projected vertically upwards with velocity V from earths surface
and attains height h, then
(A) if V = V e (where Ve is escape velocity) then h =
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V2
(B) if V is very small then h
2g
mgR
(C) if h = R, then projection kinetic energy is
2
(D) if h = R then total energy is negative.
## 3. A, B, C are vertices of an equilateral triangle imagined in free A
space (no gravity). First a mass m is brought from infinity to m
point A and the work done in doing so is W A. Then keeping m
fixed at A, another identical mass m is brought from infinity to
point B. The work done is W B. The work done by the another
identical mass m brought from infinity to C is W C keeping A and
B fixed. Then
B C
(A) W A = 0 (B) W A < W B
(C) WC > W B > W A (D) W A > W B > W C.
## 4. A particle is dropped from a height equal to the radius of the
earth above the tunnel dug through the earth as shown in the R
figure.
M : Mass of earth.
(A) Particle will oscillate through the earth to a height R on
both sides C
(B) Particle will execute simple harmonic motion
(C) Motion of the particle is periodic
2GM
(D) Particle passes the centre of earth with a speed =
R
## 5. Inside a uniform sphere of mass M and radius R, a cavity A
(A) Gravitational field inside the cavity is uniform.
(B) Gravitational field inside the cavity is nonuniform. R
(C) The escape velocity of a particle projected from point B
88GM R/3
A is .
45R C
(D) Escape velocity is defined for earth and particle system only.
6. Assuming the earth to be a sphere of uniform density the acceleration due to gravity
(A) at a point outside the earth is inversely proportional to the square of its distance from
the centre
(B) at a point outside the earth is inversely proportional to its distance from the centre
(C) at a point inside is zero
(D) at a point inside is proportional to its distance from the centre.
## 7. Gravitational potential at the centre of curvature of a hemispherical bowl of radius R and
mass M is V.
(A) gravitational potential at the centre of curvature of a thin uniform wire of mass M,
bent into a semicircle of radius R, is also equal to V.
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(B) In part (A) if the same wire is bent into a quarter of a circle then also the gravitational
potential at the centre of curvature will be V.
(C) In part (A) if the same wire mass is nonuniformly distributed along its length and it is
bent into a semicircle of radius R, gravitational potential at the centre is V.
(D) none of these
8. In a solid sphere two small symmetrical cavities are created whose centres lie on a
diameter AB of sphere on opposite sides of the centre.
(A) The gravitational field at the centre of the sphere is zero.
(B) The gravitational potential at the centre remains unaffected if cavitiesare not present
(C) A circle at which all points have same potential is in the plane of diameter AB.
(D) A circle at which all points have same potential is in the plane perpendicular to
the diameter AB.
## 9. A communications Earth satellite
(A) goes round the earth from east to west
(B) can be in the equatorial plane only
(C) can be vertically above any place on the earth
(D) goes round the earth from west to east
## 10. If a satellite orbits as close to the earth's surface as possible,
(A) its speed is maximum
(B) time period of its rotation is minimum
(C) the total energy of the earth plus satellite system is minimum
(D) the total energy of the earth plus satellitesystem is maximum
COMPREHENSION
I. The escape velocity is the least velocity required by a body to escape away from the
gravitational pull of the earth. In the escaping condition kinetic energy is equal to
potential energy, i.e., total energy of the body is zero. The escape velocity is
independent of angle of projection. Escape velocity depends on the mass of central
body as well as radius of central body. When velocity of orbiting body increases, its
kinetic energy increases and hence total energy.
1. A projectile is fired with a velocity less than escape velocity. Then sum of its kinetic and
potential energy is
(A) negative (B) positive
(C) zero (D) may be positive or zero.
## 2. As the radius of orbiting body decreases, its time period
(A) increases (B) decreases
(C) remains unchanged (D) none of these.
3. For the planets orbiting around the sun, the quantity which remains constant is
(A) linear speed (B) kinetic energy
(C) angular speed (D) angular momentum.
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## II. The gravitational field strength E and gravitational y
potential V are related as
V V V
E i j k
x y z
In the figure, transversal lines represent
x
equipotential surfaces. A particle of mass m is O 10cm 20cm 30cm
released from rest at the origin. The gravitational -30V -40V
-10V -20V
unit of potential, 1V 1cm 2 / s 2
## 4. y-component of E at the point whose co-ordinates are (4cm, 4cm) is
(A) 1 cm/s2 (B) 8 cm/s2
2
(C) 4 cm/s (D) none of these
## 5. Speed of the particle (v) (y is in cm and v is in cm/s) as function of its y-co-ordinate is
(A) v 2 y (B) v 2.y
(C) v = 2y (D) v 2y 4y2 2y 2
## 6. A particle of mass m (where m is the mass in y
gram) is released from point O with initial velocity
zero. The work done by gravitational force on the
particle, when it is taken from O to B, is 2 2
(A) 30 m, unit 1V=1cm /s
B
(B) 15 m, unit 90
(C) 30 m, unit
x
(D) + 15 m, unit O 10cm 20cm 30cm
-30V -40V
-10V -20V
## MATCH THE FOLLOWING
I.
Column A Column B
(A) Gravitational potential (p) On the surface of planets with
density ratio 1 : 2
(B) Escape velocity (q) Conservation of angular momentum
(C) Ratio of the acceleration due to (r) Varies with the reference point
gravity 1 : 2
(D) Orbiting satellites (s) Does not depend on the angle
II.
Column A Column B
(A) When v < v0 (p) The path of satellite is hyperbolic
(B) When v = vesc 2 v0 (q) The satellite will strike the earth
(C) When v > vesc (r) The orbit of satellite is elliptical
(D) vesc > v > v0 (s) The path of satellite is parabolic
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## 12. Hints (Subjective)
LEVEL- I
1. Apply COE
4
2. g = g GR ,
3
Ve = 2gM/R
G M1 m
5. The force exerted on m by M1 = 2
R1 R 2
2
2GM
7. r=
r2
g R2
8. g= 02
r
LEVEL- II
1. Apply COE
6. |m1 v1 + m2 v 2 | = 0 ; K1 + U1 = K2 + U2
7. Each portion of the ring having a mass m attracts the particle towards itself with
G( m)m
a force F given as F =
r2
8. Gravitational force provides necessary centripetal acceleration to keep the stars
in their respective orbits.
## 10. Use conservation of angular momentum and conservation of energy.
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## 13. Answers to the Subjective Assignment
LEVEL I
1. 2Gm / a 2. g, 15.84 km/sec
m 1 Gm 1
3. 4. 1
M 2 2 2
4GM1m
5. 6. 10.83 km
(R1 R2 )2
7. 42297 km 8. 0.23 N
9. 84 minute 10. (b) 9.98 km/s
LEVEL II
12GM GM
1. 8.99 103 km/sec 2. ,
25R2 R
GMm
3. r=R 4.
d L d
GM
5.
r
2G 2G 2 GMm
6. m2 , m1 . 7. (Fx)max=
m1 m2 r (m1 m2 )r 3 3 R2
G(m1 m2 ) 2 GM r1r2
8. = 10. m
R3 (r1 r2 )
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LEVEL I
1. B 2. C
3. C 4. C
5. B 6. B
7. C 8. B
9. B 10. D
11. C 12. C
13. A 14. D
15. B 16. B
17. C 18. C
19. D 20. C
LEVEL II
COMPREHENSION
1. (A) 2. (B)
3. (D) 4. (A)
5. (A) 6. (A)
## 2. (A) (r), (q); (B) (s); (C) (p); (D) (r)
***
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FLUID MECHANICS
6. Assignment (Subjective Problems)
LEVEL I
1. There are three different liquids (liquid 1, liquid 2 and liquid 3) having density
13.6 gm/cc., 1.3 gm/cc and 0.8 gm/cc. Liquid 1 is poured in a uniform U-tube,
which is kept vertical. Liquid 2 & liquid 3 are poured separately into the two arms
till upper surface of liquid 2 & 3 are same. What is the height of liquid 3 if the
height of liquid 2 is 16cm?
2. A very narrow hole exists at a height (H h) from the bottom of a tank, which
contains water upto a height H. Find the distance where stream of water coming
out from the hole will strike the floor.
3. Two cylindrical vessels of radius r = 100m are filled with water upto heights H
and 2H respectively. If the vessels are connected by a narrow tube through two
holes made at their bottom, find the work done by gravity.
## 4. A garden hose having an internal diameter of 0.75 in. is connected to a lawn
sprinkler that consists merely of an enclosure with 24 holes, each 0.050 in. in
diameter. If the water in the hose has a speed of 3.0 ft/sec. at what speed does it
leave the sprinkler holes?
## 5. A cube floating on mercury has one-fourth of its volume submerged. If enough
water is added to cover the cube, what fraction of its volume will remain
immersed in mercury? Does the answer depend on the shape of the body?
## 6. A seated tank containing a liquid of density C
A a
moves with a horizontal acceleration a as h
shown in the figure. Find the difference in
B
pressure between the points A and B.
7. Find the work done in increasing the radius of a soap bubble from initial radius r1
to final radius r2. Given T = surface tension of soap solution.
8. If n identical water droplets falling under gravity with terminal velocity v coalesce
to form a single drop which has the terminal velocity 4v. Find the number n.
9. A U-tube is partially filled with a liquid. The horizontal part of the tube is 2 m.
The tube is accelerated horizontally with a constant acceleration of 5 m/s2. What
is the difference in the heights of the liquid in the two arms of the U-tube?
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10. If the velocity gradient of water near the surface of a deep river is 6 s1, find the
shearing stress between the horizontal layers of water (coefficient of viscosity of
water = 10 2 poise)
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LEVEL II
## 1. A silver block of mass 3.15kg is connected to a string and is then immersed in a
liquid of relative density 0.82. Find the tension in the string, if relative density of
silver is 10.5.
## 2. Water is flowing continuously from a tap having a bore of internal diameter
8 10-3 m. Calculate the diameter of the water stream at a distance 2 10-1 m
below the tap. Assume that water velocity as it leaves the tap is 4 10-1 m/s.
3. There are two identical small holes (cross section area = ) opposite sides of a
tank containing a liquid. (density ) The tank is opened at the top. The difference
is height between the two holes is h. As the liquid comes out of the two holes,
find the net horizontal force. The tank will experience.
## 4. Two cylindrical tanks of cross-sectional area
A1 and A2 with their bases at the same level A1 A2
each contain a liquid of density . The height
of liquid in the tanks are H1 and H2, H1
respectively. The tanks are joined together H2
through a pipe of cross-sectional area a as
shown in the figure.
(a) Find the time taken to equalize the a
levels in the tanks.
(b) What is the work done by gravity in
equalizing the levels?
## 5. A long cylindrical tank of cross-section area
20kg
0.5m2 is filled with water. It has an opening at a
height 50 cm from the bottom, having area of
cross-section 1 10-4 m2 . A movable piston of
cross-section area almost equal to 0.5 m2 is fitted
on the top of the tank such that it can slide in the
tank freely. A load of 20 kg is applied on the top
of the water by piston, as shown in the figure.
Find the speed of the water jet with which it hits
the surface when piston is 1m above the bottom
(Ignore the mass of the piston).
6. The vertical arms of U-tube have unequal radius R and r (R > r). If a liquid of
surface tension T and density rests in equilibrium inside the U-tube, find the
level difference h between the meniscuses in the two arms.
7. A small hollow vessel which has a small hole in it is immersed in water to a depth
of 40 cm before water enters into the vessel. Calculate the radius of the hole.
[Surface tension of water = 70 10 3 N/m, density of water = 103 kg/m3]
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## 8. A rectangular tank of height 10 m filled with
water, is placed near the bottom of an incline
10 m x
of angle 300. At height x from bottom a small
hole is made (as shown in figure) such that 30
0
## the stream coming out from hole, strikes the
inclined plane normally. Find x.
9. Water flows through a tapering horizontal tube of radii of cross section of the
ends r1 = 20 cm and r2 = 10 cm. The velocity of water at the points for the radius
of cross section r1 is v1 = 1m/sec. Find the force imparted by the emerging water
at the other end of the tube.
10. Two soap bubbles of radii a and b combine under isothermal conditions to form a
single bubble of radius c without any leakage of air. If P o = atmospheric pressure and
4T a 2 b 2 c 2
T = surface tension of soap solution, show that Po =
c 3 a3 b3
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## 7. Assignment (Objective Problems)
LEVEL - I
1. A and B are two metallic pieces. They are fully immersed in water and then
weighed. Now they show same loss of weight. The conclusion therefore is:
(A) A and B have same weight in air
(B) A and B have equal volumes
(C) The densities of the materials of A and B are the same
(D) A and B are immersed to the same depth inside water.
2. An ice cube contains a large air bubble. The cube is floating on the surface of
water contained on a trough. What will happen to the water level, when the cube
melts?
(A) It will remain unchanged (B) It will fall
(C) It will rise (D) First it will and then rise
3. In a hydraulic lift, used at a service station the radius of the large and small
piston are in the ratio of 20 : 1. What weight placed on the small piston will be
sufficient to lift a car of mass 1500kg?
(A) 3.75Kg (B) 37.5Kg
(C) 7.5Kg (D) 75Kg
4. Water and mercury are filled in two cylindrical vessels upto same height. Both
vessels have a hole in the wall near the bottom. The velocity of water and
mercury coming out of the holes are v1 and v2 respectively. Thus
(A) v1 = v2 (B) v1 = 13.6v2
(C) v1 = v2/13.6 (D) v 1 = (13.6 )v 2
5. An ice cube contains a glass ball. The cube is floating on the surface of water
contained in a trough on the surface of water contained in a trough. What will
happen to the water level, when the cube melts?
(A) It will remain unchanged (B) It will fall
(C) It will rise (D) First it will fall and then rise
## 6. A square hole of side length is made at a depth of y and a circular hole is
made at a depth of 4y from the surface of water in a water tank kept on a
horizontal surface. If equal amount of water comes out of the vessel through the
holes per second then the radius of the circular hole is equal to(r, << y) :
(A) / 2 (B) / 2
(C) / (D) / 2
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## 7. In the figure shown a liquid is flowing through a P vP
tube at the rate of 0.1 m3/sec. The tube is
branched into two semi circular tubes of cross
sectional area A/3 and 2A/3. The velocity of
liquid at Q is (the cross-section of the main tube Q vQ
(A) = 10 -2 m2 and VP = 20 m/sec.):
(A) 5 m/sec (B) 30 m/sec
(C) 35 m/sec (D) None of these.
## 8. A small hole is made at a height of h = (1/ 2 ) m from the
bottom of a cylindrical water tank and at a depth of h = 2
m from the upper level of water in the tank. The distance,
where the water emerging from the hole strikes the ground
is:
(A) 2 2m (B) 1 m R
(C) 2 m (D) None of these.
9. The excess pressure inside one soap bubble is three time that inside a second
soap bubble. The ratio of the volumes of the two bubbles
(A) 1/9 (B) 9/1
(C) 1/27 (D) 27/1
## 10. An air bubble of diameter 2 mm rises steadily through a solution of density
1750 kg/m3 at the rate of .35 cm/sec. Coefficient of viscosity of the solution is
(Assume mass of the bubble to be negligible)
(A) 9 poise (B) 6 poise
(C) 11 poise (D) 4 poise
A
11. The velocity of the water flowing from the inlet pipe is less
than the velocity of water flowing out from the spin pipe B.
(A) variation of water level in vessel will be irregular. B
## (B) water level will remains constant.
(C) the water level will perform periodic oscillation motions.
(D) none of the above.
## 12. In a streamline flow of a liquid
(A) every particle has its own velocity, different from others.
(B) all particles move with a constant velocity, even if the path is curvilinear.
(C) At a point on the streamline, particle can have two velocities.
(D) At a point on the streamline, particle can have only one velocity along the
tangent.
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13. A metallic sphere floats in an immiscible mixture of water ( w = 103 kg/m3) and a
liquid ( L = 13. 5 103 kg/m3) such that 4/5 portion is in water and (1/5)th portion
is in the liquid. The density of the metal in kg/m3 is
(A) 4.5 103 (B) 4.0 103
(C) 3.5 10 3 (D) 3.0 103
14. A stream line body with relative density d1 falls into air from a height h1 on the
surface of a liquid of relative density d2, where d2 is greater than d1. The time of
immersion of the body into the liquid will be
2h1 d1 2h1
(A) (B)
g d2 d1 g
2h1 d1 2h1 d2
(C) (D)
g d2 g d1
15. A tank is filled with water to a height H. Two holes are made on its side wall, one
at a height of h from the bottom and other at a depth h from the top. The
horizontal jets starting from the two holes meet the ground or side (in level with
the bottom of the tank) at the same point. This distance of this point from the
side of the tank is
(A) [4h(H h)] (B) [h(H h)]
## (C) [2h(H h)] (D) [3h(H h)]
16. Which of the following graphs best represents the relation between the height h of
the liquid in a capillary tube and radius of the capillary tube?
h h
(A) (B)
r r
h h
(C) (D)
r r
17. A boat floating in a tank is carrying passengers. If the passengers drink water,
how will it affect the water level of the tank?
(A) It will go down
(B) It will rise
(C) It will remain unchanged
(D)It will depend on atmospheric pressure.
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18. A cylinder is filled with non viscous liquid of density d to a height h0 and a hole is
made at a height h1 from the bottom of the cylinder. The velocity of liquid issuing
out of the hole is
(A) (2gh 0 ) (B) 2g(h0 h1 )
(C) ( dgh1 ) (D) (dgh0 )
19. A spherical ball of mass m and radius r is allowed to fall in a medium of viscosity .
The time in which the velocity of the body increases from zero to 0.63 times the
terminal velocity is called time constant ( ). Dimensionally can be represented by
mr 2 6 mr
(A) (B)
6 g2
m
(C) (D) none of these.
6 r
20. A large bottle is fitted with a capillary siphon. Ratio of times taken to empty the bottle
when it is filled with (i) water (ii) petroleum of relative density 0.8.
( water = 0.001poise, water = 0.002 poise, dwater = 1000 kg/m3)
(A) 5/4 (B) 4/5
(C) 2/5 (D) 3/5
LEVEL - II
C
1. A Siphon tube is used to remove liquid from a container as
shown in figure. In order to operate the Siphon tube it must H
A
initially be filled with the liquid.
h
(A) Speed of the liquid through the Siphon is 2g(h y)
B
(B) Pressure at C is Patm g h H y y
(C) Pressure at PA is less then PD D
(D) Pressure at PA equal to PD
## 2. A circular cylinder of radius R and height H is filled with water
2
to a height H. It starts rotating about its axis with constantly
3
H 2H
increasing angular speed. Choose the correct alternatives. 3
(A) at all speeds, shape of the free surface is parabolloid
(B) the free surface touches first the brim of cylinder and then
the base of the cylinder
(C) the free surface cannot touch the base without spilling water
(D) the free surface touches the brim as well as base at the same instant
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3. The vessel shown in the figure has two sections of areas of cross
section A1 and A2. A liquid of density fills both the sections, up to A1
h
a height h in each. Neglect atmospheric pressure
(A) the pressure at the base of the vessel is 2h g A2 X
(B) the force exerted by the liquid on the base of the vessel is h
2h gA2
(C) the weight of the liquid is < 2h gA2
(D) the walls of the vessel at the level X exert a downward force h g (A2 A1) on the
liquid
4. A small body of density is dropped from rest at a height h into a lake of density ,
where > . Which of the following statement or statements is or are correct if all
dissipative effects are neglected? (neglect viscosity)
(A) the speed of the body just entering the lake is 2gh
(B) the body in the lake experiences upward acceleration equal to {( / ) 1} g
(C) the maximum depth to which the body sinks in the lake is h /( )
(D) the body does not come back to the surface of the lake
## 5. A target object is placed at the bottom of a tank of depth 20 m,
and filled with a liquid of density 500 gm/cc. Another object of
density 100 gm/cc is dropped onto it, from a height of 45 m from 45m
the upper surface of the liquid. (take g = 10 m/s2)
(A) The second object will collide with the target.
(B) The second object does not collide with the target.
(C) The least separation between object and target is 11.25 m.
20m
(D) The retardation of the object in the liquid is 40 ms2.
6 A closed vessel is half filled with water. There is a hole near the top of the vessel and air
is pumped out from this hole.
(a) The water level will rise up in the level
(b) The pressure at the surface of the water will decrease
(c) The force by the water on the bottom of the vessel will decrease
(d) The density of the liquid will decrease
7 In a streamline flow,
(a) the speed of a particle always remains same
(b) the velocity of a particle always remain same
(c) the kinetic energies of all the particle arriving at a given point are the same
(d) the momenta of all the particles arriving at a given point are the same
8 Water is flowing is streamline motion through a tube with its axis horizontal. Consider
two points A and B in the tube at the same horizontal level.
(a) The pressure at A and B are equal for any shape of the tube
(b) The pressures are never equal
(c) The pressure are equal if the tube has a uniform cross-section
(d) The pressure may be equal if tube has non-uniform cross-section
9 There is a small hole near the bottom of an open tank filled with a liquid. The speed of
the water ejected does not depend upon
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## (a) area of the hole (b) density of the liquid
(b) height of the liquid from the hole (d) acceleration due to gravity
10. Water is flowing through a long horizontal tube. Let PA and PB be the pressures at two
points A and B of the tube
(a) PA may be equal to PB
(b) PA may be greater than PB
(c) PA may be smaller than PB
(d) PA = PB only if the cross-sectional area at A and B are equal
COMPREHENSION
## I. Hydrostatic force on a submerged curved surface :
consider first a vertical flat surface of height h and width W.
The pressure at a depth y below the liquid surface
P= gy where : density of liquid
dF= ( gy) (Wdy)
= gW y dy
h
F= dF = gW y dy
0
1
= gWh 2
2
## This result may be interpreted as
h
F = (Wh) g
2
= Projected area pressure at the centroid of the projected area
Projected area can be considered as the image of the surface (flat or curved) on a
screen.
Thus,
FX = Force in X-direction
= Projected area normal to X-direction pressure at the centroid of the projected
area
## 1. For the flat surface discussed in the passage, the total
torque of the hydrostatic force about point O is
2 1
(A) g wh 3 (B) g wh 3
3 3
1
(C) g wh 3 (D) g wh 3
2
2. The line of action of the resultant horizontal force acts at a distance r below point O.
The value of r is
h h
(A) (B)
2 3
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2h h
(C) (D)
3 b
## 3. A flat plate of height 1m and width 1m is hinged at its
middle as shown in the figure. To the left of the plate is a
liquid of density . To prevent the rotation of the plate, a
force F is applied from the other side at the bottom. The
value of F is
g
(A) Zero (B)
2
g g
(C) (D)
4 6
## II. A cylindrical tank having crosssectional area A = 0.5 m2
is filled with two liquids of density 1 = 900 kg m3 and 2 = h
600 kg m3, to a height h = 60 cm each as shown in figure.
A small whole having area a = 5 cm2 is made in right h
vertical wall at a height y = 20 cm from the bottom. A F y
horizontal force F is applied on the tank to keep it in static
equilibrium. The tank is lying on a horizontal surface.
Neglect mass of cylindrical tank comparison to mass of
liquids. (take g = 10 ms-2)
## 4. The velocity of efflux
(A) 10 ms1 (B) 20 ms1
(C) 4 ms1 (D) 35 ms1
## 5. Horizontal force F to keep the cylinder in static equilibrium, if it is placed on a smooth
horizontal plane
(A) 7.2 N (B) 10 N
(C) 15.5 N (D) 20.4 N
6. Minimum and maximum values of F to keep the cylinder in static equilibrium just after
the water starts to spill through the hole. If the co-efficient of static friction between
contact surfaces is 0.01.
(A) 0, 40 N (B) 5.4 N, 52.2 N
(C) 0, 70 N (D) 0, 52.2 N
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## 1. With regard to dependence of quantities given in column a on the quantities
mentioned in column b, match columns a and column b.
column I Column II
(A) Youngs modulus of a substance (p) Depends on area of cross-section
(B) Bulk modulus of a substance (q) Depends on the nature of
material
(C) Modulus of rigidity of a substance (r) Depends on temperature
(D) Volume of a substance (s) Depends on length
2. A cube of ice edge 4 cm is placed in an empty cylindrical glass of inner base area 64
cm2. Assume that ice melts uniformly from each side so that it always remains its cubical
shape then edge of the ice cube is x and height of water formed in cylindrical glass is h
at the instant the ice cube just leaves contact with the bottom of the glass.
(density of ice = 0.9 gm/ml, density of water = 1 gm/ml). Then match the following :
Column II
Column I
(A) X (p) 0.9 cm
(B) H (q) 1 cm
(C) Volume of ice melted (r) 56.7 cm3
(D) Volume of water formed (s) 63 cm3
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8. HINTS (Subjective)
LEVEL- I
## 1. Pressure at A = Pressure at B 3gh + 1 g (H h) = 2gH
1 2
2. Hh= gt t = {2(H h)/g}1/2
2
3. Calculate the change of potential energy of the system.
4. Apply continuity equation.
5. v g = (v/4) g = /4
LEVEL- II
1. T = mg Fb = mg V g
## 2. Applying Bernoulli 's theorem, we get
1 1
v12 + gh = v 22 (pressure being same)
2 2
3. Force = rate of change of momentum
A1
4. dh2 = dh1 (continuity equation )
A2
The difference in levels decreases by
A1 A 2
dh = dh1 + dh2 = dh1
A2
20 10N
5. Pressure at the top is P1 = P0 +
0.5m2
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## 9. Answers to the subjective assignment
LEVEL I
1. 15.4 cm 2. 2 (H h) / h
3. 19.24 J 4. 29 ft/sec.
5. 0.19, NO 6. PB PA = h g = a
7. 8 T(r 22 r 12 ) 8. 8
9. 1 m. 10. 10 3 N/m2
LEVEL II
1. 29.04 N
2. 3.54 10-3 m
3. 2 gh
A1A 2 2(H1 H2 ) 1
4. (a) (b) gA1A 2 (H1 H2 )2
a( A1 A 2 ) g 2
5. 4.56 m/s
2T 1 1
6.
g r R
7. 3.6 10 5 m
8. 8.33 m.
9. 502.65 N.
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LEVEL I
1. B 2. A
3. A 4. A
5. B 6. C
7. D 8. D
9 C 10. A
11. C 12. D
13. C 14. A
15. A 16. C
17. D 18. D
19. C 20. C
LEVEL II
## 1. (A), (B), (D) 2. (A), (B), (C)
3. (A), (B), (C), (D) 4. (B), (C)
5. (B), (C), (D) 6. (B), (C)
7. (C), (D) 8. (C)
9. (A), (B) 10. (A), (B), (C), (D)
COMPREHENSION
1. (B) 2. (C)
3. (D) 4. (C)
5. (A) 6. (D)
## MATCH THE FOLLOWING
1. (A) (q), (r); (B) (q), (r); (C) (q), (r); (D) (p), (r), (s)
## 2. (A) (q); (B) (p); (C) (s); (D) (r)
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## SIMPLE HARMONIC MOTION
14. Assignment (Subjective Problems)
LEVEL I
## 1. The equation of a simple harmonic motion is given by x = 6 sin 10 t + 8 cos 10 t,
where x is in cm, and t is in seconds. Find the resultant amplitude.
## 2. A particle of mass 4 g performs S.H.M. between x = - 10 cm and x = + 10 cm
along x-axis with frequency 60 Hz, initially the particle starts from x = +5 cm. Find
(a) equation of motion of the particle.
(b) the initial phase, amplitude and time period of the particle.
(c) velocity- displacement and acceleration displacement curve of this motion.
(d) plot the graphs of (i) KE vs displacement
(ii) PE vs displacement
(iii) Total energy vs displacement
3. A cubical body (side .1 m and mass 0. 02 kg) floats in water. It is pressed and
then released so that it oscillates vertically. Find the time period. (density of
water = 1000 kg/m3).
m
4. Find the time period of the motion of a particle shown
in figure. Neglecting the small effect of the bend near 10 cm
0
45 60
0
the bottom.
## 5. Consider a situation shown in the figure. Show that if k
2m 3m
the blocks are displaced slightly in opposite directions
and released, they will execute S.H.M. calculate the
time period.
6. A uniform rod of mass m and length is pivoted at one end. It is free to rotate in
a vertical plane. Find the time period of oscillation of rod if it is slightly displaced
from vertical and released.
7. A particle is executing SHM. A and B are the two points at which its velocity is
zero. It passes through a certain point P at intervals of 0.5 and 1.5 sec with a
speed of 3 m /s. Determine the maximum speed and also the ratio AP/PB.
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Q
8. A ball is suspended by a thread of length L at the point O
on the wall PQ which is inclined to the vertical by a small O
## angle to the thread with the ball is now displaced
through a small angle away from the vertical and also
from the wall if the ball is released, find the period of C
P
oscillation of the pendulum when B
(a) <
(b) > . Assume the collision on the wall to be perfectly elastic.
M
9. A small solid cylinder of mass M attached to a horizontal k
R
massless spring can roll without slipping along a
horizontal surface. Show that if the cylinder is displaced
and released, if executes S.H.M. Also find its time
period.
## 10. The friction coefficient between the two blocks shown in
figure is and the horizontal plane is smooth (a) If the k
m
system is slightly displaced and released find the
M
magnitude of the frictional force between the blocks when
the displacement from the mean position is x. (b) what
can be the maximum amplitude if the upper block does not
slip relative to the lower block ?
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LEVEL II
## 1. Determine the period of oscillation of
mercury of mass m = 200g poured into a x
## bent tube whose right arm forms an angle
= 300 with the vertical. The cross-sectional x
area of the tube S = 0.50 cm2. The viscosity
of mercury is to be neglected.
## 2. A disk of mass m is connected to k2
two springs of stiffness k1 and k2
k1
as shown in the figure. Find the
angular frequency of the system
for small oscillation. Disc can roll
on the surface without slipping.
l l
3. A uniform bar with mass m lies symmetrically
across two rapidly rotating fixed rollers, A and B
A B
with distance ' ' between the bars centre of mass
mg
and each roller. The rollers whose direction of
rotation are shown in figure slip against the bar
with coefficient of friction . Suppose the bar is
displaced horizontally by a small distance 'x' and
then released, find the time period of oscillation.
## 4. A rod of length L, cross-sectional area A and density is hanging from a rigid
support by spring of stiffness k. A small sphere of mass m is rigidly attached at
the bottom of the rod. The rod is partially immersed in a liquid of density . Find
the period of small oscillations.
## 5. A vertical uniform tube of length , density is floating in a system of two
immiscible liquids of densities 1 and 2. Its bottom end is at the interface of the
liquids. Find the period of small oscillation.
## 6. Find the angular frequency of
small oscillation of block m in the k1
m
arrangement shown. Rod is
k2
massless. [Assume gravity to be
absent]
k3 l/2 l/2
## 7. A disc is free to rotate with an angular velocity , about an axis
passing through centre and perpendicular to the plane. A r
## massless spring of force constant k = 10 N/m is attached to a
k m
particle of mass m = 0.1 kg which can slide without friction,
initially at the centre of the disc and spring in its natural length.
Find the period of the particle is is (a)10 rad/hr (b) 6 rad/hr
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## 8. A string of natural length 2 and modulus of elasticity Y is stretched between two
fixed points A and B on a smooth horizontal table such that AB = 3 . A mass m is
attached to the mid point of the string and makes a small oscillation in a
horizontal line perpendicular to AB. Find the frequency of oscillation of m.
## 9. A system consisting of a smooth movable wedge of
k
angle and a block A of mass m are connected
together with a massless spring of spring constant k, m
as shown in the figure. The system is kept on a
frictionless horizontal plane. If the block is displaced M
## slightly from equilibrium and left to oscillate, find the
frequency of small oscillations.
## 10. Find out the angular frequency of m
small oscillations of the system
shown. The T - structure is b
massless. The springs are initially
relaxed. a a
k k
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LEVEL I
## 1. A particle moves in x-y plane according to the equation r ( i 2 j )A cos t the
motion of the particle is
(A) on a straight line (B) on an ellipse
(C) periodic (D) simple harmonic
2. Which of the following quantities are always positive in a simple harmonic motion?
(A) F. a (B) v. r
(C) a. r (D) F. r
## 3. The magnitude of average acceleration in half time period in a simple harmonic
motion is
2 2
2A A
(A) (B)
2
2
A
(C) (D) Zero
2
4. A small block oscillates back and forth on a smooth concave surface of radius R.
The time period of small oscillation is
R 2R
(A) T = 2 (B) T = 2
g g
R
(C) T = 2 (D) None of these
2g
## 5. A particle of mass 10 gm lies in a potential field v = 50 x2 + 100. The value of
frequency of oscillations in Hz is
5
(A) 5 Hz (B) Hz
10
(C) Hz (D) none of these.
3
## 6. When two mutually perpendicular simple harmonic motions of same frequency,
amplitude and phase are superimposed
(A) the resulting motion is uniform circular motion.
(B) the resulting motion is a linear simple harmonic motion along a straight line
inclined equally to the straight lines of motion of component ones.
(C) the resulting motion is an elliptical motion, symmetrical about the lines of
motion of the components.
(D) the two S.H.M. will cancel each other.
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## 7. The angular frequency of small oscillations of the system
shown in the figure is
K 2K K = spring constant
(A) (B)
2m m Massless
K 4K smooth
(C) (D) pulley
4m m
m
8. A particle executes SHM with a frequency f. The frequency with which it's KE
oscillates is
(A) f/2 (B) f
(C) 2f (D) 4f
9. A simple pendulum has some time period T. What will be the percentage
change in its time period if its amplitudes is decreased by 5 % ?
(A) 6 % (B) 3 %
(C) 1.5 % (D) 0 %
10. The work done by the string of a simple pendulum during one complete
oscillation is equal to
(A) total energy of the pendulum (B) KE of the pendulum
(C) PE of the pendulum (D) Zero
## 10. Two uniform rods are welded together to form a letter T as l
shown in the figure. Each rod is of mass M and length l. If this A B
combination is hinged at A and kept in vertical plane then time
period of small oscillations about A is equal to
l 3l l
(A) 2 (B) 2
6 3g 2 2g
l 11l C
(C) 2 (D) 2
2g (6 5 )(g )
h
2. A cylindrical piston of mass M slides smoothly M
inside a long cylinder closed at one end, enclosing P A
a certain mass of a gas.
The cylinder is kept with its axis horizontal. If the piston is slightly compressed
isothermally from its equilibrium position, it oscillates simple harmonically, the
period of oscillation will be
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Mh MA
(A) T = 2 (B) T = 2
PA Ph
M
(C) T = 2 (D) T = 2 MPhA
PAh
## 3. The period of the free oscillations of the system shown here if
mass M1 is pulled down a little and force constant of the spring is
k and masses of the fixed pulleys are negligible, is
M2
M1 M2 M1 4M2
(A) T = 2 (B) T = 2
k k M1
k
M2 4M1 M2 3M1
(C) T = 2 (D) T = 2
k k
## 4. The period of small oscillations of a simple pendulum of length l if its point of
suspension O moves a with a constant acceleration = 1 i + 2 j with respect
to earth is
(A) T = 2 2 2 1/ 2
(B) T = 2 2 2 1/ 2
{(g 2) 1} {( g 1) 2}
(C) T = 2 (D) T = 2
g {g2 2 1/ 2
1}
5. A particle moves along the X-axis according to the equation x = 10 sin 3( t).
The amplitudes and frequencies of component SHMs are
(A) amplitude 30/4, 10/4 ; frequencies 3/2, 1/2
(B) amplitude 30/4, 10/4 ; frequencies 1/2, 3/2
(C) amplitude 10, 10 ; frequencies 1/2, 1/2
(D) amplitude 30/4, 10 ; frequencies 3/2, 2
## 6. A pendulum makes perfectly elastic collision with block
of m lying on a frictionless surface attached to a spring
of force constant k. Pendulum is slightly displaced and k
m
released. Time period of oscillation of the system is m
m m
(A) 2 (B)
g k g k
m
(C) 2 (D) 2
g k
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## 7. Three springs of each force constant k are connected as
shown figure. Point mass m is slightly displaced to compress A
and released. Time period of oscillation C 90 B
m m m
(A) 2 (B) 2
2k 3k
A
m m
(C) 2 (D) 2
k k 2 k 1
## 8. Two blocks each of mass m are connected with
springs of force constant k. Initially springs are A B
relaxed. Mass A is displaced to left and B is displaced
towards right by same amount and released then time
period of oscillation of any one block (Assume
collision to be perfectly elastic)
m m
(A) 2 (B) 2
k 2k
m m
(C) (B)
k 2k
## 9. If for a particle moving in SHM, there is a sudden increase of 1% in restoring
force just as particle passing through mean position, percentage change in
amplitude will be
(A) 1% (B) 2%
(C) 0.5% (D) zero.
## 10. S1 and S2 are two identical springs. The oscillation k k
frequency is f. If one spring is removed, frequency S2
M S1
will be
(A) f (B) 2f
(C) 2 f (D) f 2
LEVEL II
1. Equation of SHM is x = 10 sin 10 t. Find the distance between the two points
where speed is 50 cm/sec. x is in cm and t is in seconds.
(A) Zero (B) 20 cm
(C) 17.32 cm (D) 8.66 cm.
## 2. A simple pendulum of length L and mass M is oscillating in a plane about a
vertical line between angular limits and + . For an angular displacement
(| | < ) the tension in the string and velocity of the bob are T and v respectively.
The following relations hold good under the above condition.
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(A) T cos = Mg
Mv 2
(B) T + Mg cos =
L
(C) The magnitude of tangential acceleration of he bob | aT | = g sin
(D) T = Mg (3 cos 2 cos )
## 3. A coin is placed on a horizontal platform, which undergoes vertical simple harmonic
motion of angular frequency . The amplitude of oscillation is gradually increased. The
coin will leave contact with the platform for the first time
(A) at the highest position of the platform (B) at the mean position of the platform
(C) for an amplitude of g/ 2 (D) for an amplitude of g /
## 4. The position vector of a particle that is moving in three dimensions is given by
r (1 2 cos 2 t)i (3 sin2 t) j (3t)k
in the ground frame. All units are in SI. Choose the correct statement (s) :
3
(A) The particle executes SHM in the ground frame about the mean position 1, ,3t .
2
(B) The particle executes SHM in a frame moving along the zaxis with a velocity of
3 m/s.
5
(C) The amplitude of the SHM of the particle is m .
2
4 3
(D) The direction of the SHM of the particle is given by the vector i j .
5 5
5. Which of the following will have a different time period, if taken to the moon ?
(A) A simple pendulum.
(B) A spring mass system oscillating vertically in the gravitational field.
(C) A torsion pendulum.
(D) An hourglass clock, which is essentially a vessel filled with a liquid, emptying
through a hole in the bottom.
## 1. Select the correct alternatives
(A) A simple harmonic motion is necessarily periodic.
(B) A simple harmonic motion is necessarily oscillatory
(C) Oscillation motion may be periodic
(D) A periodic motion is necessarily oscillatory
2. Which of the following is/are essential for simple harmonic motion?
(A) Inertia (B) Restoring force
(C) Material Medium (D) gravity
3. Which of the following is/are the characteristics(s) of SHM?
(A) projection of uniform circular motion on any straight line
(B) periodic nature
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## (C) displacement time graph is a sine curve
(D) acceleration is zero at the mean position
5. Which of the following will change their time period if they are taken to moon ?
(A) a simple pendulum (B) a physical pendulum
(C) a torsional pendulum (D) a spring pendulum
6. Which of the following statements are true for the oscillations of the mass suspended
with a spring?
(A) Time period varies directly as the square root of the suspended mass.
(B) A stiffer spring gives lesser time period
(C) The mass can execute oscillations in the state of weightlessness
(D) The system will have the same time period both on the moon and earth
COMPREHENSION
## I. The physical pendulum is just a rigid body, of whatever P
shape, capable of oscillating about a horizontal axis
passing through it. For small oscillations the motion of a
physical pendulum is almost as easy as for a simple l CM
pendulum. Figure shows a rigid body pivoted about point L
P and displaced from equilibrium through an angle . CM
The gravitational force provides a restoring torque of
magnitude mgl sin about point P and time period is
given by
I M
T 2 Stick
mgl
1. The period of oscillation for small angular displacement of a stick of length L pivoted
L 2L
(A) 2 (B) 2
2g 3g
L 3L
(C) 2 (D) 2
6g 4g
2. If the stick is pivoted about a point P, distance x from the center of mass, the period of
oscillation is
L2 12x2 3L2 2x 2
(A) 2 (B) 2
12 gx 2gx
12L2 x2 2L2 3x2
(C) 2 (D) 2
12gx 2gx
## 3. For what value of x period of oscillation is minimum
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L 2
(A) (B) L
12 3
L 2L
(C) (D)
3 3
II. If a mass attached to a spring (massless) is pulled down twice as far, the force will be
twice as much, the resulting acceleration is twice as great, the velocity it acquires in a
given time is twice as great, the distance covered in a given time is twice as great, but it
has to cover twice as great a distance in order to get back to the origin. In a linear
equation the time pattern does not change. If there were no friction the oscillation stays
on with same amplitude. However when there exists a friction of magnitude smaller for
small oscillation and larger for larger oscillation the amplitude gets dropped by the same
fraction in every cycle.
4. When the displacement from the mean position is doubled, the quantity that will vary will
be
(A) time period only (B) velocity only
(C) acceleration only (D) both (B) and (C).
## 5. If amplitude of an oscillation is a relative to the preceeding one in a system with initial
maximum amplitude A0 oscillating in a damping media, the amplitude just after n
oscillations will be
a a
(A) A 0 (B) A 0 a A 0
n n
(C) A0 a (n 1) (D) A0 an.
6. To solve the S.H. equation with friction, one must consider each half cycle separately
since.
(A) friction reverses its direction in each cycle
(B) the S.H. equation will not be linear
(C) friction is independent of the displacement
(D) all of these.
## 1. Match the Following :
Column A Column B
(A) A body is moving along a straight (p) This will may be a uniform linear
line and accelerating uniformly motion
(B) A body is moving along a straight (q) This will be a non-uniform linear
line. It covers a distance 72 m motion.
during the first six seconds of its
motion and another 72 m during
the next six seconds
(C) A body is thrown vertically upward. (r) During the motion, linear
It rises to some height and then momentum is not conserved.
falls down along the same line.
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(D) A bullet is fired into air from a gun. (s) Position-time graph of the motion
will may be a straight line that is
parallel neither to x-axis nor the y-
axis.
## 2. A particle of mass 2 kg is moving on a straight line under the action of force
F = (8 2x) N.
Particle is released from x = 7 m
For the subsequent motion match the following (all values in Column II are in S.I. units)
Column II
Column I
(A) Amplitude of SHM is (p) 0.5
(B) Time taken to move from x = 2.5 m to (q) 3
x = 4 m (approx)
(C) Total energy of SHM system (r) 6
(D) Velocity of particle at x = 4m (s) 9
## 3. A block is executing SHM on a rough horizontal fext
surface under the action of an external variable II
force. I
The external force (fext) is plotted against x
position (x) from mean position on shown in III
graph.
Match which graph is possible for different
IV
motions.
Column II
Column I (Nature of graph)
x = position
v = velocity of particle
(A) x positive, v positive (p) I
(B) x positive, v negative (q) II
(C) x negative, v positive (r) III
(D) x negative, v negative (s) IV
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LEVEL- I
3. F = -A gx
## 5. The relative acceleration = ar =a1 + a2
6. When the rod is rotated through an angle about O, the restoring torque about
the suspension point is 0 = mg( /2) .
## 8. The motion of simple pendulum is angular simple harmonic, so its equation of
g
motion is given by = 0 sin t, where =
L
kx
a = - 2x = M m
LEVEL- II
## 1. The difference of level in the two tubes is given by
= x + xcos = x (1+cos )
2. Total energy = Kpot + Ktran + U
3. The net horizontal force acting on the rod is given as f1 and f2 will not be equal
F = f1 f2 = (N1 N2)
4. When the block is pushed in by a distance x in the liquid, the net force acting on
the M and m system is
F = LA g + mg kx A(h0 + x) g
5. Initially the cylinder is just touching the liquid of density 2. When the cylinder is
dipped by a distance x1 further,
F = mg x1 A 2 g - 1 A 1 g
## 7. Restoring force = (kx m 2x)
Stress
8. Y=
Strain
9. Apply COE & COM.
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## 17. Answers to the Subjective Assignment
LEVEL I
1. 10 cm.
1
2. (a) x = 10 sin (120 t + /6) (b) /6, 10 cm, sec.
60
## 3. 0.088 sec 4. 0.73 s
6m 2
5. 2 6. 2
5k 3g
2 1
7. 3 2 m/sec ;
2 1
L L 1
8. (a) T1 = 2 (b) [ 2 sin / ]
g g
3M mkx m(M m)
9. 2 10. (a) (b)
2k m M Mk
LEVEL II
2(k1 4k 2 )
1. 0.8 sec 2.
3m
m L A
3. 2 4. T2
g k A g
4k 1k 3 k 1k 2 4k 2k 3
5. 6.
g
1 2 g 4k 3 k 2 m
2Y
7. (a) 0.785 sec (b) no oscillation. 8.
3m
k mM
9. 2 2
where mred =
mred cos m sin m M
2ka 2 g
10.
mb 2 b
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LEVEL I
1. (A) 2. (A)
3. (A) 4. (A)
5. (B) 6. (B)
7. (C) 8. (C)
9. (D) 10. (D)
11. (D) 12. (A)
13. (C) 14. (A)
15. (B) 16. (B)
17. (A) 18. (C)
19. (C) 20. (C)
LEVEL II
## 1. (A), (C) 2. (C), (D)
3. (A), (C) 4. (B), (C), (D)
5. (A), (D) 6. (A), (B), (C)
7. (A), (B) 8. (B), (C), (D)
9. (A), (B) 10. (A) (B), (C), (D)
COMPREHENSION
1. (B) 2. (A)
3. (A) 4. (D)
5. (D) 6. (D)
## MATCH THE FOLLOWING
1. (A) (q), (r); (B) (p), (s); (C) (q), (r); (D) (q), (r)
## 3. (A) (p); (B) (s); (C) (q); (D) (r)
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## ELASTICITY & WAVES
11. Assignment (Subjective Problems)
LEVEL I
1. If a wave form has the equation y1 = A1 sin ( t - kx) & y2 = A2 cos ( t - kx),
find the equation of the resulting wave on superposition.
2. A wave train has the equation y = 4 sin (30 t + 0.1x) where x is in cm and t in
seconds. What is the frequency of the source? How much time does a wave
pulse take to reach a point 30 cm from it?
3. When the stretching force of a wire in increased by 25 kg-wt, the frequency of the
note emitted is changed in the ratio 2/3. Calculate the original stretching force.
4. A policeman on duty detects a drop of 10% in the pitch of the horn of a moving
car as it crosses him. If the velocity of sound is 330 meters per second, calculate
the speed of the car.
5. A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person
can hear sound of maximum frequency 15 KHz. What is the highest harmonic
that can be played on this string which is audible to the person?
6. A wire of length L is fixed at both ends such that F is tension in it. Its mass per
unit length is given from one end to other end as = ox where o is constant.
Find time taken by a transverse pulse to move from lighter end to its mid point.
## 7. The transverse displacement of a string, fixed at both of its ends, is given as
y(x, t) = 0.06 sin 2 x cos 120 t where x, y are in meters
and t is in seconds. The length of the string is 1.5 m & its mass m = 3 10-2 kg.
Find the
(a)wavelength (b) frequency. (c) amplitude of the component of waves.
(d) maximum velocity of the particle. (e) amplitude at a distance x = 0.375 m.
## 8. Two closed organ pipes A and B have lengths 50 cm and 70 cm respectively.
The 5th harmonic (=2nd overtone) of A resonates with nth harmonic of B. Find n.
## 9. The displacement of the medium in a sound wave is given by the equation
y = A cos(ax+bt) (M.K.S. system)
where A, and b are positive constant.
(a) What is the wave length and frequency of the incident wave?
(b) Find the wave speed and maximum particle speed.
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10. The intensity of sound from a point source is 1.0 x 10 -8 W/m2 at a distance of 5.0
m from the source. What will be the intensity at a distance of 25 m from the
source?
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LEVEL II
## 1. A loop of rope is whirled at a high angular velocity , so that it becomes a taut
(i) Find the tension in the rope if the linear mass density of the rope is .
(ii) A kink develops in the whirling rope. Under what condition does the kink
remain stationary relative to an observer on the ground?
2. AB is cylinder of length 1.0 m, fitted with a flexible diaphragm C at the middle and
two thin flexible diaphragms A and B at the ends. The portions AC and BC contain
hydrogen and oxygen respectively. The diaphragms A and B are set into vibration
of the same frequency. What is the minimum frequency of these vibrations for
which the diaphragm C is a node? Under the conditions of the experiment, the
velocity of sound in hydrogen is 1100 m/s and in oxygen is 300 m/s.
3. A long tube contains air at a pressure of P and temperature T. The tube is open
at one end and closed at the other end by a movable piston. A tuning fork near the
open end is vibrating with a frequency f. Resonance is produced when the piston is
at distance L1 and L2 from the open end. Mean molecular mass of the air is M.
(a) Find the speed of sound in air.
(b) Find the adiabatic constant of the air.
4. A piston is fitted in a cylindrical tube of small cross section with the other end of
the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The
piston is gradually pulled out of the tube and it is found that a second resonance
occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the
speed of sound in the air of the tube.
5. Three tuning forks with unknown frequencies f 1, f2 and f3 are vibrated. 5 beats per
second are heard when f1 and f2 are vibrated, 6 beats per second for f1 and f 3,
while 7 beats per second for f2 and f3. If f2 is loaded with wax, number of beats for
f2 and f3 decreases while for f1 and f2 increases.
Find tuning forks having maximum frequency and minimum frequency in terms of f2.
## 6. Two rods AB and CD of equal cross- sectional area A and A
Youngs constant y1 and y2 are joined and suspended from
1
a fixed support. A block of mass M is attached to the lowest
point C. The density of the rods is negligibly small. Find B
displacement of the point C.
2
C
M
7. A string vibrates according to the equation y = 5 sin ( x/3) cos (40 t) where x and
y are in cm and t is in second.
(a) What are the amplitude and velocity of the component waves whose
superposition can give rise to this vibration?
(b) What is the distance between two successive nodes?
(c) What is the velocity of a particle of the string at position x = 1.5 cm and
t = 9/8 second?
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8. How long will it take sound waves to travel distance between the points A and B
if the air temperature between them varies linearly from T1 to T2? The velocity of
sound propagation in air is equal v = T , where is a constant.
9. The first overtone of an open organ pipe beats with the first overtone of a closed
organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the
closed organ pipe is 110 Hz. Find the lengths of the pipes.
10. A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length
and rotated with an angular velocity of 20 rad/s in the horizontal plane. Calculate
the range of frequencies heard by an observer stationed at a large distance from
the whistle.
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LEVEL I
## 1. The amplitude of resulting wave dues to superposition of y1 = A sin ( t kx) &
y2 = A sin ( t kx + ) is
(A) 2A cos (B) 2A tan ( /2)
(C) A cos sin (D) none
2. A sine wave has an amplitude A and wavelength . The ratio of particle velocity
and the wave velocity is equal to (2 A = )
(A) 1 (B) = 1
(C) 1 (D) data insufficient.
3. The equation of a wave pulse moving with a speed 1 m/sec at time t = 0 is given
1
as y = f(x) = . Its equation at time t = 1 second can be given as
1 x2
1 1
(A) y = 2
(B) y =
1 (1 x ) 1 (1 x )2
1 1
(C) y = 2
(D) y =
1 (x 1) 1
1
1 x2
4. The velocity of a transverse wave in a string does not depend on
(A) tension (B) density of material of string
(C) radius of string (D) length of string
## 5. The frequency of a tuning fork with an amplitude A = 1 cm is 250 Hz. The
maximum velocity of any particle in air is equal to
(A) 2.5 m/s (B) 5 m/s
(C) 3.30 / m/sec (D) none of these
6. In a resonance column experiment, the first resonance is obtained when the level
of the water in tube is 20 cm from the open end. Resonance will also be obtained
when the water level is at a distance of
(A) 40 cm from the open end. (B) 60 cm from the open end.
(C) 80 cm from the open end. (D) data insufficient.
7. A wire of length having tension T and radius r vibrates with natural frequency f.
Another wire of same metal with length 2 having tension 2T and radius 2r will
vibrate with natural frequency
(A) f (B) 2f
f
(C) 2 2f (D)
2 2
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8. Under the same conditions of pressure and temperature, the velocity of sound in
oxygen and hydrogen gases are v0 and vH, then
(A) vH= vo (B) vH = 4vo
(C) vo = 4 VH (D) vH = 16 vo
## 9. A tuning for of frequency 600 Hz produces a progressive travelling wave having
wave velocity 300 m/s. Two particles of a medium, separated by 1.5 m, vibrate
being affected by the wave
(A) in phase (B) in opposite phase.
(C) 45 out of phase. (D) none of these
10. At t=0 source starts falling under gravity and a detector is projected S
upwards with a velocity 10 m/s. For the vertical upward motion of
detector
(A) apparent frequency received by detector = source frequency.
(B) initially apparent frequency > source frequency and finally less 5m
## than source frequency.
(C) apparent frequency depends only on the detector velocity. D
(D) date insufficient.
11. A string is clamped on both ends. Which of the following wave equations is valid
for a stationary wave set up on this string? (Origin is at one end of string.)
(A) y = A sin kx. sin t (B) y = A cos kx sin t
(C) y = A cos kx. cos t (D) None of the above.
## 12. A string is hanging from a rigid support. A transverse wave pulse is
set up at the bottom. The velocity v of the pulse related to the
distance covered by it is given as
(A) v x (B) v x
(C) v 1/x (D) none of these x
13. The third overtone of a closed organ pipe is equal to the second harmonic of an
open organ pipe. Then the ratio of their lengths is equal to
(A) 7/4 (B) 3/5
(C) 3/2 (D) none of these
## 14. Standing waves can be produced in
(A) solid only (B) liquid only
(C) gases only (D) all of the above
15. If the temperature of the medium drops by 1 %, the velocity of sound in that
medium
(A) increases by 5 % (B) remains unchanged
(C) decreases by 0.5 % (D) decreases by 2 %
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16. The velocity of sound through a diatomic gaseous medium of molecular weight M
at 0C is approximately.
R 3R
(A) (B)
M M
382 R 273 R
(C) (D)
M M
## 17. The amplitude of a wave disturbance propagating in the positive x direction is
1 1
given by y = 2
at time t = 0 and by y = at time t = 2 seconds
(1 x ) 1 ( x 2)2
where x and y are in meters. The shape of the wave disturbance does not
change during the propagation. The velocity of the wave is
(A) 0.5 m/s (B) 1 m/sec
(C) 2 m/s (D) 1.5 m/sec
## 18. A wave is represented by the equation y = [A sin {10 x + 15 t+ ( /3)}] where x
is in meters and t is in seconds. The expression represents
(A) A wave travelling in positive x-direction with a velocity 1.5 m/s.
(B) A wave travelling in negative x-direction with a velocity 1.5 m/s.
(C) A wave travelling in the negative x-direction having a wavelength 2 m.
(D) A wave travelling in positive x-direction having a wavelength 2 m.
19. A transverse wave is given by A sin( t x) where and are constants. The
ratio of wave velocity to maximum particle velocity is
(A) A (B) 1/ A
(C) 1 (D) none of the above.
## 20. Two blocks, each of mass m, are connected by a
massless thread Y and A represent Youngs modulus
and cross sectional area of wire respectively. The strain
developed in the thread is m
m
mg 1 sin mg
(A) (B)
2 yA yA
mg sin 2mg
(C) (D)
yA yA
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LEVEL II
## 1. A wave going in a solid
(A) Must be longitudinal (B) May be longitudinal
(C) Must be transverse (D) May be transverse
2. A wave is represented by the equation
y = (0.001 mm) sin [(50 s 1)t + (2.0 m 1)x]
(A) The wave velocity = 100 m/s (B) The wavelength = 2.0 m
(C) The frequency = 25/ Hz (D) The amplitude = 0.001 mm
3. An electrically maintained tuning fork vibrates with constant frequency and constant
amplitudes. If the temperature of the surrounding air increases but pressure remains
constant, the sound produced will have
(A) Larger wavelength (B) Larger frequency
(C) Larger velocity (D) Larger time period
4. The fundamental frequency of a vibrating organ pipe is 200 Hz.
(A) The first overtone is 400 Hz (B) The first overtone may be 400
Hz
(C) The first over tone may be 600 Hz (D) 600 Hz is an overtone
5. A listener is at rest with respect to the source of sound. A wind starts blowing along the
line joining the source and the observer. Which of the following quantities do not
change?
(A) Frequency (B) Velocity of sound
(C) Wavelength (D) Time period
6. The figures represent two snaps of a travelling wave on a string of mass per unit length,
1
= 0.25 kg/m. The two snaps are taken at time t = 0 and at t s. Then
24
(A) speed of wave is 4 m/s
(B) the tension in the string is 4 N
(C) the equation of the wave is y = 10 sin ( x 4 t )
6
(D) the maximum velocity of the particle m/s
25
y(mm) y(mm)
10 10
5 5
5 x(m) 5 1 x(m)
10 10 1
t=0 t=24 s
Figure I Figure II
7. As a wave propagates,
(A) the wave intensity remains constant for a plane wave
(B) the wave intensity decrease as the inverse of the distance from the source for a
spherical wave
(C) the wave intensity decreases as the inverse square of the distance from the
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## source for spherical wave
(D) total intensity of the spherical wave over the spherical surface remains constant
at all times, while source is at the centre of spherical surface.
8. A source is moving across a circle given by the equation x 2+y2 = R2, with constant speed
330
v m/s, in anti-clockwise sense. A detector is at rest at point (2R, 0) w.r.t. the
6 3
centre of the circle. If the frequency emitted by the source is f and the speed of sound,
C = 330 m/s. Then
(A) the position of the source when the detector
3 R
records the maximum frequency R,
2 2
(B) the co-ordinate of the source when the detector records minimum frequency is (0,
R)
6 3
(C) the maximum frequency recorded by the detector is f
6 3
6 3
(D) the minimum frequency recorded by the detector is f
6 3
## 9. A wave is represented by the equation :
1
y (1mm) sin 50 s t (2.0m 1 )x + (1mm) cos 50 s 1
t (2.0m 1 )x
(A) The wavevelocity is zero, since it is a standing wave.
3
(B) A node is formed at x m.
8
(C) The amplitude of the oscillation at the antinode is 2 mm.
(D) Energy transfer occurs along the positive xaxis.
## 10. A very light rod AB is initially hung from a point P by P
means of two identical copper wires of the same
length as the rod as shown in the figure. Particles of
masses 1 kg and 4 kg are then attached to the ends A
and B of the rod. The ratio of the fundamental
A B
frequencies of vibration of the wires AP and BP, i.e.,
fA
=
fB
1
(A) 4 (B)
2
(C) 16 (D) 2.
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COMPREHENSION
## I. A narrow tube is bent in the form of a circle of radius R, as
shown in the figure. Two small holes S and D are made in the
tube at the positions right angle to each other. A source R
placed at S generates a wave of intensity I0 which is equally S
divided into two parts: one part travels along the longer path,
while the other travels along the shorter path. Both the part
waves meet at the point D where a detector is placed.
D
## 1. If a maxima is formed at a detector then, the magnitude of wavelength of the wave
produced is given by
R
(A) R (B)
2
R
(C) (D) all of these
4
2. If a minima is formed at the detector then, the magnitude of wavelength of the wave
produced is given by
3
(A) 2 R (B) R
2
2
(C) R (D) None of these
5
(A) 4I0 (B) 2I0
(C) I0 (D) 3I0
## II. Speed of a transverse wave depends on mass Tension = constant
and tension. Two strings of equal lengths are v 1 2
joined at B. Mass of string BC is four times mass B
C
of string AB. If a wave pulse is generated in string A
AB, which travels towards boundary at B with x=0 x=L x=2L
speed v. Equation of incident pulse is given as
yi A i sin( t kx)
Based on above information, answer the following
questions.
Ai 2Ai
(A) (B)
3 3
Ai 2A i
(C) (D) .
3 3
## 5. Speed of transmitted wave on string BC is
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v
(A) v (B)
2
(C) 2v (D) None of these.
## 6. Equations of reflected and transmitted waves respectively are :
Ai Ai
yr sin( t kx) yr sin( t kx)
(A) 3 (B) 3
2A i 2Ai
yt sin( t 2kx) yt sin( t 2kx)
3 3
Ai Ai
yr sin( t kx) yr sin( t kx)
(C) 3 (D) 3 .
2Ai 2A i k
yt sin( t 2kx) yt sin( t x)
3 3 2
## 1. The figure shows a string at a certain moment as a A B
transverse wave passes through it. Three particles A, B
and C of the string are also shown. Match the physical C
quantities in the left column with the description in the
column on the right.
Column A Column B
(A) Velocity of A (p) Downwards, if the wave is travelling towards right.
(B) Acceleration of A (q) Downwards, if the wave is travelling towards left.
(C) Velocity of B (r) Downwards, no matter which way the wave is
travelling.
(D) Velocity of C (s) Zero.
## 2. A source of sound in moving along a circular orbit of N
sound detector located for away is executing linear M
A B C D
SHM with amplitude 6 m on line BCD as shown. The
5 P
frequency of detector for oscillation is per
## second. The source is at A when detector at B at t =
0. Source emits a continuous sound wave of
frequency 340 Hz. (velocity of sound = 330 ms1).
Match the column A with B.
Column A Column B
(A) The frequency of sound recorded by (p) 255 H.
detector at t = 3T/4.
(B) The frequency of sound recorded by (q) 1 : 1.
detector at t = T/4.
(C) The ratio of the time period of source (r) 442 Hz.
and detector (circular motion and
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SHM).
(D) Maximum velocity of (s) 2 : 1.
detector/maximum velocity of source.
(T is time period of oscillation).
## 3. The diagrams in Column I show transverse sinusoidal standing/travelling waveforms on
stretched strings. In each case, the string is oscillating in a particular mode, and, its shape
and other characteristics are shown at time t = 0. The maximum amplitude (in all the
cases) is A, the velocity of the waveform on the string is e, the mass per unit length of the
string is and the frequency of vibration is f (angular frequency = ).
The kinetic energy of the string (of length L) is represented by the functions in Column
II. Match the correct entries in Column II.
Column II
Column I
(A) Fixed Fixed (p) 2
end end
9 c2 2
A
A 4 L
L
(B) Fixed Free (q) 2
end
9 c2 2 2
end A sin t
A 4 L
L
(C) Free (r)
end
Free 9 2 c2 2 2
end A sin t
16 L
A
A
L
(D) (s) 2
c2 2 2
A A sin t
4 L
L
Travelling wave
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LEVEL- I
Ar = A12 A 22
## 2. Compare it with the standard wave equation
y = A sin ( t + kx)
T T 2 .5
3. f1 = , f2 =
1 T
8. f1 =
2L m
P T
f2 = ; P = number of loops.
2L m
LEVEL- II
2
1. y = 2a sin x cos t
## compare it with the given equation.
(2n 1)v
2. f= n = 1, 2, 3 . . .. . for Ist, 2nd, 3rd . . .. . .overtone.
4
## 3. The centripetal force
v2
2T sin = (dm)
R
v1 v
4. y = A cos (ax + bt) 5. n1 = , n2 = 2
4 4
v RT
6. (a) L2 L1 = (b) v =
2f M
8. P = I1A1 = I2 A2 = I1 (4 r12 ) = I2 (4 r22 )
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## 14. Answers to the Subjective Assignment
LEVEL I
1
1. A12 A 22 sin { t kx + tan-1 (A2/A1)}. 2. (a) 15 Hz. (b) sec.
10
## 3. 2 kg wt. 4. 17.4 m/s
1 o 32
5. 75 6. L
3 2 F
7. (a)1 m (b) 60 Hz (c) 0.03 m (d) 7.2 m/s (e) 4.2 cm.
8. 7
2
9. (a) = (b) f = b/2 (b) b/a, Ab
a
10
10. 4 10 W/m2.
LEVEL II
4f 2M(| L 2 L1 |) 2
1. (a) 2f(L2 L1) (b) 2. f1 = f2 + 5, f3 = f2 7
RT
Mg 1 2
3. 4. 327.68 m/s.
A y1 y2
Mg 1 2
5. f1 = f2 + 5, f3 = f2 7 6.
A y1 y2
2
7. (a) 5/2 cm, 120 cm/s (b) 3 cm (c) Zero 8.
T1 T2
9. 1 = 0.993 m or 1.006 m
## 10. Range = 403 Hz to 484 Hz
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LEVEL I
1. (D) 2. (A)
3. (B) 4. (D)
5. (B) 6. (B)
7. (D) 8. (B)
9. (A) 10. (B)
11. (A) 12. (A)
13. (A) 14. (D)
15. (C) 16. (C)
17. (B) 18. (B)
19. (B) 20. (A)
LEVEL II
## 1. (B), (D) 2. (C), (D)
3. (A), (C) 4. (B), (C), (D)
5. (A), (D) 6. (A), (B), (C), (D)
7. (A), (C), (D) 8. (A), (B), (C), (D)
9. (B), (C) 10. (D)
COMPREHENSION
1. (D) 2. (A)
3. (B) 4. (A)
5. (B) 6. (C)
## 3. (A) (q); (B) (r); (C) (s); (D) (p)
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## HEAT AND THERMODYNAMICS
SECTION I
Level I
(Single Choice Questions)
1. A meter washer has a hole of diameter d1 and an external diameter d 2 , where d 2 3d1 .
On heating, d 2 increases by 0.3%. Then d1 will
(a) decrease by 0.1% (b) decrease by 0.3%
(c) increase by 0.1% (d) increase by 0.3%.
## 2. If I is the moment of inertia of a solid body, the change in I corresponding to a small
change in temperature T is
1
(a) I T (b) I T
2
(c) 2 I T (d) 3 I T .
## 3. If is the coefficient of linear expansion, the change in the period t of a physical
pendulum with temperature change of T is
1 1
(a) t T (b) t T
2 4
3 1
(c) t T (d) t T.
4 3
4. A steel scale is to be prepared such that the millimeter intervals are to be accurate within
6 10 5 mm. The maximum temperature variation during the ruling of the millimeter marks
( 12 10 6 C 1 ) is
(a) 4.0C (b) 4.5C
(c) 5.0C (d) 5.5C.
5. A block of ice at 10C is slowly heated and converted to steam at 100C. Which of the
following curves represents the phenomenon qualitatively ?
(a) (b)
(c) (d)
## -10C Heat supplied -10C Heat supplied .
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6. A point source of heat of power P is placed at the center of a spherical shell of mean
radius R. The material of the shell has thermal conductivity k. If the temperature
difference between the outer and the inner surface of the shell is not to exceed T, then
the thickness of the shell should not be less than
2 R 2 kT 4 R 2 kT
(a) (b)
P P
R 2 kT R 2 kT
(c) (d) .
P 4P
7. Three rods A, B and C have the same dimensions. Their thermal conductivities are
k A , k B and kC respectively. A and B are placed end to end, with their free ends kept at a
certain temperature difference. C is placed separately, with its ends kept at the same
temperature difference. The two arrangements conduct heat at the same rate. kC must
be equal to
k A kB
(a) k A k B (b)
k A kB
1 k A kB
(c) (k A kB ) (d) 2. .
2 k A kB
8. A cyclic process is shown in the p-T diagram. Which of the curves show the same
process on a V-T diagram ?
P
A C
O T
B C B A
V V
(a) (b)
A C
O T O T
C C
V V
(c) (d)
A B A B
O T O T .
## 9. A solid at temperature T1 is kept in an evacuated chamber at temperature T2 T1 . The
rate of increase of temperatrue of the body is proportional to
(a) T2 T1 (b) T22 T12
(c) T24 T14 (d) T23 T13
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10. 70 calories of heat are required to raise the temperature of 2 moles of an ideal gas at
constant pressure from 30C to 35C. The amount of heat required in calories to raise
the temperature of same gas through the same range (30C to 35C) at constant volume
is
(a) 30 (b) 50
(c) 70 (d) 90.
11. A black body radiates power P and maximum energy is radiated by it around a
wavelength 0 . The temperature of the black body is now changed such that it radiated
3 0
maximum energy around the wavelength . The power radiated by it now is
4
256 16
(a) P (b) P
81 9
64 4
(c) P (d) P.
27 3
## 12. Five rods having thermal conductivities k1 , k2 , k3 , k4 and k1 k3
k5 are arranged as shown. The points A and B are
A k5 B
maintained at different temperature such that no thermal
current flows through the central rod. k2 k4
(a) k1k4 k 2 k3 (b) k1 k3 , k 2 k4
k k3
(c) k1k3 k2 k4 (d) 1 .
k 4 k2
13. An ideal gas expands according to the law pV 2 = const. The molar heat capacity C is
(a) CV R (b) CV R
(c) CV 2 R (d) CV 3R .
14. If W1 is the work done in compressing an ideal gas from a given initial state through a
certain volume isothermally and W2 is the work done in compressing the same gas from
the same initial state through the same volume adiabatically, then
(a) W1 W2 (b) W1 W2
(c) W1 W2 (d) W1 2W2 .
15. The rate of emission of a black body at 0C is R. Its rate of emission at 273C is
(a) 4R (b) 8R
(c) 16R (d) 32R.
## 16. A body emits radiation when its temperature is
(a) >0C (b) >100C
(c) > surrounding temperature
(d) it emits radiation at all temperatures (T > 0K)
## 17. Radiation from a black body at the thermodynamic temperature T1 is measured by a
small detector at distance d1 from it. When the temperature is increased to T2 and the
distance to d 2 , the power received by the detector is unchanged. What is the ratio
d 2 / d1 ?
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2
T2 T2
(a) (b)
T1 T1
2 4
T1 T2
(c) (d) .
T2 T1
18. Two identical containers joined by a small pipe initially contain the same gas at pressure
p0 and absolute temperature T0 . One container is now maintained at the same
temperature while the other is heated to 2T0 . The common pressure of the gases will be
2 4
(a) p0 (b) p0
3 3
5
(c) p0 (d) 2 p0 .
3
## 19. An ideal gas changes from state a to state b as shown in b
figure. What is the work done by the gas in the process ?
(a) zero (b) positive
T
(c) negative (d) infinite.
a
P
20. The weight of a person is 60 kg. If he gets one kilo-calorie of heat through food and the
efficiency of his body is 28%, then upto how much height he can climb ? Take g = 10 m
s-2
(a) 100 cm (b) 196 cm
(c) 400 cm (d) 1000 cm.
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Level II
(Single Choice Questions)
## 1. Three rods of the same dimensions have thermal 50C
conductivities 3k , 2k and k . They are arranged as
shown, with their ends at 100C, 50C and 0C. The 2k
100C
temperature of their junction is
200 3k
(a) 75C (b) C k
3
100 0C
(c) 40C (d) C.
3
## 2. Steam at 100C is passed into 1.1 kg of water contained in a calorimeter of water
equivalent 0.02 kg at 15C, till to 80C. The mass of steam condensed (in kg) is (Take
latent heat of steam = 540 cal g-r)
(a) 0.130 (b) 0.065
(c) 0.260 (d) 0.135.
3. An ideal monoatomic gas is taken round the cycle (3P, V) (3P, 3V)
ABCDA as shown in following P-V diagram. The work C D
done during the cycle is
(a) PV (b) 2PV P
(c) 4PV (d) zero. B A
(P, V) (P, 3V)
O V
## 4. Heat energy absorbed by a system in going through a 30
cyclic processshown in figure, is
(a) 107 J (b) 104 J
10
(c) 102 J (d) 10 3
J.
10 30
P in kPa
5. A cyclic process is shown on the V T diagram. The same process
C
on a P T diagram is shown by
V
D
B
A
O T
A B D A
(a) P (b) P
D C C B
O T O T
C
C B
P P
(c) (d) D
D B
A
A
O T O T .
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6. Two containers of equal volume contain the same gas at pressure p1 and p2 and
absolute temperature T1 and T2 respectively. On joining the vessels, the gas reaches a
p
common pressure p and a common temperature T. The ratio is equal to
T
p1 p2 1 p1 p2
(a) (b)
T1 T2 2 T1 T2
p1T2 p2T1 p1T2 p2T1
(c) (d) .
T1 T2 T1 T2
## 7. A cyclic process ABCD is shown in the p V diagram. Which of A B
the following curves represent the same process ?
p C
D
V
A B D C
(a) T D (b) B
V
C A
p T
B A B
(c) p A (d)
C V
D
D C
T T .
## 8. A cyclic process is shown in the p-T diagram. Which of the curves
C B
show the same process on a V-T diagram ?
C B C B p
A
p p
(a) (b) O T
A A
V V
C B
p
(c) (d) p
B A A C
V V .
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## 9. An ideal gas is taken through the cycle A B C A,
C B
as shown in figure. If the net heat supplied to the gas in 2
the cycle is 5 J, the work done by the gas in the process
C A is
(a) - 5 J (b) - 10 J 1 A
(c) - 15 J (d) - 20 J.
P(N/m )
2
10
10. If there is no heat losses, the heat released by the condensation of x grams of steam at
100C into water at 100C converts y grams of ice at 0C into water at 100C. The ratio
y/x is
(a) 1 : 1 (b) 2 : 1
(c) 3: 1 (d) 4 : 1
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SECTION II
(Other Engg Examination Questions)
1. 125 ml of gas A at 0.60 atmosphere and 150 ml of gas B at 0.80 atmosphere pressure at
same temperature are filled in a vessel of 1 litre volume. The pressure of mixture at the
same temperature will be
(a) 0.195 atmosphere (b) 0.212 atmosphere
(c) 0.120 atmosphere (d) 0.140 atmosphere
## 2. The length of a metallic rod is 5 m at 0C and becomes 5.01 m on heating up to 100C.
The cubical expansion of the metal will be
(a) 2.0 10 5 / C (b) 4.0 10 5 / C
(c) 6.0 10 5 / C (d) 2.33 10 5 / C
3. How much work has to be done in decreasing the volume of an ideal gas by an amount
of 2.4 10 4 m3 at normal temperature and constant normal pressure of 1 105 N / m 2 ?
(a) 24 joule (b) 25 joule
(c) 27 joule (d) 28 joule
4. Helium gas is filled in a closed vessel (having negligible expansion coefficient). When it
is heated form 300 K to 600 K then average kinetic energy of helium atoms will be
(a) half (b) unchanged
(c) two times (d) 2 times
5. The total area of the walls of a room is 137 m2. An electric heater is used to maintain the
temperature inside the room at + 20C, while the outside temperature is 10C. Walls
are made of three layers of different materials. The innermost layer is made of wood 2.5
cm thick middle layer is made of cement 1 cm thick and outermost layer is made of
bricks 25 cm thick. What will be the power of electric heater ? Assume that there is no
loss of heat from the roof and the floor. The coefficient of thermal conductivity of wood,
cement and brick are 0.125, 1.5 and 1 watt/mC respectively.
(a) 9000 watt (b) 8000 watt
(c) 7000 watt (d) 5000 watt
6. When the temperature of a rod is increased by 10C, then its length increases by 1%.
When a cube is made of the material of the rod and its temperature is increased by
10C, then its volume will increase by
(a) 1% (b) 5.6%
(c) 3% (d) 30%
7. The ratio of amount of heat necessary to heat a known mass of water from 0C to 50C
and convert the same mass of ice into steam will be
(a) 5/6 (b) 1/8
(c) 16/31 (d) 5/72
8. The lengths and radii of two rods made of same material are in the ratio 1 : 2 and 2 : 3
respectively. If the temperature difference between the ends for the two rods be the
same, then in the steady state, the amount of heat flowing per second through them will
be in the ratio
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(a) 1 : 3 (b) 4 : 3
(c) 8 : 9 (d) 3 : 2
9. A hot liquid takes 5 minutes to cool from 70C to 60C. How long will it take to cool from
60C to 50C ?
(a) 5 minutes
(b) more than 5 minutes
(c) less than 5 minutes
(d) less or more than 5 minutes depending on the nature of liquid
10. Two metal A and B are having their initial length in the ratio 2 : 3 and coefficients of
linear expansion in the ratio 3 : 4. When they are heated through same temperature
difference, the ratio of their linear expansions is
(a) 1 : 2 (b) 2 : 3
(c) 3 : 4 (d) 4 : 3
11. The length of a metal rod at 0C is 0.5 m. When it is heated, its length increases by 2.7
mm. The final temperature of the rod is (Coefficient of linear expansion of the metal = 90
10 6/C)
(a) 20C (b) 30C
(c) 40C (d) 60C
12. The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40C
respectively. If 1/4th of the gas is released from the vessel and the temperature of the
remaining gas is raised to 353C, the final pressure of the gas is
(a) 1440 kPa (b) 1080 kPa
(c) 720 kPa (d) 540 kPa
13. A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of
5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball,
the rise in temperature is (Specific heat of steel = 460 JKg 1C 1, g = 10 ms 2)
(a) 0.01C (b) 0.1C
(c) 1C (d) 1.1C
14. If a metallic sphere gets cooled from 62C to 50C in 10 minute and in the next 10
minute gets cooled to 42C, then the temperature of the surroundings is
(a) 30C (b) 36C
(c) 26C (d) 20C
15. One end of a metal rod of length 1.0 m and area of cross-section 100 cm2 is maintained
at 100C. If the other end of the rod is maintained at 0C, the quantity of heat transmitted
through the rod per minute is (Coefficient of thermal conductivity of material of rod = 100
Wkg 1K Y)
(a) 3 103 J (b) 6 103 J
(c) 9 103 J (d) 12 103 J
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SECTION III
(More than One Correct Choice Questions)
1. Which of the following expressions for an ideal gas undergoing adiabatic expansion
is/are correct
(a) The expression of workdone is given by p1V1 p 2 V2 / 1
(b) The adiabatic pV lines are steeper than the isothermal lines
(c) p1V1 p 2 V2
(d) T1 1V1 T2 1V2
2. Figure shows is the P-V diagram for a Carnot cycle. In this diagram,
A
B
P
D
C
V
(a) curve AB represents isothermal process and BC adiabatic process
(b) curve AB represents adiabatic process and BC isothermal process
(c) curve CD represents isothermal process and DA adiabatic process
(d) curve CD represents adiabatic process and DA isothermal process
## 3. For an ideal gas,
(a) the change in internal energy in a constant-pressure process from temperature T1 to
T2 is equal to nCV (T2 T1), where CV is the molar heat capacity at constant volume
and n is the number of moles of the gas
(b) the change in internal energy of the gas and the work done by the gas are equal in
(c) the internal energy does not change in an isothermal process
4. A spherical black body of radius r radiates power P, and its rate of cooling is R
2
(a) P r (b) P r
1
(c) R r2 (d) R
r
5. Two rods of length L1 and L2 are made of materials of co-efficients of linear expansions
1 and 2 respectively such that L1 1 = L2 2. The temperature of the rods is increased by
T and correspondingly the change in their respective lengths are L1 and L2.
(a) L1 L2
(b) L1 = L2
(c) the difference in the length (L1 L2) is a constant and is independent of rise of
temperature
(d) data is insufficient to arrive at a conclusion
6. A metal rod of length L0, made of material of Youngs modulus Y, area A is fixed
between two rigid supports. The coefficient of linear expansion of the rod is . The rod is
heated such that the compressive force in the rod is T
(a) T L0 (b) T 1/
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(c) T A (d) T Y
7. The molar specific heat for a gas may have a value given by
dU dQ
(a) C V (b) C P
dT dT P
dU dV
(c) C P (d) data insufficient
dT dT
## 8. CP is always greater than CV due to the fact that
(a) no work is being done on heating the gas at constant volume
(b) when a gas absorbs heat at constant pressure its volume must change so as to do
some external work
(c) the internal energy is a function of temperature only for an ideal gas
(d) for the same rise of temperature, the internal energy of a gas changes by a smaller
amount at constant volume than at constant pressure
## 9. The first law is Thermodynamics is based on
(a) the law of conservation of energy (b) the law of conservation of heat
(c) the law of conservation of work (d) the equivalence of heat and work
10. The following are the p-V diagrams for cyclic processes for a gas. In which of these
processes is heat absorbed by the gas ?
p V
(a) V (b) p
p V
(c) V (d) p
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SECTION IV
(Match the Columns)
1. Match the physical quantities given in Column I with their dimensional formulas given in
Column II
Column I Column II
A. Specific heat (p) ML2T 2K 1
B. Latent heat (q) M 0 L2T 2 K 1
C. Molar specific heat (r) ML2 T 2 mol 1K 1
D. Thermal capacity (s) M 0 L2 T 2
2. Match the thermodynamic processes on an ideal gas given in Column I with the
corresponding change in internal energy (dU), heat transfer (dQ) and work done (dW)
given in Column II.
Column I Column II
A. Isothermal process (p) dQ = dU + PdV
B. Adiabatic process (q) dW = dU
C. Isochoric process (r) dU = 0
D. Isobaric process (s) dQ = dU
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SECTION V
(Passage Based Problems)
[Write-up I : Questions 1 to 3]
One mole of a mono-atomic gas is taken through the cycle shown in the given figure :
B C : cooling at constant volume
D A : heating at constant volume
The pressure and temperature at A, B etc. are denoted by PA, TA, PB, TB etc., respectively.
Given that
2/5
2 1 2 1
TA = 1000 K, PB PA and PC PA . Given : 0.85 , R = 8.3 J mol K 1.
3 3 3
P A
1. The work done by the gas in the process A B is : B
(a) 1260 Joules (b) 1560 Joules
(c) 1867 Joules (d) 2160 Joules D
C
## 2. The heat lost by the gas in the process B C is : V
(a) 1271 Joules (b) 1371 Joules
(c) 5291 Joules (d) 1871 Joules
3. The temperature TD is :
(a) 300 K (b) 400 K
(c) 500 K (d) 600 K
[Write-up II : Questions 4 to 6]
A gaseous mixture enclosed in a vessel of volume V consists of one gram mole of a gas A with
(= CP/CV) = 5/3 and another gas B with = 7/5 at a certain temperature T. The gram molecular
weights of the gases A and B are 4 and 32 respectively. The gases A and B do not react with
each other and are assumed to be ideal. The gaseous mixture follows the equation PV19/13 =
4. Find the number of gram moles of the gas B in the gaseous mixture.
(a) 2 (b) 4
(c) 6 (d) 8
## 5. Compute the speed of sound in the gasesous mixture at T = 300 K
(a) 200 m/s (b) 400 m/s
(c) 600 m/s (d) 800 m/s
6. If T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the
gaseous mixture.
(a) 0.17% (b) 0.27%
(c) 0.37% (d) 10.47%
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## [Write-up III : Questions 7 to 9]
The rectangular box shown in the figure has a partition which can slide without friction along the
length of the box. Initially each of two chambers of the box have one mole of a mono-atomic
ideal gas ( = 5/3) at a pressure P0, volume V0 and temperature T0. The chamber on the left is
slowly heated by an electric heater. The walls of box and partition are thermally insulated. Heat
loss through lead wire of heater is negligible. The gas in left chamber expands, pushing the
partition until the final pressure in both chambers becomes 243p0/32.
## 7. Final temperature of the gas in right chamber is
(a) 2.25 T0 (b) 4.5 T0
(c) 8.75 T0 (d) 12.93 T0
## 8. Final temperature of the gas in left chamber is
(a) 2.25 T0 (b) 4.5 T0
(c) 8.75 T0 (d) 12.93 T0
## 9. The work done by the gas in the right chamber is
(a) 5.5 T0 J (b) 10.5 T0 J
(c) 15.5 T0 J (d) None of these
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SECTION VI
LEVEL 1
(Subjective Questions)
1. The ratio of specific heat of a gas at constant pressure to that at constant volume is .
Find the change in internal energy of a mass of gas when the volume changes from V to
2V at constant pressure P ?
2. Find the molar heat capacity in a process of an ideal diatomic gas if it does a work of
Q/4 when a heat of Q is supplied to it?
3. In a room where the temperature is 30C, a body cools from 61C to 59C in 4 minutes.
Find the time taken by the body to cool from 51C to 49C .
4. A sphere of mass 164 g and diameter 6.0 cm floats in a bath in liquid at 0C. The
temperature of the liquid is gradually raised. When the temperature is 50C, the sphere
just begins to sink. If the density of the liquid at 0C is 1.49 g cm 3,find the coefficient of
cubical expansion of the liquid. Neglect the expansion of the sphere.
5. One mole of a certain ideal gas is contained under the weightless piston of a vertical
cylinder at a temperature T. The space over the piston opens into the atmosphere of
pressure p0 . What work has to be done in lifting the piston slowly so that the volume of
the gas under the piston increases isothermally n times?
## 6. The operating temperature of a tungsten filament in an incandescent lamp is 2000 K and
its emissivity is 0.30. Find the surface area of the filament of a 25 watt lamp. Stefan
constant 5.67 10 8 W m 2 K 4 .
7. A tungsten heater wire of 1 meter is rated at 3 kW m-1 and is 5.0 10 4 m in diameter. It is
embedded along the axis of a ceramic cylinder of diameter 0.12 m. When operating at
the rated power, the wire is at 1500C; the outside of the cylinder is at 20C. Find the
thermal conductivity of the ceramic.
## 8. A blackened solid copper sphere of radius 2 cm is placed in an evacuated enclosure
whose walls are kept at 100C. At what rate must energy be supplied to the sphere to
keep its temperature constant at 127C ? Stefan constant 5.67 10 8 J m 2 K 4
## 9. When a system is taken from state i to state f along the P
path iaf, it is found that Q = 50 cal and W = 20 cal. Along a f
the path ibf, Q = 36 cal (figure) (a) What is W along the
path ibf ? (b) If W = - 13 cal for the curved return path f i, i b
what is Q for this path ? (c) Take U i = 10 cal. What is
O V
Uf ?
(d) If U b = 22 cal, what is Q for the process ib and for the process bf ?
10. The ends of a meter stick are maintained at 100C and 0C. One end of a rod is
maintained at 25C. Where should its other end be touched on the meter stick so that
there is no heat current in the rod in steady state ?
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Level II
(Subjective Questions)
1. An ideal gas, whose adiabatic exponent is equal to , is expanded so that the
amount of heat transferred to the gas is equal to the decrease of internal energy.
Find
(a) The molar heat capacity of the gas in this process,
(b) The equation of the process in the variables T, V;
(c) The work performed by one mole of the gas when its volume increases
times if the initial temperature of the gas is T0 .
## 2. Find temperature as a function of radius r in case of spherical shell. Inner and
outer surfaces temperature are fixed at 1 and 2 respectively. Inner and outer
radius of shell are r1 and r2 respectively.
3. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The
outer surface areas of the two bodies are same. The two bodies emit total radiant
power at the same rate. The wavelength B corresponding to maximum spectral
radiancy in the radiation from B is shifted from the wavelength corresponding to
maximum spectral radiancy in the radiation from A by 1.00 m . If the
temperature of A is 5802 K calculate:
(a) The temperature of B and (b) wavelength B .
## 4. 20,000 J of heat energy is supplied to a metal block of mass 500 g at
atmospheric pressure. If the initial temperature of the block is 30C, find (a) the
final temperature of the block, (b) work done by the block on the surroundings
and (c) the change in internal energy of the block. Given specific heat of metal =
400 JKg 1C 1, relative density of metal = 8.0, coefficient of volume expansion of
metal = 8 10 5C 1 and atmospheric pressure = 105Pa.
CP
5. The initial pressure and volume of a given mass of gas are P0 and V0 .
CV
The gas can exchange heat with the surrounding.
V
(a) It is slowly compressed to a volume 0 and then suddenly compressed to
2
V0 / 4 . Find the final pressure.
(b) If the gas is suddenly compressed from the volume V0 to V0 / 2 and then
slowly compressed to V0 / 4 , what will be the final pressure ?
## 6. A thermally insulated vessel is divided into two
parts by a heat-insulating piston which can move in
the vessel without friction. The left part of the
vessel contains one mole of an ideal monatomic
gas, and the right part is empty. The piston is
connected to the right wall of the vessel through a
spring whose length in free state is equal to the
length of the vessel (figure). Determine the heat
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## capacity C of the system, neglecting the heat
capacities of the vessel, piston and spring.
## 7. A vertical cylinder of cross-sectional area S contains one mole of an ideal
monatomic gas under a piston of mass M. At a certain instant, a heater which
transmits to a gas an amount of heat q per unit time is switched on under the
piston. Determine the established velocity of the piston under the condition
that the gas pressure under the piston is constant and atmospheric pressure is
equal to p0 , and the gas under piston is thermally insulated.
## 8. Three rods of material x and three rods of C
material y are connected as shown in figure. All x x
the rods are of identical length and cross- 60C y 10C
x E
sectional area. If the end A is maintained at A B
60C and the junction E at 10C, calculate y y
temperature of junctions B, C and D. The D
thermal conductivity of x is 0.92 cal/cm-sC and
that of y is 0.46 cal/cm-sC.
## 9. Ice at 0C is added to 200 g of water initially at 70C in a vacuum flask. When 50
g of ice has been added and has all melted the temperature of the flask and
contents is 40C. When a further 80 g of ice has been added and has all melted
the temperature of the whole becomes 10C. Find the specific latent heat of
fusion of ice.
## 10. A hot body placed in a surrounding of temperature 0 obeys Newtons law of
d
cooling k( 0 ) . Its temperature at t = 0 is 1 . The specific heat capacity
dt
of the body is s and its mass is m. Find
(a) The maximum heat that the body can lose and
(b) The time starting from t = 0 in which it will lose 90% of this maximum
heat.
*****
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## PREVIOUS YEARS IIT-JEE PROBLEMS
(JUDGE YOURSELF AT JEE-LEVEL)
1. A solid body X of heat capacity C is kept in an atmosphere whose temperature is
TA 300 K. At time t 0 the temperature of X is T0 400 K. It cools according to
Newtons law of cooling. At time t1 , its temperature is found to be 350 K.
At this time (t1 ) , the body X is connected to a large body Y at atmospheric temperature
TA , through a conducting rod of length L, cross-sectional area A and thermal conductivity
K. The heat capacity of Y is so large that any variation in its temperature may be
neglected. The
cross-sectional area A of the connecting rod is small compared to the surface area of X.
Find the temperature of X at time t 3t1 .
## 2. Two moles of an ideal monatomic gas is taken through a P
2P1 B C
cycle ABCA as shown in the P-T diagram (figure). During
the process AB, pressure and temperature of the gas vary
such that PT = constant. If T1 300 K, calculate P1 A
(a) The work done on the gas in the process AB and T
T1 2T1
(b) The heat absorbed or released by gas in each of the process.
Give answers in terms of the gas constant R.
## 3. An ice cube of mass 0.1 kg at 0C is placed in an isolated container which is at 227C.
The specific heat capacity c of the container varies with temperature T according to the
empirical relation c = A + BT, where A = 100 cal/kg-K and B 2 10 2 cal/kg-K2. If the
final temperature of the container is 27C, determine the mass of the container.
(Latent heat of fusion for water 8 104 cal/kg, specific heat capacity of water
3
10 cal/kg-K).
4. A 5 m long cylindrical steel wire with radius 2 10 3 m is suspended vertically from a rigid
support and carries a bob of mass 100 kg at the other end. If the bob gets snapped,
calculate the change in temperature of the wire ignoring radiation losses.
(For the steel wire: Youngs modulus = 2.1 1011 Pa; Density = 7860 kg/m3; Specific heat
capacity = 420 J/kg-K).
## 5. A monatomic ideal gas of two moles is taken through a D
VD C
cyclic process starting from A as shown in the figure. The
V V V
volume ratios are B 2 and D 4 . If the temperature TA
VA AA VB B
at A is 27C, Calculate VA
A
O TA TB
T
(a) The temperature of the gas at point B,
(b) Heat absorbed or released by the gas in each process,
(c) The total work done by the gas during the complete cycle.
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## 6. The top of an insulated cylindrical container is covered by a
disc having emissivity 0.6 and conductivity 0.167 W/Km
and thickness 1 cm. The temperature is maintained by Oil out
circulating oil as shown :
(a) Find the radiation loss to the surroundings in J/m2 s Oil in
if temperatrue of the upper surface of disc is 127C,
and temperature of surroundings is 27.
(b) Also find the temperature of the circulating oil. Neglect the heat loss due to
convection.
17
Given 10 8 Wm 2 K 4
3
## 7. A diatomic gas is enclosed in a container by a movable piston of
cross-sectional area A = 1 m2 at 300 K, as shown in the figure.
The length of the gas column is 1 m. The gas is now heated to
400 K isobarically.
h=1m
(i) Find the new height of the piston.
(ii) Now the gas is compressed to its initial volume
Find the final temperature of the gas.
## 8. In the figure shown temperature of the furnace is
Insulation Ts
maintained constant at T1. Temperature of the surrounding
medium is constant at Ts. Heat comes out of the furnace T1
through a solid cylinder of thermal conductivity K, cross T2
sectional area A and length L. The temperature of the other Furnace
Insulation
end of the cylinder is T2 = Ts + T ( T < < Ts). If T (T1
Ts), find the proportionally constant. The cylinder losses the
heat to the surrounding only through radiation. Emisivity of
the cylinder is .
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SECTION I
Level I
(Single Choice Questions)
1. (d) 2. (c)
3. (a) 4. (c)
5. (a) 6. (b)
7. (b) 8. (c)
9. (c) 10. (b)
11. (a) 12. (a)
13 (b) 14. (b)
15. (c) 16. (d)
17. (b) 18. (b)
19. (a) 20. (b)
Level II
(Single Choice Questions)
1. (b) 2. (a)
3. (c) 4. (c)
5. (a) 6. (b)
7. (b) 8. (b)
9. (a) 10. (c)
SECTION II
(Other Engg Examination Questions)
1. (a) 2. (c)
3. (a) 4. (c)
5. (a) 6. (c)
7. (d) 8. (c)
9. (b) 10. (a)
11. (d) 12. (b)
13 (b) 14. (c)
15. (b)
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SECTION III
(More than One Correct Choice Questions)
1. (a, b, c) 2. (a, c)
3. (a, b, c, d) 4. (b, d)
5. (b, c) 6. (c, d)
7. (a, b, c) 8. (a, b)
## 9. (a, d) 10. (a, b, c)
SECTION IV
(Match the Columns)
## 2. (a) (p, r) ; b (p, q) ; (c) (p, s) ; (d) (p)
SECTION V
(Passage Based Problems)
1. (c) 2. (c)
3. (c) 4. (a)
5. (b) 6. (a)
7. (a) 8. (d)
9. (c)
SECTION VI
Level I
(Subjective Questions)
P(2V) P(V) PV
1. U
1 1
10
2. C R
3
3. 6 min
4. 5.5 10 4 / C
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5. (n 1) RT RT n n
6. 0.918cm 2
7. 1.77J s 1m 1 K 1
8. 1.78 J s 1
## 9. (a) 6 cal (b) 43 cal
(c) 40 cal (d) 18 cal., 18 cal
10. 25 cm from cold end
Level II
(Subjective Questions)
R 1 /2 2 RT0 [1 1/ ( 1) / 2
]
1. (a) C (b) TV constant (c) W
1 ( 1)
1 1
r1 r
2. 1 ( 1 2 )
1 1
r1 r2
3. (a) 1934 K (b) B = 1.5 m.
## 4. (a) 130C (b) 0.05 J (c) 19999.95 J
5. 2 1
P0 (both cases)
6. C 2R
2 q
7.
5 p0 S Mg
8. TB 30 C ; TC TD 20 C
9. 90 cal/g
n10
10. (a) ms( 1 0 ) (b)
k
## SUBJECTIVE UNSOLVED (IIT-JEE) LEVEL
1. Tx 300 K (12.5)e 2 KAt1 / CL
## 2. (a) Work done on gas = (1200 mol K) R
(b) QCA = (1200) Rln2
Q AB 2100R , Q BC 1500 R .
3. m 0.495 kg
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4. T = 4.568 10 3 C
5. (a) TB 600 K
(b) 162.6C
Q AB 1500 R
Q BC 831.6R
Q CD 900R ; QP A = 831.6 R
(c) Q = 600 R
6. (a) 595 watt/m2
(b) 162.6C
7. (i) 4 3 m, (ii) 448.8 K
K
8. Proportionality constant =
4 LTs3 K
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ELECTROSTATICS
7. Assignment (Subjective Problems)
LEVEL I
## 1. Point charges of magnitude q, 2q and 8q are to be placed on a 9 cm long straight
line. Find the positions where the charges should be placed such that potential
energy of this system is minimum.
## 2. Water from a metal vessel maintained at a potential of 3 volt falls in spherical
drops 2 mm in diameter through a small hole into a thin walled isolated metal
sphere of diameter 8 cm placed in air until the sphere is filled with water. Ignoring
the thickness of the metal calculate the final potential of the sphere and its
electrical energy.
3. An infinite number of charges each equal to `q are placed along the x-axis at
x = 1, x = 2, x = 4, x = 8, and so on. Find the potential and electric field at the
point x = 0 due to this set of charges. What will be the potential and electric field
if in the above set up the consecutive charges have opposite sign?
## 4. A uniform electric field of strength 106v/m is directed vertically downwards. A
particle of mass 0.01 kg and charge 10-6 coulomb is suspended by an
inextensible thread of length 1m. The particle is displaced slightly from its mean
position and released.
(a) Calculate the time period of its oscillation.
(b) What minimum velocity should be given to the particle at rest so that it
completes a full circle in a vertical plane without the thread getting slack?
(c) Calculate the maximum and minimum tensions in the thread in this situation.
5. Two equal charges q are kept fixed at a and +a along the x-axis . A particle of
q
mass m and charge is brought to the origin and given a small displacement
2
along the (a) X-axis and (b) Y-axis. Describe quantitatively the motion in two
cases.
## 6. A strip of length having linear charge density is placed near a negatively
charged particle P of mass m and charge -q (as shown in the figure) at a
distance d from the end A of the strip. Find the velocity of P as it reaches a
point at the distance d/2 from end A.
P
-q + + + + + + + + + + +
A B
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7. A thin fixed ring of radius R and positive charge Q is placed in a vertical plane.
A particle of mass m and charge q is placed at the centre of ring. If the particle
is given a small horizontal displacement, show that it executes SHM also find the
time period of small oscillations of this particle, about the centre of ring. (Ignore
gravity)
8. A non-conducting sphere having a cavity as shown R/2
in figure is uniformly charged with volume charge P
density . Find the potential at a point P which is at C R
a distance of x from C.
x
9. A particle of charge q and mass m moves along the x-axis under the action of an
electric field E = k cx, where c is a positive constant and x is distance from the
point, where particle was initially at rest.
Calculate :
(a) distance travelled by the particle before it comes to rest.
(b) acceleration at the moment, when it comes to rest.
+q
10. Charges +q and q are located at the q
## corners of a cuboid as shown in the a
figure. Find the electric potential q
energy of the system. +q
q
a +q
q +q
3a
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LEVEL II
1. (a) Two similar point charges q 1 and q2 are placed at a distance r apart in air. If a
dielectric slab of thickness t and dielectric constant k is put between the
charges, calculate the coulomb force of repulsion between them.
(b) If the thickness of the slab covers half the distance between the charges, the
coulomb repulsive force is reduced in the ratio 4 : 9. Calculate the dielectric
constant of the slab.
## 2. Calculate the electric field at the centre of a y
non-conducting semicircular ring of linear R
charge density as shown in the figure.
O x
O
## 3. A small ball of mass 2 10 3 kg having a charge of 1 C is suspended by a string
of length 0.8 m. Another identical ball having the same charge is kept at the point
of suspension. Determine the minimum horizontal velocity which should be
imparted to the lower ball so that it can make complete revolution in a vertical
circle. Take g = 10 m/s2.
## 4. Four charges are placed as shown in the cartesian y
co-ordinate system. Calculate the electric field at +q
(0,+a)
point P(0, 0, a)? B
(-a, 0) A +q
C -q (0,0, 0) (+a, 0) x
D
(0,-a) -q
## 5. Two hollow concentric conducting spheres of
30 cm
radius 10 cm and 30 cm are insulated. If the inner
sphere is given a charge 10-4 C and outer 10-6 C, P Q R
O
then calculate the intensity at the points P, Q and 10 cm
R which are situated at distance of 5 cm, 20 cm
and 90 cm from the centre respectively.
## 6. Some equi-potential y (cm)
10V 20 V 30 V 40 V
## surfaces are shown in 20 cm
30 cm
300
figure (1) and (2). 10 20 30 40 x (cm)
What can you say
10 cm
Figure 1 60 V
and direction of the
30 V
electric field? Figure 2
20 V
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7. A circular ring of radius R with uniform positive charge q is located in the y-z
plane with its centre at the origin O. A particle of mass m and positive charge q is
projected from the point P 3 R ,0,0 on the positive x-axis directly towards O,
with initial speed V. Find the smallest (non zero) value of the speed such that the
8. A very long uniformly charged thread oriented along the axis of a circle of radius
R rests on its centre with one of the ends. The charge of the thread per unit
length is . Find the flux of the vector E across the area enclosing the circle.
## 9. An Aluminium rod of length having net charge Q is
placed in front of a small charged Aluminium ball of mass /2
## m and charge -q at a distance /2 from centre. If the
space between the strip and ball offers negligible
resistance to the motion of ball then find the velocity with
which it reaches a distance x from the strip. (The charge
distribution on Aluminium rod is uniform and it is fixed).
## 10. In the arrangement shown in figure,
three concentric conducting shells are
shown. The charge on the shell of
radius b is q0 . If the innermost and
outermost shells are connected to the a
earth, find their charge densities and the b
potential on the shell of radius b in terms
of a and q0. Given that a : b : c = 1 : 2 : 4 c
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LEVEL - I
## 1. An electron of mass me, initially at rest, moves through a certain distance in a
uniform electric field in time t1. A proton of mass mp, also initially at rest, takes
time t2 to move through an equal distance in this uniform electric field. Neglecting
t
the effect of gravity, the ratio 2 is equal to
t1
1/ 2
me
(A) 1 (B)
mp
1/ 2
mp mp
(C) (D)
me me
## 2. If positively charged pendulum is
oscillating in a uniform electric field as
shown in figure. Its time period as
compared to that when it was
uncharged: + ++ + + ++
(A) will increase (B) will decrease
(C) will not change (D) will first increase then decrease
## 3. A and B are two concentric spheres If A is given a B
+ +
charge Q while B is earthed as shown in figure : ++ +
(A) The charge density of A and B are same + A +
(B) The field inside and outside A is zero +
+ + +
(C) The field between A and B is not zero
(D) The field inside and outside B is zero
4. The maximum electric field intensity on the axis of a uniformly charged ring of
charge q and radius R will be
1 q 1 2q
(A) (B)
4 0 3 3R 2 4 0 3R2
1 2q 1 3q
(C) 2
(D)
4 0 3 3R 4 0 2 2R 2
## 5. The figure is a plot of lines of force due to two
charges q1and q2. Find out the sign of charges
q1
(A) both negative
q2
(B) Upper positive and lower negative
(C) both positive
(D) upper negative and lower positive
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6. There are two concentric metal shells of radii r1 and r2(>r1). If the outer shell has
a charge q and the inner shell is grounded, the charge on the inner shell is
(A) zero (B) (r1/r2)q
(C) r1r2q (D)
## 7. Electric charge q, q and 2q are placed at the corners of an equilateral triangle
ABC of side L. The magnitude of electric dipole moment of the system is
(A) qL (B) 2qL
(C) ( 3)qL (D) 4qL
8. Three identical particles of charge Q and mass m are placed such that they form
an equilateral triangle of side . If they are released simultaneously. The
maximum speed attained by any one of the particles will be
1 1
(A) Q (B) Q
2 0m 6 0m
(C) Zero. (D) none of these.
## 9. A point charge q is placed at a point on the axis of a non-conducting circular
plate of radius r at a distance R (>>r) from its center. The electric flux associated
with the plate is
qr 2 qr 2
(A) (B)
4 0 R2 4 0R2
qR2 q
(C) (D)
4 0r 2 4 0
## 10. In the electric field due to a point charge q, a test B
charge is carried from A to the points B, C, D and E
C
lying on the same circle around q. The work done is
(A) the least along AB A
+q
D
(C) zero along any one of the paths AB, AD, AC E
and AE
(D) the least along AE.
11. Find the charge on an iron particle of mass 2.24 mg, if 0.02 % of electrons are
removed from it.
(A) -0.01996 (B) 0.01996 C
(C) 0.02 C (D) 2.0 C
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## 12. A thick shell with inner radius R and outer radius 3R
has a uniform charge density c/m3. It has a spherical
cavity of radius R as shown in the figure. The electric
R
field at the centre of the cavity is
(A) zero. (B) 2 R/ o
3R
(C) 3 R/4 o (D) 7 R/12 o
Cavity
13. The electric potential energy of a uniformly charged thin spherical shell of radius
R having a total charge Q is
(A) Q2/4 0R (B) Q2/6 0R
2
(C) Q / 8 0R (D) Q2/16 0R
14. Two small metallic spheres each of mass m are
suspended together with strings of length and placed
together. When a quantum of charge q is transferred
to each the strings make an angle of 900 with each
other. The value of q is
(A) 0 mg (B) 2 0 mg
(C) 4 0 mg (D) 8 0 mg
15. Two concentric spheres of radii R and r have similar charges with equal surface
density ( ). The electric potential at their common centre is
(A) / 0 (B) (R r )
0
0
## 16. Two connected charges of +q and -q respectively are
at a fixed distance AB apart in a non uniform electric +q
field whose lines of force are shown in the figure
The resultant effect on the two charges is A
(A) a torque in the plane of the paper and no
resultant force -q
(B) a resultant force in the plane of the paper and no
B
torque
(C) a torque normal to the plane of the paper and no
resultant force
(D) a torque normal to the plane of the paper and a
resultant force in the plane of the paper
17. An electron is accelerating in gravity free region in the absence of an electric
field. It will lose its energy in form of
(A) thermal energy (B) loss in gravitational potential energy
(C) electromagnetic radiations (D) none of the above
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18. A charge +10 9 C is located at the origin in free space and another charge Q at
(2, 0, 0). If the X-component of the electric field at (3, 1, 1) is zero. The value of
Q is
(A) + 2.4 10 10 C (B) - 6.2 10 20 C
(C) - 4.3 10 10C (D) - 1.2 10 20 C
B
19. The figure shows an infinite line charge of density A
## C/m. The work done by the electrostatic force on r
a `unit charge, when it is moved along the path
ABC, is (plane of the curve ABC contains the line C
r
charge)
(A) zero. c/m
(B) ( /2 o)ln2
(C) ( /2 o)ln3
(D) ( /2 o)ln3
## 20. A charge Q is placed at the centre of a hemispherical surface of radius R. The
flux of electric field due to charge Q, through the surface of hemisphere is
(A) Q/4 0 (B) Q/4 0
(C) Q/2 0 (D) Q/2 0
LEVEL II
## 1. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the
electric field due to the sphere at a distance r from its centre
(A) increases as r increases for r < R
(B) decreases as r increases for 0 < r <
(C) decreases as r increases for R < r <
(D) is discontinuous at r = R
2. A parallel plate capacitor with plate area A and separation d has charge Q. A slab of
dielectric constant k is inserted in space between the plates almost completely fills the
space. If E 0 and C0 be the electric field and capacitance before inserting the slab, then
E0
(A) the electric field after inserting the slab is
k
(B) the capacitance after inserting the slab is k C0
1
(C) the induced charge on the slab is Q 1
k
U0
(D) the energy stored in the capacitor becomes , U0 being the energy of the
k
capacitor before inserting the slab
## 3. In the circuit shown, a potential difference of 100 V is A 2C
applied across AB. Then C
(A) the potential difference between points C and D is 100V C C
50 V 2C
(B) the potential difference between A and C is 25 V D
B
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## (C) the potential difference between B and D is 25 V
(D) all the capacitors have equal charge
4. Three very large plates are given charges as shown in the Q 5Q 10Q
figure. If the cross-sectional area of each plate is the same,
then the final charge distribution on the surface of the plates
a, b, c, d, e, f is a b c d e f
(A) 7Q on surface e and 3Q on surface f
(B) -2Q on surface b and 3Q on surface a
(C) -7Q on surface d and 2Q on surface c A B C
(D) the magnitude of charges at all surfaces b, c, d, e is equal
5. S1 and S2 are two equipotential surfaces on which the potentials are not equal
(A) S1 and S2 cannot intersect
(B) S1 and S2 cannot both be plane surfaces
(C) in the region between S1 and S2, the field is maximum where they are closest to
each other
(D) a line of force from S1 to S2 must be perpendicular to both
6. X and Y are large, parallel conducting plates close to each other. Each face has an area
A. X is given a charge Q. Y is without any charge. Point A, B and C are as shown in the
figure
A B C
X Y
Q
(A) the field at B is
2 0A
Q
(B) the field at B is
0A
## (C) the fields at A, B and C are of the same magnitude
(D) the fields at A and C are of the same magnitude, but in opposite directions
7. In the circuit shown, the potential difference across the 3mF capacitor is V, and the
equivalent capacitance between A and B is CAB
3 F 6 F
2 F
A 60V B
18
(A) CAB = 4 F (B) CAB = F
11
(C) V = 20 V (D) V = 40 V
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8. In the circuit shown, each capacitor has a capacitance C. The emf of the cell is e. If the
switch S is closed
C S
C
C
+
(A) some charge will flow out of the positive terminal of the cell
(B) some charge will enter the positive terminal of the cell
(C) the amount of charge flowing through the cell will be Ce
(D) the amount of charge flowing though the cell will be 4/3Ce
## 9. If the flux of the electric field through a closed surface is zero,
(A the electric field must be zero everywhere on the surface
(B the electric field may be zero everywhere on the surface
(C the net charge inside the surface must be zero
(D the charge in the vicinity of the surface must be zero
10. Figure shows a closed surface which intersects a conducting sphere. If a positive
charged is placed at the point P, the flux of the electric field through the closed surface
closed conducting
surface sphere
(A) will remain zero (B) will become positive
(C) will become negative (D) will become undefined
COMPREHENSION
q1 q2
I. Two points charges q1 = + 1 C and q2 = 2 C are placed at
A and B respectively as shown in the figure. The distance A B
between q1 and q2 is 4 cm. 4cm
1. A line of force emanates from q1 making an angle 90 with AB. This line of force
(A) enters q2 at an angle 90 (B) enters q2 at an angle 60
(C) enters q2 at an angle 45 (D) does not enter q2 but goes off to
## 2. The net electric flux will be zero.
(A) over any surface that encloses a volume including A and B, but having very large
(B) over any surface that includes A twice and B once
(C) over any surface that encloses a volume excluding A and B
(D) only over a surface that encloses zero volume
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## 3. The electrostatic potential is zero at
(A) a point on the line AB between q1 and q2 but closer to q 2
(B) a point on the line AB but not between A and B
(C) infinitely many point in space
(D) no point in space
II. When an excess charge is placed on an isolated conductor, it will distribute itself on the
surface of that conductor so that all points of conductor whether on the surface or inside
the conductor becomes to same potential. This is true even if there is a cavity inside the
conductor.
If an isolated conductor is placed in an external electric field, all points of conductor still
come to a single potential regardless of whether the conductor has an excess charge.
The free electrons distribute themselves on the surface in such a way that the electric
field they produce at interior point cancels the external field that would otherwise be
there.
## 4. Choose the incorrect charge distribution
+
+ + +
+ + + + +
+ + +
+ + +
+ + +
(A) + + (B) +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+ + + + +
(C) + + (D) + + +
## 5. In electrical field, equipotential surfaces must
(A) be plane surfaces
(B) be tangential to the direction of field
(C) be spaced such that surfaces having equal difference in potential are separated by
equal distance
(D) have decreasing potential in the direction of field
## 6. At a point inside a charged conductor,
(A) the electric field and potential must be zero
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## (B) the electric field is non-zero but constant
(C) potential is zero
(D) potential is constant
## 1. A charge q is placed inside a spherical cavity, made in X
an uncharged conducting sphere of radius R as shown. R r
A point charge Q is placed at a separation d from the
centre of solid sphere as shown in the figure. Q
Oa d
q r
Column A Column B
(A) Electrostatic potential at point X inside (p) kQ kq kq kQ
the conductor
d R r' r
(B) Electrostatic potential at point O (q) kQ
r2
(C) The magnitude of the electrostatic (r) kQ
field at point X inside the conductor
r
due to the charges induced at the
outermost surface only
(D) Electrostatic potential due to all the (s) kQ kq
induced charges at point X inside the
d R
conductor
2. A parallel plate air capacitor is charged by connecting its plates to a charged capacitor
the terminals of a battery. The battery is disconnected and a
dielectric slab is introduced partially between the plates, as shown
in the figure. Consider the change in the value of each quantity
mentioned in the first column below from the time when no
dielectric slab was introduced to the time when it was, and match
it with the nature of change in it as mentioned in column on the
right.
dielectric slab
Column A Column B
(A) Surface density of charge, (p) Increases.
(B) Electric field intensity, E (q) Decreases.
(C) Charge on the capacitor, q (r) Remains same.
(D) Net force acting on either plate, F (s) Increases at some points decreases
at others.
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## 3. Three concentric conducting shells are arranged as
shown. The charges on the shells are Q, 2Q and Q C
B
respectively. A
2Q
1 RQ
Here K .
4 Q
0 2R
3R
Column I Column II
(A) Potential of B (p) KQ
3R
(B) Potential of C (q) KQ
2R
(C) Potential difference between A and C (r) KQ
6R
(D) Potential difference between B and C (s) Zero
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9. Hints (Subjective)
LEVEL- I
kq1q2
1. Potential energy =
r
dU
For U to be a minimum, =0
dx
2. q=4 0Vr
kq kq kq
3. V=
1 2 4
kq kq kq
E=
12 22 42
qE
4 (a) g g
m
(b) For minimum velocity consider the extreme case when tension just becomes
zero at the highest point A.
1 1
Conserving energy: mv B2 mv 2A mg qE 2
2 2
(c) Maximum tension is at the lowest point
q q
qx qx
1 2 2
=F 2 2
4 0 a x a x
## (b) The net force on the particle = 2 Fcos
6. Find the potential at any point and apply energy conservation law.
7. Find the field E at a distance x from the centre of charged ring on the axis.
Consider x << R.
8. Consider negative and positive charge of same volume charge density in the
cavity.
dv dv
9. F = qE, F = v
dt dx
kq1q2
10. Take all possible combination of charges and use .
r12
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LEVEL- II
## 2. Consider two symmetrical elemental charge and calculate their field at O.
kq
4. E r . Vector sum of field at P due to all the charges.
r3
k
5. Electric field due to an arc at its centre is 2 sin .
R 2
Intensity at the centre due to the circular wire is zero.
Apply principle of superposition.
6. FA FAB FAC
1 q2 60o
2 cos in the direction D to A
4 o a2 2
For equilibrium
Tcos = mg
Tsin = FA
## 7. By symmetry, horizontal components will cancel among themselves.
8.
9. Find the electric field at a distance x on the perpendicular bisector. Use
symmetry and apply kinematics principle.
10. Potential at the surface of one shell will be due to combination of other shells too.
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LEVEL I
## 1. 2q, 8q at the two ends and q at 3cm from 2q
4kq 2kq 4
2. 4800 V, 512 10-7 J 3. 2kq, , , kq
3 3 5
## 5. (a) SHM, (b) continue to move up along the Y-axis
3
q d 2 4 0 mR
6. 7. 2
2 0m d qQ
4 1 1 2k qk
8. R3 k 9. (a) (b)
3 x R 2 c m
x2
4
kq2 1 1
10. Potential energy = 2 2 4 1
a 3 5
LEVEL II
q1q 2 q
1. (a) F = 2
(b) k = 4 7.
4 0 r t t k 4 0 mR
2k R
2. 8.
R 2 0
## 3. 5.86 m/s Qq ln( / 2 x )
9.
kq i j 2 m
4. Enet 0
2a2 q0
5. 0, 2.25 x 107N/C, 1.12 x 106 N/C 10. inner ,
12 a 2
6
6. (i)200 V/m (ii) 2 q0 q0
r outer = 2
,
96 a 24 0a
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LEVEL - I
1 C 2. A
3. C 4. C
5. A 6. B
7. C 8. A
9. B 10. C
11. B 12. D
13. C 14. D
15. C 16. D
17. C 18. C
19. D 20. C
LEVEL - II
## 1. (A), (C) 2. (A), (B), (C), (D)
3. (A), (B), (C) 4. (A), (B), (C)
5. (A), (C), (D) 6. (A), (C), (D)
7. (A), (D) 8. (A), (D)
9. (B), (C) 10. (B)
COMPREHENSION
1. (B) 2. (C)
3. (C) 4. (D)
5. (D) 6. (D)
## MATCH THE FOLLOWING
1. (A) (s); (B) (s); (C) (q); (D) (p)
## 3. (A) (p); (B) (s); (C) (p); (D) (r)
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CAPACITOR
5. Assignment (Subjective Problems)
LEVEL I
## 1. Find the equivalent k1 k1
up to
P
capacitance between k2 k2
C1 C1
the ends P and Q. The
plates are of area A,
k1 k1
and the distance k2
C2 k2 C2
between them is d. The
k1 k1
dielectric constants are Q up to
k1 and k2 where k1 = 2 k2 k2
C1 C1
and k2 = 4 of material.
## 2. The plates of a parallel plate capacitor, having
V
area A, are maintained at constant potential d
difference V. If the initial separation between
the plates is d, find the work done in
increasing the separation of plates to 2d.
## 3. A 1 F and a 2 F capacitor are connected in series across a 1200 V supply.
(a) Find the charge on each capacitor and the voltage across each capacitor.
(b) The charged capacitors are disconnected from the line and from each other,
and are now reconnected with terminals of like charge connected together.
Find the final charge on each capacitor and the voltage across each
capacitor.
## 4. A capacitor is filled with two dielectrics
2 d/2 2 3
of the same dimensions but of dielectric
d
constant 2 and 3 respectively. Find the 3 d/2
ratio of capacities in the two possible A/2 A/3
arrangement.
## 5. A battery of 10 V is connected to a capacitor of capacity 0.1F. The battery is now
removed and this capacitor is connected to a second uncharged capacitor. If the
charge is equally distributed on these two capacitors, find the total energy stored
in the two capacitors. Find the ratio of final energy to the initial energy.
6. The distance between the plates of a parallel plate capacitor is 0.05m. A field of
3 x 104 V/m is established between the plates and an uncharged metal plate of
thickness 0.01 m is inserted into the capacitor parallel to its plate. Find potential
difference
(a) Before the introduction of the metal plate.
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## (b) After its introduction.
(c) What would be the potential difference if a plate of dielectric constant K = 2 is
introduced in place of metal plate?
7. Two parallel plate capacitors A and B having capacitance 1 F and 5 F are
charged separately to the same potential of 100 volt.
Now the positive plate of A is connected to the negative plate of B and negative
plate of A to the positive plate of B.
Find the final charge on each capacitors and total loss of electrical energy in the
given system.
8. Two spherical conductors of radius R and 2R, having potential 4V, and 2V are
kept isolated. Find the loss in electrostatic energy if they are connected by a
conducting wire.
A
9. Find the equivalent capacitance
between A and B, if the plates have
equal area A. d/2 k1
d
k2 k3
B
## 10. In the given circuit diagram, find the charge
which will flow through direction 1 and 2 when k C1
the key is closed.
C2
1 2
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LEVEL II
## 1. If the area of parallel plates shown in the
figure is A and they are placed at
distance d apart form each other, then
find the equivalent capacitance between
k1 k2
A and B. The two outer plates are
connected with a conducting wire.
A B
d d d
## 2. A capacitor of capacitance 0.1 F is charged until the difference in potential
between its plates is 25 V. Then the charge is shared with a second capacitor
which has air as dielectric. The potential difference falls to 15 V. If the experiment
is repeated with dielectric introduced between the plates of the second capacitor,
the potential difference is 8 V. What is the dielectric constant of the material
introduced?
## between points A and B of the circuit shown in
figure. Under what condition is it equal to zero?
C3 B C4
## 4. Two metal plates form a parallel plate k1
Metal plate
capacitor. The distance between the
d
plates is given as d. A metal plate of
thickness (d/2), and two dielectric slabs k2
of thickness (d/4) is introduced between v
the plates as shown in the figure. If the
metal plate is removed find the work
done in slowly removing it. (The plates
of capacitor is connected to a battery
having potential difference v)
C
5. Find the equivalent capacitance A
## between A and B in the circuit shown C C
C
below. If the ends A and B are C
connected across a 12 V cell, find the C C
C
electrostatic potential energy of the B
C
system. (the capacitance of each
capacitor is 100 F)
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## 6. Each capacitor has a capacitance of C
S
5 F. Find the charge that will flow through
MN when the switch S is closed.
C C
M N 50 V
## 7. In the figure shown, determine the E1
potential differences on the plates of C2
C1
capacitors C1 = 3 F, C2 = 7 F, if value of
E1 = 12kV, E2 = 13kV.
E2
## 8. Find the equivalent capacitance between A A B
and B, if the plates have equal area A and
the separation between the plates is d.
9. A uniform electric field E exists between the plates of a capacitor. The plate
length is and the separation of the plates is d.
(a) An electron and a proton start from the negative plate and positive plate
respectively and go to the opposite plates. Which of them wins this race?
(b) An electron and a proton are projected parallel to the plates from the midpoint
of the separation of plates at one end of the plates. Which of the two will have
(i) same initial velocity
(ii) same initial kinetic energy, and
(iii) same initial momentum?
## 10. Figure shows a parallel plate capacitor having
square plates of edge a and plate separation d.
The gap between the plate is filled with a
dielectric of dielectric constant k which varies
from the left plate to the right plate as k = ko + k
x, where ko and are positive constants and x
is the distance from the left end. Calculate the
capacitance.
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LEVEL - I
## 1. The equivalent capacitance between A C
and B is C
(A) C A
B
(B) 2C C
C
(C) 1.5 C
(D) none of the above C
## 2. In the circuit shown in the figure, the capacitor C is
charged to a potential Vo. The heat generated in the C
+ S
circuit when the switch S is closed, is
(A) C Vo2 (B) 2C Vo2
(C) 4C Vo2 (D) 8C Vo2 3Vo
3. The plates of a parallel plate charged capacitor are not parallel, the interface
charge density is
(A) is higher at the closer end (B) is non-uniform
(C) is higher at inclined plate. (D) none of the above
## 4. There are n identical capacitors, which are connected in parallel to a potential
difference V. These capacitors are then reconnected, in series. The potential
difference between the extreme ends is :
(A) zero (B) nV
(C) (n 1) V (D) none of the above
5. The force with which the plates of a pa0rallel plate capacitor having a charge Q
and area of each plate A, attract each other is
(A) directly proportional to Q2 and inversely to A.
(B) inversely proportional to Q2 and directly to A.
(C) does not depend upon Q2 and is inversely proportional to A.
(D) none of the above
## 6. The equivalent capacitance between points A and B 3 F 3 F 3 F
A
for the given figure is
(A) 1 F (B) 2 F 2 F 2 F 3 F
(C) 3 F (D) 4 F
B
3 F 3 F 3 F
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A
7. The equivalent capacitance between A and B is
C
(A) 6 C C
(B) 4C C
(C) 2C C C C
B
(D) none of the above
## 8. A dielectric slab of thickness 4 mm is placed between the plates of a parallel
plate capacitor. If the distance between plates is reduced by 3.5 mm, the
capacity of the capacitor remains same. Find the dielectric constant of the
medium.
(A) 2 (B) 4
(C) 6 (D) 8
## 9. The effective capacitance between A and B will be 2 F
(A) 0.5 F (B) 1.5 F A 1 F
(C) 2 F (D) 2.5 F 1 F 2 F
B
2 F
## 10. If the capacitance between two
successive plates is C, then the
B
capacitance of the equivalent system A
between A and B is
C
(A) (B) 3C
3
2 3
(C) C (D) C
3 2
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LEVEL II
1. Two capacitors are once connected in parallel and then in series. If the
equivalent capacitance in two cases are 16F and 3F respectively, then
capacitance of each capacitor is
(A) 16 F, 3F (B) 12 F, 4 F
(C) 6F, 8F (D) none of these
## 2. Two dielectrics of equal size are inserted inside a
parallel plate capacitor as shown. With what factor A
the effective capacitance increases ?
d k1 k2
k1k 2 k k2
(A) (B) 1
k1 k 2 2
2k1k 2
(C) (D) none of above
k1 k 2
10 V
3. What is the energy stored in the capacitor between
terminals a and b of the network shown in the figure?
(Capacitance of each capacitor C = 5 F). C C
(A) 1 J (B) 0.25 J a
b
C C
(C) zero. (D) 15.6 J
## 4. One of the plates of a charged parallel plate capacitor is connected to a non
conducting spring of stiffness K and the other plate is fixed. The other end of the
spring is also fixed. In equilibrium distance between the plates is d, which is twice
of the elongation in the spring. If length of the spring is halved by cutting it, the
distance between the plates in equilibrium will be (Consider that in both the
cases spring is in nature length, if the capacitor is uncharged)
3d 5d
(A) (B)
4 4
3
(C) 2d (D) d
2
## 5. Two identical parallel plate capacitors of same dimensions are connected to a
DC source in series. When one of the plates of one capacitor is brought closer to
other plate
(A) the voltage on the capacitor whose plates came closer is greater than the
voltage on the capacitor whose plates are not moved.
(B) the voltage on the capacitor whose plates came closer is smaller than the
voltage on the capacitor whose plates are not moved.
(C) the voltage on the two capacitors remain equal.
(D) the applied voltage is divided equally between the two capacitors.
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6. You are given 32 capacitors of 4 F capacitance each. How do you connect all of
them so that the effective capacitance becomes 8 F?
(A) 4 capacitors in series and 8 such groups in parallel.
(B) 2 capacitors in series and 16 such groups in parallel.
(C) 8 capacitors in series and 4 such groups in parallel.
(D) All of them in series.
## 7. Figure shows a spherical capacitor with inner sphere
earthed. The capacitance of the system is
a b
4 ab 4 b2
(A) (B)
b a b a
(C) 4 0 (b + a) (D) none of these
## 8. The charge flowing across the circuit on closing K
the key K is equal to
C C
(A) CV (B) V
2 1 2C
(C) 2CV (D) zero
V
+
## 9. The potential difference across the capacitor 3 F
of 2 F is 2 F
6 F
(A) 10 V (B) 60 V
(C) 28 V (D) 56 V 3 F
70V
## 10. What is the potential at point O? +5v
(A) 4.27 V (B) 17 V
(C) zero (D) 34V 4 F 1 F
O
+6v +3v
3 F 3 F
+3v
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7. Hints (Subjective)
LEVEL- I
1. Find the value of C1 and C2 and analyse it as the series parallel combination.
2. The capacitance changes so the energy changes.
3. The charges in series combination are equal.
4. In first case, capacitors in series and in second case they are in parallel.
5. Total charge remains conserved.
6. If capacitance changes, the electrostatic energy changes.
7. Find the equivalent capacitance and charge distribution.
8. Find common potential and electrostatic potential energy.
9. Capacitance with dielectrics k 2 , k 3 are in parallel and capacitor with dielectric
## k1 is in series with the above combination.
10. Compare the charge distribution in both the cases.
LEVEL- II
## 1. Find individual capacitance and arrange it in the circuit.
2. The total charge has to be conserved.
3. Charge on C1 and C2 will be equal and C3 & C4 will be equal.
4. Find out total energy in both the cases and difference between them.
5. Apply series parallel concept.
6. Find the equivalent capacitance
7. Apply KVL and charge conservation.
8. Potential on extreme plates will be equal and potential on inner plates will be
equal.
9. Force on the charges is due to electric field. Find accelerations. Use kinematics.
10. Take an elemental capacitance of thickness dx. Write capacitance for this then
integrate.
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## 8. Answers to the subjective Assignment
LEVEL I
2
1 0 AV
1. Ceq = 1.07 0A /d 2. W=
4 d
3200
3. (a) 800 C, 800V, 800 C, 400V. (b) 1600/3V,1600/3 C, C
3
24 1
4. 5. 2.5J,
25 2
6. (a) 1500 Volt (b) 1200 Volt (c) 1350 Volt
2
5 16 0Rv
7. 0.6 x 10-4C; 3.33 x 10-4C; x 10 2
J 8.
3 3
2A 0 (k 2 k 3 )(k1 )
9.
d 2k1 k 2 k 3
c 1c 2
10. Charge flown through path 2= -
c1 c2
Charge flown through path 1= c2
LEVEL II
A 0 k1k 2 k1 k 2
1. 2. 3.2
d (k1 k 2 )
C2C3 C1C4
3. A B E , when C1/C2 = C3/C4.
C1 C2 C3 C4
4 A 0 v 2k12k 22
4.
d(k1 k 2 )(k1 k 2 2k1k 2 )
## 5. 125 F, 9000 J 6. 333.3 C
2 0A
7. 700 V, 300V 8.
d
9. (a) Electron (b) (i) Electron (ii) Both equal deviation (iii) Proton
2
0a
10.
d
ln 1
K0
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## 9. Answers to the Objective Assignment
LEVEL - I
1. A 2. D
3. A 4. B
5. A 6. A
7. C 8. D
9. C 10. B
LEVEL - II
1. B 2. B
3. C 4. B
5. B 6. A
7. B 8. B
9. B 10. A
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CURRENT ELECTRICITY
10. Assignment (Subjective Problems)
LEVEL I
## 1. A wire of resistance 15 is bent to form a regular hexagon ABCDEFA. Find the
equivalent resistance of the loop between the points (a) A and B, (b) A and C
and (c) A and D.
## 2. Find the potential difference Va - Vb E1 R1
in the circuits shown in figure
a b
R3
E2 R2
2 4A 2
3. Find the P.D. between points A and
B in the branch of a circuit shown 5v
9v
in figure. Which point is at higher
potential A B
## 4. In the circuit shown in figure 4000 6000
V1 and V2 are two voltmeters having
S
resistances 6000 and 4000 respectively
V1 V2
E.M.F. of the battery is 250 volts, having
6000 4000
negligible internal resistance. Two
resistances R1 and R2 are 4000 and E=250V
6000 respectively. Find the reading of the
voltmeters V1 and V2 when
(i) Switch S is open
(ii) Switch S is closed
## 5. A galvanometer of resistance 95 , shunted by a resistance of 50 ohm gives a
deflection of 50 divisions when joined in series with a resistance of 20 k and a
2 volt battery, what is the current sensitivity of galvanometer (in div/ A) ?
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## 6. A part of a circuit in steady state 1 amp
3
along with current flowing in the 2 6 D
branches, with value of each A I1
2 amp
resistance is shown in figure. 4 F
2
3 I3
Calculate the energy stored in the 2 amp I2
B C 4
capacitor C. 3
6
1 amp
## 7. Calculate the current through 3 resistor 3
and the power dissipated in the entire
4
circuit shown in figure. The emf of the 8
battery is 1.8 V and its internal resistance 2 6
is 2/3 .
6
## 8. A capacitor of capacitance 10 F is connected to a battery of emf 2V. It is found
that it takes 50 ms for the charge on the capacitor to become 12.6 C. Find the
resistance of the circuit.
## 9. Three 60 W 120 V light bulbs are A
connected across a 120 V power line
shown in figure. Find (a) the voltage
B C
across each bulbs (b) the total power
dissipated in the three bulbs.
## 10. A heater is designed to operate with a 10 B C
power of 1000 watts in a 100 volt line.
It is connected in combination with a
R
resistance R, to a 100 volt mains as
shown in figure what should be the
100 V
value of R so that the heater may
operate with a power of 62.5 watts.
11. Two resistors 400 ohm and 800 ohm are connected in series with a 6V battery. It
is desired to measure the current in the circuit. An ammeter of10 ohm resistance
is used for this purpose. What will be the reading in the ammeter? Similarly if a
voltmeter of 10, 000 ohm resistance is used to measure the potential difference
across 400 ohm, what will be the reading of the voltmeter?
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## 12. Two cells, having emf. of 10 V and 8V respectively, are r2
10 r1 8V
connected in series with a resistance of 24 in the V
external circuit.
If the internal resistances of each of these cells in ohm
are 200% of the value of their emf respectively, find the
R = 24
current in the circuit.
## 13. A galvanometer having 50 divisions provided with a variable shunt S is used to
measure the current when connected in series with a resistance of 90 and a
battery of internal resistance 10 . It is observed that when the shunt resistances
are10 and 50 respectively, the deflection are respectively 9 and 30 divisions.
What is the resistance of the galvanometer ?
## 14. Find the current flowing through the branch B 10 V
AC in the steady state as also the charge on
the capacitor C. If the externally applied R
potential are now withdrawn, how will the R
charge on the capacitor vary as a function of
A C
time? (R = 1k , C = 10 F) 5V R 0V
## 15. In the circuit shown in figure, R1 = 1 , R2 = 2 , R2
C1 = 1 F, C2 = 2 F and E = 6V. Calculate charge R1 C1
## on each capacitor in steady state. C2
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LEVEL II
## 1. A battery of emf 1.4 V and internal resistance 2 is connected to a
100 resistor through an ammeter. The resistance of the ammeter is 4/3 . A
voltmeter is also connected to find the potential difference across the resistor.
(i) Draw the circuit diagram.
(ii) The ammeter reads 0.02A. What is the resistance of the voltmeter?
(iii The voltmeter reads 1.10 V. What is the error in reading?
## 2. Calculate the potential difference 3 F 1 F
B
between the points A and B between
the points B and C of figure in steady 3 F 1 F
state.
1 F
A 20 10 C
100 V
3. Find the equivalent resistance of the circuits shown in figure between the points a
and b. Each resistor has a resistance r.
a b
a b
## 4. In network shown in figure below 6 6V
A
calculate potential difference
between A and B. 4V 2 3 1
4V
B
## 5. In the network of resistors each of B
resistance R as shown in the figure,
calculate the equivalent resistance F
between the junctions A and E.
E G
A C
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## same resistance R = 11 and C = 2 F. They R
R
are connected through a battery of 10 V. R R
R
When cell is switched on, find
(i) maximum current in the circuit R R
(ii) energy stored in capacitor after time t.
V
## 7. In the given circuit (see fig.), E1 = 3 volts, R1 E1
E2 = 2 volts, E3 = 6 volts, R1 = 6 , R2 = 2
R3 = 4 , R4 = 3 and C = 5 F. Find the C R2 R3
current in R3 and energy stored in the
R4
E3
## 8. Find how the voltage across the capacitor C
S
varies with time t (figure) after the shorting
of the switch S at the moment t = 0 E R C
## 9. In the circuit shown in figure, a voltmeter 30 V
reads 30 V when it is connected across V
300
400 resistance. Calculate what the
same voltmeter will read when it is 400
connected across the 300 resistance.
60 V
10. An ammeter and a voltmeter are connected in series to a battery with emf
E = 6.0 V. When a certain resistance is connected in parallel with the voltmeter,
the reading of the latter decreases = 2.0 times, whereas the reading of the
ammeter increases by the same factor. Find the voltmeter reading after the
connection of the resistance.
## 11. Three equal resistances each of R ohm R
A C B
are connected as shown in figure. A D
R R
battery of 2 V and internal resistance
0.1 is connected across the circuit.
Calculate the value of R for which the
2V, 0.1
heat generated in the circuit is maximum.
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## equal to R1 = 10 . R2 = 20 and R3 = 30 and 1
R1
O
the potentials of the points 1, 2, and 3 are equal R3
to 1 = 10V, 2 = 6V and 3 = 5V.
3
## 13. Find a potential difference A - B between the plates R3 C
of a capacitor C in the circuit shown in figure. If the A B
sources have emf's E1 = 4.0 V and E2 = 1.0 V and the E2
R2 R1
resistances are equal to R1 = 10 , R2 = 20 , and R3
= 30 . The internal resistances of the sources are
negligible E1
## 14. A constant voltage V = 25 V is maintained R1 C R2
between points A and B of the circuit (figure).
Find the magnitude and direction of the current A B
flowing through the segment CD if the
R3 R4
resistances are equal to R1 = 1.0 , R2 = 2.0 ,
R3 = 3.0 and R4 = 4.0 . D
## 15. Consider an "alternator chain" shown below
R R
A
2R 2R
B
R R
Find the equivalent resistance between A and B.
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LEVEL I
## 1. By increasing the temperature the resistance of a conductor
(A) increases. (B) decreases.
(C) remains constant. (D) initially increases then decreases.
## 2. A battery of emf V volts, resistance R1
R1 R2
& R2, a condenser C and switches S1
and S2 are connected in a circuit as C
V
shown in figure. The capacitor C gets S2
S1
fully charged to V volts when
(A) S1 & S2 are both closed (B) S1 & S2 are both open
(C) S1 closed and S2 open (D) S2 closed & S1 open.
## 3. The equivalent resistance of
a circuit between A and B is A B
3 6 6
## (A) 3 (B) zero.
3
(C) 6 (D)
2
4. In the circuit shown in figure, the voltmeter reading
would be
3V
(A) 0 volt. (B) 0.5 volt. A V
(C) 1 volt. (D) 2 volt. 1 2
## 5. The V-I graph for a conductor at temperatures T1 and
T2
T2 are as shown in the figure, (T2 T1) is proportional
to V T1
(A) cos 2 (B) sin 2
(C) cot2 (D) tan 2
I
## 6. If a copper wire is stretched to make it 0.1 % longer. The percentage change in
its resistance is
(A) 0.2 % increase (B) 0.2% decrease
(C) 0.1 % increase (D) 0.1 % decrease
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R1 R2
7. In the circuit shown in figure E1 = 7V, E2 = 7V R1 = R2 =
1 and R3 = 3 respectively. The current through the
E1 E2
resistance R3 is R3
(A) 2A
(B) 3.5A
(C) 1.75A
(D) none of these
## 8. The figure shows in apart of an electric circuit, then 5A
I
the current I is
6A 1A
(A) 1 A (B) 3A
(C) 2 A (D) 4 A 2A
3A
## 9. A cell of e.m.f E and internal resistance r is connected across a resistance r. The
potential difference between the terminals of the cell must be
(A) E (B) E/2
(C) E/4 (D) 3E/2
## 10. A conductor carries a current of 50 A if the area of cross-section of the
conductor is 50 mm2 then value of the current density in Am 2 is
(A) 0.5 (B) 1
1
(C) 10 (D) 3/10
500
11. In the adjoining circuit, the battery E1 has an G
E.M.F. of 12 volts and zero internal resistance.
E1 X E2
While the battery E2 has an E.M.F. of 2 volts if
the galvanometer G reads zero than the value
of the resistance X in ohms is
(A) 10 (B) 100
(C) 14 (D) 200
## 12. Nine similar resistors of resistance R are connected as A
shown in the figure. Equivalent resistance between points A
and B is
3 4
(A) R (B) R B
5 3
9
(C) R (D) R
5
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1A
13. Referring to the adjoining circuit which of the following is/are A
R
true
(A) R = 80 ohms
4
(B) R = 6 ohms
(C) R = 10 ohms
(D) Potential difference between points A & E is 2V
14. In a gas discharge tube if 3 1018 electrons are flowing per sec from left to right
and 2 1018 protons are flowing per second from right to left a given cross
section the magnitude and direction of current through the cross section
(A) 0.48A, left to right (B) 0.48 A, right
(C) 0.80A, left to right (D) 0.80 A, right to left
15. In a portion of some large electrical network, current in certain branches are
known. The values of (VA - VB) and (VC - VD) are X and Y respectively. Where x
and y are
7A 5V
2 4 6
A
3V 4V 2A B
8 10
9V 3A
D
C
## (A) X = 29 V, Y = 26 V (B) X = 58V, Y = 52 V
(C) X = 58 V, Y = 52 V (D) X = 29 V, Y = 26 V
16. A 50 V battery is connected across a 10 resistor and a current of 4.76 A flows. The
internal resistance of the battery is
(A) 0.5 (B) 0.1
(C) 0.2 (D) 0.3
17. For the circuit shown which of the following statements is true ?
V 1 = 30V V2 = 20V
S1 S3 S2
+ +
C1 = 2pF C2 = 3pF
## (A) with S1 closed, V1 = 1 5 V, V2 = 20 V
(B) with S3 closed V1 = V2 = 25 V
(C) with S1 & S2 closed, V1 = V2 = 0
(D) with S1 and S3 closed, V1 = 30 V, V2 = 20 V
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18. In the given circuit ammeter and voltmeter are ideal and 3 6 V,1
## battery of 6V has internal resistance 1 . The reading of
V 3
voltmeter and ammeter is
3
4 4 A
(A) zero, (B) V, zero
3 3
(C) 6 A, 0.1 A (D) 3.6 V, 0.6A
19. The maximum power dissipated by external resistance R by a cell of an external
emf E and internal resistance r is E2/4R which is obtained for
(A) R < r (B) R > r
(C) R = r (D) any value of R.
## 20. An electric current of 16 A exists in a metal wire of cross section 10 6 m2 and
length 1m. Assuming one free electrons per atom. The drift speed of the free
electrons in the wire will be
(Density of metal =5 103 kg/m3, atomic weight = 60)
(A) 5 10 3 m/s (B) 2 10 3 m/s
(C) 4 10 3 m/s (D) 7.5 10 3 m/s
## 21. A circuit consists of a resistance R connected to n similar cells. If the current in
the circuit is the same whether the cells are connected in series or in parallel
then the internal resistance r of each cell is given by
(A) r = R/n (B) r = nR
(C) r = R (D) r = 1/R
22. A cell of e.m.f. E and internal resistance r is connected in series with an external
resistance nr then the ratio of the terminal potential difference to E.M.F. is
1
(A) 1/n (B)
n 1
n n 1
(C) (D)
n 1 n
## 23. To measure a potential difference across a resistor of resistance R a voltmeter
of resistance Rv is used. To measure the potential with a minimum accuracy of
95 % then
(A) Rv = 5R (B) Rv = 15 R
(C) Rv = 10 R (D) Rv 19 R
24. A cell of E.M.F. E and internal resistance r supplies currents for the same time t
through external resistance R1 and R2 respectively. If the heat produced in both
cases is the same then the internal resistance is
1 1 1 R R2
(A) (B) r = 1
r R1 R 2 2
(C) r = R1R 2 (D) r = R1 + R2
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## 25. A battery of 10 volt is connected to a resistance of 20 ohm through a variable
resistance R, the amount of charge which has passed in the circuit in 4 minutes,
if the variable resistance R is increased at the rate of 5 ohm/min.
(A) 120 coulomb (B) 120 loge2 coulomb
120 60
(C) coulomb (D) coulomb
loge 2 loge 2
## 26. ABCD is a square of side 'a' metres and is made of I2 ax
A B
wires of resistance x ohms/metre. Similar wires are I1
connected across the diagonals AC & BD. The ax ax
effective resistance between the corners A & C will be
a 2x
(A) (2 2) ax (B) 6ax a 2x
## (C) 3ax (D) (3 2) ax D ax C
27. To get maximum current in a resistance of 3 , one can use n rows of m cells
(connected in series) connected in parallel. If the total no. of cells is 24 and the
internal resistance is 0.5 ohm then
(A) m = 12, n = 2 (B) m = 8, n = 3
(C) m = 2, n = 12 (D) m = 6, n = 4
28. In the circuit below the resistance R has a value that
depends on the current. Specifically, R is 20 ohms
R
when I is zero and the increase in resistance in ohms is 250 V
numerically equal to one half of the current in amperes.
What is the value of current I in circuit ?
(A) 8.33 amp (B) 10 amp
(C) 12.5 amp (D) 18.5 amp
29. The potential difference between points A & B in a section of a circuit shown is
(A) 5 volts (B) 1 volts
(C) zero volts (D) 13 volts
1 amp. 2 amp.
2 2 2 2
1 B
3 amp. A 1
3V 2V
## 30. Two cells of the same e.m.f. e but different internal + - + -
resistances r1 and r2 are connected in series with an e, R1 e, R2
external resistance 'R'. the potential drop across the
first cell is found to be zero. The external resistance
R is R
(A) r1 - r2 (B) r1 / r2
(C) r1 r2 (D) r1 + r2
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LEVEL II
## 1. In the circuit shown in the figure the point F is 5
A G C
grounded as shown. Which of the following is
a wrong statement? 2 3
(A) Potential at E is zero
(B) Potential at D is zero A E
10 V 3V
(C) The current in the circuit will be 0.5 A G 4
F
(D) The current in the circuit is same whether or D
not F is grounded
## 2. Ohms law may be expressed as
(A) potential difference, V I (electric current)
(B) electric current density, J E (electric field)
v
(C) , symbols having usual meaning
I A
v
(D) resistance, R constant
I
## 3. A wire frame in the form of a tetrahedron ABCD is connected to
a dc source. The resistance of all the edges of the tetrahedron
are equal. Choose the correct statement (s)
(A) the electric current through DC is zero
(B) if we remove anyone of the edges AD, AC, DB or BC, the
change in current remain same
(C) the current through AB is maximum
(D) the change in current will be same if AB or DC is removed
## points A, B, C and D are marked on the ring as shown in the figure.
A battery of emf E may be connected across the ring in two ways : B D
Case I : Battery is connected between B and D
Case II : Battery is connected between A and B.
Choose the correct statement(s)
(A) In each case same current flows out of the battery A
(B) In case I more current comes out of the battery
(C) In case II more current comes out of the battery
(D) In case I resistance is more than in case II
5. A piece of germanium (material used for making semiconductors) and a piece of copper
(material used for making conducting wires) are cooled from room temperature to 85 K.
R1 and R2 be the resistance of the pieces respectively and 1 and 2 are the temperature
coefficients of resistance of materials respectively.
(A) R1 decreases and R2 increases (B) R1 increases and R2 decreases
(C) 1 < 0, 2 >0 (D) 1 > 0, 2 >0
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6. A number of resistors R1, R2, R3, are connected in series such that Rs is the
equivalent resistance of series combination. A current I is flowing in the circuit due to a
potential V applied across the circuit. V 1, V2, V3, .. are potentials across R1, R2,
R3, respectively.
(A) same current I will flow through each resistor
(B) V1 + V2 + V3 + .. = V
R1 R2 R3
(C) V1 = V, V2 V, V3 V ;
RS RS RS
(D) data insufficient
7. A circuit has an equivalent resistance R0. A voltmeter of resistance Ru is applied across
the circuit to measure the potential drop across R 0. The new equivalent resistance of the
circuit is
R 0R
(A) R0 (for R0 < < R ) (B)
R0 R
(C) R0 + R (D) data insufficient
## 8. In the circuit shown,
2
0.2 F
6V 2.8
(A) a current of 0.9 A flows through 2 resistor when steady state is reached
(B) a potential drop of 4.2 V appears across the resistance 2.8
(C) a potential drop of 1.8 V appears across the capacitor C
(D) a potential 4.2 V appears across the capacitor C
9. In the given circuit the point A is 9 V higher than point B
A B C D
6V 15V 24V
1 2 1
R
(A) R = 1 (B) R = 7
(C) potential difference between B and D is 30 V
(D) potential difference between B and C is 15 V
## 10. Two identical fuses are rated at 10 A. If they are joined
(A) in parallel, the combination acts as a fuse of rating 20 A
(B) in parallel, the combination acts as a fuse of rating 5 A
(C) in series, the combination acts as a fuse of rating 10 A
(D) in series, the combination acts as a fuse of rating 20 A
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COMPREHENSION
## I. A frame is formed by nine identical wires of C R F
R R
resistance R each as shown in the figure. R R
A R B
R R R
D E
5 1
(A) R (B) R
11 11
15 6
(C) R (D) R
11 11
## 2. Choose the incorrect statement.
(A) the ratio of current through CF and DE is 1
(B) the ratio of current through CD and FE is 1
(C) the ratio of current through AC and AD is 6/5
(D) the ratio of current through AC and AD is 5/6
## 3. Choose the correct statement.
(A) C and D are at same potential (B) VC + VE = VA + VB
(C) VD + VF < VA + VB (D) E and C are at same potential
## II. The figure shows an potentiometer arrangement. The E = 10V
uniform wire AB (shown in the figure) has a total r=1
resistance of 99 . (The galvanometer shown in figure
has zero resistance) 99 cm
A B
1 Jockey
G
cell
4. When jockey is touchal with wire AB at point C such that AC = 40 cm. There is no
current in galvanometer. So emf of cell is
(A) 4 V (B) 5/4 V
(C) 5 V (D) 6 V
## 5. If jockey is touched at point B then potential difference across uniform wire AB is
(A) 6.93 volt (B) 8.93 volt
(C) 9.93 volt (D) None of the above
## 6. If jockey is touched at point A then potential difference across the cell is
(A) 3 volt (B) 2.5 volt
(C) 2 volt (D) None of the above
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## 1. A battery has an emf E and internal resistance r. A variable resistor R is connected
across the terminals of the battery.
Column I Column II
(A) Current in the circuit is maximum (p) R
when
(B) Potential difference across the (q) R=0
terminals is maximum when
(C) Power delivered to the resistor is (r) E
maximum when i
r
(D) Power delivered to the load is zero (s) r=R
when
(Item of Column I can match one or more than one item in Column II)
2. Assume a tetrahedron ABCD made up of conducting wires, r D
C
each side of which has resistance r. Now match the following r
column : r r
r
A r B
Column I Column II
(A) Two dimensional equivalent of the (p) D
tetrahedron r
r C r
r r
A r B
(B) Equivalent resistance between C and (q) D
r r r
D r
2
A r r
B
C
r
(C) If a battery is connected between any (r) C
two points, the potentials of the other A B
two points are always equal
D
r
(D) If a battery is connected between any (s) C
two points, the electric current coming
out is same
A B
3. Column I shows some circuits and Column II its equivalent resistance between 1 and 2.
(All resistances are equal to R)
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Column I Column II
(A) 1 (p) 8
R
7
2
(B) (q) R
1 2
(C) (r) R
2
1 2
(D) (s) 2R
1 2
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LEVEL- I
## 1. Apply series parallel concept.
2. Apply KVL and KCL.
3. Apply KVL from A to B.
4. When switch is open V1 and V2 are in series.
5. Find current through Galvanometer
6. Apply KVL, KCL
7. Note that 3, 8 + 6, 2 + 4 and 6 ohm are in parallel.
8. Apply equation for charging of capacitor.
9. Find out resistance of each bulb.
10. Find out resistance of heater, then current through it.
11. Total resistance of the circuit equals to (400 + 800 + 10) ohm.
13. Apply Galvanometer formula. Deflection is proportional to current.
14. For the capacitor charging resistance equals to 2R R / 2R + R, Pd = 5 V.
LEVEL- II
## 1. First draw diagram.
2. Total potential difference across 3 F and 1 F in the upper branch which are in
series equals to 100 V.
3. Is there a hidden Wheatstone bridge somewhere ?
4. Current flowing in a each loop is independent of the other loop. Find the current
then apply KVL.
5. Note symmetry of the circuit and equipotential points.
6. Write basic differential equation for charging of each capacitors by 18 KVL. First
find the potential difference across C at steady state, time constant equals to
RC
.
2
7. At steady state no current in C. Then apply KVL to remaining circuit.
9. From the data it is clear that voltmeter resistance is such that equivalent
resistance of 400 ohm and Rv is 300 ohm.
10. Apply the concept of shunt resistance.
11. Make A and D as a single point and B and C as common point.
12. Assume potential of point O is 0 and write KCL for point O.
13. No current through C. Apply KVL to remaining circuit.
14. CD has some resistance. Current through CD equals to current through R1
current through R2.
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## 13. Answers to the Subjective Assignment
LEVEL I
E1 E2
R1 R2
1. (a) 2.08 , (b) 3.33 , (c) 3.75 2.
1 1 1
R1 R2 R3
## 3. (i) vA-vB= 12 v, (ii) A 4. (i) 150 V, 100 V (ii) 125 V, 125 V
1 DIv
5. 6. 1.8 10 3 J
2 A
7. 0.4 A, 1.62 W 8. 5k
## 11. 4.96 mA, 1.95 V 12. 0.3 A.
13. 233
t
14. 5 mA, 50 C, q(t) = 50 C e 6. 67ms 15. q1=2 C, q2 = 12 C
LEVEL II
## 1. 200 ohm, 0.23 V 2. 25 V, 75 V
r 4r
3. , 4. - 0.5 V
2 5
7R 1
5. 6. .666 A , CV 2 (1 e t / RC 2
)
12 2
2t
-6 E RC
7. 1.5 A, 14.4 10 J. 8. 1 e
2
9. 22.4 V 10. 2V
11. 0.3 12. 0.202 A
13. - 1V 14. 1A
15. ( 5 + 1)R
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LEVEL I
1. A 2. C
3. D 4. A
5. C 6. A
7. A 8. B
9. B 10. B
11. B 12. A
13. B 14. D
15. C 16. A
17. D 18. D
19. C 20. B
21. C 22. C
23. D 24. C
25. B 26. A
27. A 28. B
29. D 30. A
LEVEL II
## 1. (A), (B) 2. (A), (B), (D)
3. (A), (B), (C) 4. (C), (D)
5. (A), (C) 6. (A),(B), (C)
7. (A), (B) 8. (A), (B), (C)
9. (B), (C) 10. (A), (C)
COMPREHENSION
1. (C) 2. (C)
3. (B) 4. (A)
5. (A) 6. (D)
MATCH THE FOLLOWING
1. (A) (q), (r); (B) (p); (C) (s); (D) (q), (p), (r)
2. (A) (p), (q), (r), (s); (B) (p), (q), (r), (s); (C) (p), (q), (r), (s); (D) (p), (q), (r), (s)
3. (A) (q); (B) (r); (C) (r); (D) (p)
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MAGNETICS
8. Subjective Problems
LEVEL I
1. Two straight infinitely long and thin parallel wires are spaced 0.1 m apart and
carry a current of 10 ampere each. Find the magnetic field at a point which is at a
distance of 0.1 m from both wires in the two cases when the currents are in the
(a) same and (b) opposite directions (Given o = 4 10-7 Tm/A).
2. A beam of protons with a velocity 4 105 m/sec enters a uniform magnetic field of
0.3 Tesla at an angle of 60o to the magnetic field. Find the radius of the helical path
taken by the proton beam. Also find the pitch of the helix, which is the distance
travelled by a proton in the beam parallel to the magnetic field during one period of
rotation. [Mass of proton =1.67 10-27 Kg, charge on proton = 1.6 10-19 C]
3. A long horizontal wire P carries a current of 50A. It is rigidly fixed. Another fine
wire Q is placed directly above and parallel to P. The weight of wire Q is
0.075 N/m and carries a current of 25A. Find the position of wire Q from P so that
the wire Q remains suspended.
## 4. A circular coil of average radius 6 cm has 20 turns. A current of 1.0 A passes
through it. Calculate the magnetic induction at
(a) the centre of the coil
(b) at a point on the axis 8 cm away from the centre.
5. (a) A proton is moving in a magnetic field. The field B is into the plane of the
page. The velocity vector v lies in the plane of the page, perpendicular to B .
Describe the motion of proton.
(b) In part (i), if the radius of the circle is 0.5m and the magnitude of the magnetic
field is 1.2 Wbm-2 , find the frequency of revolution and the kinetic energy of
the proton. Charge of the proton = 1.60 10-19 C. Mass of the
-27
proton=1.67 10 kg.
B
6. In the framework of wires shown in figure, a i
## current of i amperes is allowed to flow. A
R2
Calculate the magnetic induction at the centre
O. If angle is equal to 90 , then what will be O
the value of magnetic induction at O ? R1
E i
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7. A loop of flexible conducting wire of length 0.5 m lies in a magnetic field of 1.0T
perpendicular to the plane of the loop. Show that when a current is passed
through the loop, it opens into a circle. Also calculate the tension developed in
the wire if the current is 1.57 amp.
8. A beam of protons move undeviated through a region of space having uniform
transverse electric and magnetic fields. These fields are mutually perpendicular
and their values are 120kv/m and 50 mT respectively. If this beam strikes a
grounded target, then what will be the force exerted by the beam on the larger.
Given that beam current is equal to I = 0.8mA and mass of the
proton= 1.673 10 27 kg .
## 9. A wire is bent in the form of a circular arc with a
straight portion AB. If current flowing in the wire is i O
i, find the magnetic induction at the centre O. 2 R
A B
C
## 10. Show that the force on a wire between a y
and b of arbitrary shape of figure is the F
x
same as force on the straight wire between
b
the same two points when they carry the
same current from a to b and are placed in
the same magnetic field. Also find the force.
a
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LEVEL II
## 1. A framework AKLA forming three sides of a square is made B
from a copper wire having cross-sectional area 2.5 10-6 A A
2 i
meter . The framework can turn about a horizontal axis AA, i
## as shown in figure. The wire is placed in a vertical uniform
magnetic field. If on passing a current of 16 amp, through the
wire, the framework deflects through an angle = 20o, then K i
what is the value of magnetic induction? Given that density of L
K L
copper = 8.9 gm/c.c.
x
2. The magnetic field existing in a region is given by B Bo 1 k . A square
l
loop of edge and carrying a current i, is placed with its edges parallel to the
X-Y axes. Find the magnitude of the net magnetic force experienced by the loop.
## 3. A regular polygon of n sides is formed by bending a wire of total length 2 r which
carries a current i. (a) Find the magnetic field B at the centre of the polygon. (b)
By letting n , deduce the expression of the magnetic field at the centre of a
circular current.
y
4. A wire loop carrying a current I is placed in the x-y plane as M
shown in the figure. a v x
## (a) If a particle with charge +Q and mass m is placed at 120
o
I R +Q
the centre P and given a velocity v along NP, find its P
o
30
instantaneous acceleration.
(b) If an external uniform magnetic induction field B B i N
is applied, find the force and the torque acting on the
loop due to this field.
## 5. Two long straight parallel wires are 2 meters apart, perpendicular A
x
to the plane of the paper. The wire A carries a current of 9.6A,
1.6m
directed into the plane of the paper. The wire B carries a current 2m
## such that the magnetic field of induction at the point P, at a
S
distance of (10/11) m from the wire B, is zero. Find B
1.2m
## (a) the magnitude and direction of the current in B, (10/11)m
(b) the magnitude of the magnetic field induction at the point S, P
(c) the force per unit length on the wire B.
## 6. A very long wire bent at right angle at O. Now current P
d
I = 10 A is passed in the wire. Find the magnetic
induction at a point P lying on the perpendicular to O
the wire at O at a distance d = 35 cm.
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## 7. In the adjoining diagram, a current-carrying loop pqrs is i1
q
p
placed with its sides parallel to a long current-carrying wire.
The currents i1 and i2 in the wire and loop are 20 A and 16 A a
i2
respectively.
If a = 15 cm, b = 6 cm and d = 4 cm, what will be the force
s r
on current-loop pqrs? What will be the difference in the d b
force, if the current i2 in the loop becomes clockwise instead
of anticlockwise?
y
8. A current i = 5 amp flows through a thin wire as shown in
D
the figure. C
## (a) Find the magnetic field produced by the current at point E
G 10cm
O in the figure. F O A
x
## (b) If there exists an external magnetic field 20 cm J
B 14 i 14 j T, calculate the torque acting on the wire. H I
## 9. Two long parallel wires carrying currents 2.5 A P Q
i2=IA
and I A in the same direction (directed into i2=2.5A
x
R v
x x
plane of the paper) are held at P and Q
respectively such that they are perpendicular to r2=2m
the plane of paper. The points P and Q are r1=5m B
located at distance of 5 meters and 2 meters
respectively from a collinear point R.
(a) An electron moving with a velocity of 4 105 m/s along the positive x-
direction experiences a force of magnitude 3.2 10-20 N at the point R. Find
the value of I.
(b) Find all the positions at which a third long parallel wire carrying a current of
magnitude 2.5 amperes may be placed so that magnetic induction at R is zero.
## 10. The diagram shows combination of three E
(2)
y
R (1)
cuboidal spaces (1), (2) and (3). Space 1 V
Q
0
O x
and 3 contain electric field E as shown 90 P
## while space 2 has magnetic field B. A L R/2
(3)
particle of charge q and mass m is
E
projected as shown with velocity V0cos i +
R
V0 sin j .
Find the value of E, so that this particle enters the magnetic field parallel to the x-
axis and just passes through point P along the electric field at that point. Find its
speed at P. (Neglect the effect of gravity)
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9. Objective Problems
LEVEL I
1. Two thin long parallel wires separated by a distance b are carrying a current i
amp each. The magnitude of the force per unit length exerted by one wire on the
other is:
(A) 0(i2/b2) (B) 0i2/2 b
(C) 0i/2 b (D) 0i/4 b
## 2. A rectangular loop carrying a current i is situated near i
a long straight wire such that the wire is parallel to
one of the sides of the loop and is in the plane of the
loop. If a steady current I is established in the wire as I
shown in the figure, the loop will:
(A) rotate about an axis parallel to the wire (B) move away from the wire
(C) move towards the wire (D) remain stationary
5 cm
3. The resulting magnetic field at the point O due w x
## to the current carrying wire shown in the figure:
5 cm
(A) points out of the page
u v O z
(B) points into the page 10A O y 10A
(C) is zero
(D) is the same as due to the segment WX along.
## 4. A particle enters the region of a uniform
magnetic field as shown in figure. The path
V
of the particle inside the field is shown by
dark line.
The particle is:
(A) electrically neutral
(B) positively charged
(C) negatively charged
## 5. In the given figure, what is the magnetic field induction at I
point O?
r O
I I 0I
(A) 0 (B) 0
4 r 4r 2 r
I 0I I 0I
(C) 0 (D) 0
4r 4 r 4r 4 r
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## 6. An electron is revolving around a proton in a circular orbit of diameter 1A . If it
produces a magnetic field of 14 wb/m2 at the proton, then its angular velocity will
## 7. Electrons at rest are accelerated by a potential of V volt. These electrons enter
the region of space having a uniform, perpendicular magnetic induction field B.
The radius of the path of the electrons inside the magnetic field is:
1 mV 1 2mV
(A) (B)
B e B e
V 1 V
(C) (D)
B B e
8. Two long parallel wires carry currents i2 and i2 (i1 > i2) when the currents are in
opposite direction, the magnetic field at a point midway between the wires is
30 T. If the direction of i2 is changed, the field becomes 10 T. The ratio i1/i2 is
(A) 1 (B) 3
(C) 2 (D) 4
## 9. An infinitely long straight conductor is bent into
shape as shown in figure. It carries a current I A. r
and the radius of circular loop is r metre. Then the O
magnetic induction at the centre of the circular
loop is:
(A) 0 (B)
i i
(C) 0 ( 1) (D) 0 ( 1)
2 r 2 r
10. A charged particle is released from rest in a region of steady and uniform electric
and magnetic fields which are parallel to each other. The particle will move in a
(A) straight line (B) circle
(C) helix (D) cycloid
C
11. A conductor of mass m and length , carrying current
i (direction as shown in the figure) is placed on
B
smooth inclined making angle with horizontal. A
magnetic field B is directed vertically upwards. Then
for equilibrium of conductor tan is given by
2mg mg
(A) (B)
Bi Bi
mg Bi
(C) (D)
2Bi mg
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## 12. The magnetic field at centre of a hexagonal coil of r/2
r
side carrying a current i is
3 0I I O
(A) (B) 0
4
I
(C) 0
(D) zero
3
## perpendicular to a long straight conductor x-y carrying a
current I2, as shown in the figure. The force on AB has
magnitude is I1
I2
II II A B
(A) 0 1 2 log(2) (B) 0 1 2 log(3) L
2 2 L/2
3 0I1I2 2 0I1I2
(C) (D) X
2 3
## 14. A current i flows along a thin wire shaped as shown in
figure. The radius of the curved part of the wire is r. The
field at the centre O of the coil is : O
i 0
(A) 0 (B) 0 90
4 r 2 r
A B
i i
(C) 0 (D) 0 (3 +4)
2 r 8 r
15. A particle of mass m and charge q moves with a constant velocity v along the
positive x direction. It enters a region containing a uniform magnetic field B
directed along the negative z direction, extending from x = a to x = b. The
minimum value of v required so that the particle can just enter the region x>b is
qbB qb a B
(A) (B)
m m
qaB q b a B
(C) (D)
m 2m
## 16. A circular loop of mass m and radius r is kept in a horizontal y
position (X Y plane) on a table as shown in figure. A
uniform magnetic field B is applied parallel to x-axis. The I
current I in the loop, so that its one edge just lifts from the
O x
table, is:
r
(A) mg/ r2 B (B) mg/ rB
(C) mg/2 rB (D) rB/mg
17. In figure there exists uniform magnetic field B into the plane of paper. Wire CD is in the
shape of an arc and is fixed. OA and OB are the wires rotating with angular velocity as
shown in figure in the same plane as that of the arc about point O. If at some instant
OA = OB = and each wire makes angle = 30 with yaxis, the current through
resistance R is (wires OA and OB have no resistance)
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D C
y
B R A
B
30 30
O
2
B
(A) Zero (B)
R
2
B B 2
(C) (D) .
2R 4R
## shown. The magnetic field at the centre O is:
i 1 1
(A) zero (B) 0 R2 R1
4 R1 R 2
S R O Q P
0i 1 1 0i 1 1
(C) (D)
4 R2 R1 2 R1 R2
## 19. The magnetic field strength at a point P distant r due to an
infinite straight wire as shown in the figure carrying a current i is: r P
(A) 0 (B) 0 i/ 2 2 r i r
i
(C) ( 0 i/ 2 r) (D) 0 2 2
4 r
## 20. A wire bent in the form of a sector of radius r subtending an
angle o at centre, as shown in figure is carrying a current i. The
O
magnetic field at O is:
r
i i
(A) 0 (B) 0 ( / 180 )
2r 2r
i
(C) 0 ( / 360) (D) zero
2r
LEVEL II
## 1. A particle of charge + q and mass m moving under
Y E
the influence of a uniform electric field E i and a P V B
## uniform magnetic field B k follows trajectory from P
a
to Q as shown in figure. The velocities at P and Q
are V i and 2V j respectively. Which of the Q
2a
following statement(s) is/are correct? 2V
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3 mv 2
(A) E
4 qa
3 mv 3
(B) Rate of work done by electric field at P is
4 a
(C) Rate of work done by electric field at P is zero
(D) Rate of work done by both the fields at Q is zero.
2. A long, straight wire carries a current along the Z-axis. One can find two points in the X-
Y plane such that
(A) the magnetic fields are equal
(B) the directions of the magnetic fields are the same
(C) the magnitudes of the magnetic fields are equal
(D) the field at one point is opposite to that at the other point.
3. A long striaght wire of radius r carries a current distributed uniformly over its cross-
section. The magnitude of the magnetic field is
(A) maximum at the axis of the wire (B) minimum at the axis of the wire
(C) maximum at the surface of the wire (D) minimum at the surfae of the wire
4. A charged particle of mass 2 kg and charge 2 C moves with a velocity v 8i 6j m/s in
a magnetic field B 2k T. Then
(A) The path of particle may be x2 + y2 = 25.
(B) The path of particle may be x2 + z2 = 25.
(C) The time period of particle will be 3.14 s.
(D) None of these.
## 5. In the figure, there is a uniform conducting structure in i
which each small square has side a. The structure is kept in G D
A
uniform magnetic field B.
(A) The magnetic force on the structure is 2 2 iBa . E F
O
(B) The potential of point B = potential of point D. a
(C) Potential of point O = potential of point B. B
i
H C
(D) The magnetic force on the structure is 2 iBa . a
6. A charged particle moves in a uniform magnetic field. The velocity of the particle at some
instant makes an acute angle with the magnetic field. The path of the particle will be
(A) a circle (B) a helix with uniform pitch
(C) a helix with non uniform radius (D) a helix with uniform radius
7. A striaght conductor carries a current along the x-axis. Consider the pionts A(0, a, 0),
B(0, 0, a), C(0, a, 0) and D (0, 0, a). Then
(A) all four points have magnetic fields in different directions
(B) the magnetic fields at A and C are in opposite direction
(C) the magnetic fields at A and B are mutually perpendicular
(D) all four points have same value of magnetic field in magnitude
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8. Select the correct alternative(s) : Two ions of equal masses, one singly-ionised and the
other doubly-ionised, are projected from the same point with the same velocity
perpendicular to a uniform magnetic field. Then
(A) both ions will describe circles of equal radii
(B) the singly-ionised one will describe a circle of radius double than that of the order
(C) the two circles do not touch each other
(D) the two circles touch each other
9. Two circular coils A and B with their centers lying on the same number of turns and cary
equal currents in the same sense. They are separated by a distance, have different
diameters but subtend same angle at a point P lying on their common axis. The coil B
lies exactly midway between coil A and the point P. The magnetic field at point P due to
coils A and B is B1 and B2 respectively
(A) B1 > B2 (B) B1 < B2
B1 B 1
(C) 2 (D) 1
B2 B2 2
10. When a currentcarrying coil is placed in a uniform magnetic field with its magnetic
moment antiparallel to the field.
(A) Torque on it is maximum (B) Torque on it is zero
(C) Potential energy is maximum (D) Dipole is in unstable equilibrium.
COMPREHENSION
## I. A current carrying loop behaves as a small bar magnet. M y
If there is a moving charged particle nearby this loop, a v
the charged particle will experience a magnetic force x
given by Fm q(v B) . I +Q
P
If now a current carrying loop is placed in a uniform
magnetic field B , the loop experiences a torque given N
by I(A B) .
A wire loop carrying a current I is placed in the x-y plane as shown in figure.
## 1. The direction of magnetic field produced by the loop at point P is
(A) and into the plane (B) and outward to the plane
(C) parallel to the plane (D) none
2. If a particle with charge +Q and mass m is placed at the centre P and given a velocity v
along NP, the instantaneous acceleration of particle is
0.109 0IQv 109 0IQv
(A) (B)
ma ma
10.9 0IQv 1.09 0IQv
(C) (D)
ma ma
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3. If an external uniform magnetic induction field B B i is applied, the torque acting on the
loop due to field is
(A) 0.6136 a2IB (B) 6.136 a2IB
2
(C) 61.36 a IB (D) 613.6 a2IB
II.
Statement :
Magnetic force on a charged particle is given by Fm q(v B) and electrostatic force
Fe qE .
Question:
A particle having charge q = 1C and mass m = 1 kg is released from rest at origin. There
are electric and magnetic fields given by :
N
E (10i ) for x 1.8m and
C
B ( 5k)T for 1.8 m x 2.4m
A screen is placed parallel to yz plane at x = 3.0 m. Neglect gravity forces.
4. The speed with which the particle will collided the screen (in m/s) is
(A) 3 (B) 6
(C) 9 (D) 12
## 5. ycoordinate the particle where it collides with the screen is m.
0.6( 3 1) 0.6( 3 1)
(A) (B)
3 3
1.2( 3 1)
(C) 1.2( 3 1) (D)
3
6. Time after which the particle will collide the screen is .. second.
1 1 1
(A) 3 (B) 6 3
5 6 3 5 3
1 1 1
(C) 5 (D) 6 3
3 6 3 3 18
## MATCH THE FOLLOWING
1. Match the Following :
Column I Column II
2
(A) Unit of magnetic field (p) Am
(B) Unit of magnetic permeability ( 0) (q) N/Am
(C) Unit of magnetic flux ( ) (r) N/A2
(D) Unit of magnetic dipole moment (s) Nm/A
## 2. Match the following :
Column I Column II
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(A) A light conducting circular flexible loop (p) Magnetic field B is doubled.
of wire of radius r carrying current I is
placed in uniform magnetic field B, the
tension in the loop is doubled if
(B) Magnetic field at a point due to a long (q) Inductance is increased by four
straight current carrying wire at a point times.
near the wire is doubled if
(C) The energy stored in the inductor will (r) Current I is doubled.
become four times
(D) The force acting on a moving charge, (s) Radius r is doubled.
moving in a constant magnetic field will
be doubled if
## 3. Match the following :
Column I Column II
(A) (p) B
O x
(perpendicular
to wire) x
Straight current
carrying wire
(B) (q) B
x
O (axis of wire)
Circular current x
carrying wire
(C) (r) B
x
O (perpendicular to the x
i plane of the wires;
O being equidistant)
i
Parallel current carrying
wires in the same plane
(D) (s) B
i
x
(parallel to one
i of the wires) x
## Two perpendicular current
carrying wires in the same plane
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## 10. Hints (Subjective)
LEVEL- I
1. Draw a diagram showing the position of the point where the field is to be
determined in relation to the wires.
2. Find the time period of revolution of the proton, and then find the pitch by
multiplying by v|| to B field. Use v = r to find the radius of the helix.
4. Use the expression for B for a circular coil.
5. The proton moves along a circular path ; write Newtons law for this motion.
6. Use the equation for magnetic field due to a circular arc at the centre. Field due
to both the segments are directed inward.
7. Force on a current element i in magnetic field B is F i B . Resolve the
tension radially and equate with F.
9. Find the field at the centre O due to circular part and straight part. Both the fields
add up, being in the same direction (outward).
10. Find the force acting on a small element on then resolve it ab and ot it. Then
add separately for the whole wire.
LEVEL- II
1. Calculate the magnetic force acting on the horizontal section of the wire, the
torque due to the magnetic force balances the torque due to gravity.
2. The magnetic force on all the four sections should be calculated by integration
and then find the vector sum.
3. The B-field at the centre of the polygon is the sum of the fields due to each
individual side.
4. (a) The B - field is calculated at P and then the force is calculated by using
F Q( v B ).
(b) The torque is non-zero and calculated from the expression B , where
iA .
5. The current in B is found by using the condition that the net field at P is zero.
6. Find the filed at P due to two mutual perpendicular portions separately and then
7. Forces on ps and qr are unequal and oppositely directed whereas that on pq and
rs are equal and opposite.
9. Apply the equations for magnetic field B due to straight current. Use the Lorentz
force F q( v B) where q = - 1.6 x 10-19 C.
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LEVEL I
## 1. (i) 3.46 10-5 T (ii) 2 10 5 T
2. 1.205 10 2 m, 4.4 10 2 m
3. 3.33 mm
4. (a) 2.09 10-4 T (b) 4.5 10-5 T
5. (a) circular motion (b) 1.83 x 107 Hz, 2.76 x 10-12 J.
0i 2 0i 1 3
6. ,
4 R2 R1 8 R2 R1
7. 0.125 N
8. 2.0 10 5 N
i
9. tan
2 R
10. F = Bi
LEVEL II
3
1. 99 10 T
2. iB0 i
n2 i 0 i
3. (a) 0
tan sin (b)
2 2r n n 2r
## 0.109 0IQv 0.109 0IQv
4. (a) at an angle 300 to ve x-axis (b) , 0.614 a2 IB
ma a
130
5. (a) 3 A (b) T (c) 2. 88 10-6 N/m
4
6. B=4 10 6 T, = 45 to the horizontal.
4
7. 1.44 10 N.
8. (a) 22.78 T in the direction k (b) 6.7 j i N-m
9. (a)4 A (b)x = 1 m w.r.t R
mV0 2 sin2
10. , V0 cos
2qL
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LEVEL - I
1. (B) 1. (D)
2. (C) 2. (A)
3. (B) 3. (B)
4. (A) 4. (D)
5. (C) 5. (C)
6. (A) 6. (B)
7. (B) 7. (B)
8. (C) 8. (B)
9. (D) 9. (D)
10. (A) 10. (C)
LEVEL - II
## 1. (A), (B), (D) 2. (B),(C),(D)
3. (B), (C) 4. (A), (C)
5. (A), (B), (C) 6. (B), (D)
7. (A), (B), (C), (D) 8. (B), (D)
9. (B), (D) 10. (B), (C), (D)
COMPREHENSION
1. (A) 2. (A)
3. (A) 4. (B)
5. (D) 6. (A)
MATCH THE FOLLOWING
1. (A) (q); (B) (r); (C) (s); (D) (p)
2. (A) (p), (r), (s); (B) (r); (C) (q), (r); (D) (p)
3. (A) (s); (B) (q); (C) (p); (D) (r)
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ELECTROMAGNETIC INDUCTION
9. Assignment (Subjective Problems)
LEVEL I
## 1. An inductor coil stores 32 J of magnetic energy and dissipates energy as heat at
the rate of 320 W when a current of 4A is passed through it. Find the time
constant of the circuit when this coil is joined across an ideal battery.
2. It is desired to set up an undriven L-C circuit in which the capacitor is originally
charged to potential difference of 100.0 V. The maximum current is to be 1.0 A, and
the oscillation frequency is to be 1000 Hz. What are the required values of L and C?
3. A circular ring of diameter 20 cm has a resistance of 0.01 . How much charge
will flow through the ring if it is turned from a position perpendicular to a uniform
magnetic field of 2.0 T to a position parallel to the field?
8.4 10 3H
4. A coil of inductance 8.4 10 3 H and a resistance of 6
## 6 are connected to a 12V battery as shown in the
figure. Switch is closed at t = 0. At what time is the
magnetic energy in the coil 4.2 10 3Joule? K
12V
## 5. An inductor-coil of inductance 20 mH having resistance 10 is joined to an ideal
battery of emf 5.0 V. Find the rate of change of the induced emf at t = 0.
6. An inductor-coil carries a steady-state current of 2.0 A when connected across
an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of
the circuit.
7. An average emf of 20 V is induced in an inductor when the current in it is
changed from 2.5 A in one direction to the same value in the opposite direction in
0.1 s. Find the self-inductance of the inductor.
## 8. The figure shows a wire sliding on two x x x x x x x x x x x x x x x
parallel, conducting rails placed at a x x x x x x x x x x x x x x x
separation . A magnetic field B exists in a x xR x x x x x x x x x x x l x x
direction perpendicular to the plane of the x x x x x x x x x x x x x x x
rails. What force is necessary to keep the x x x x x x x x x x x x x x x
wire moving at a constant velocity v?
9. A semicircular copper rod of radius R rotates about an axis passing through one
of its ends and lying in the plane of the rod with an angular speed in a uniform
magnetic field B. Find the emf developed between the two ends of the rod. The
field is perpendicular to the motion of the rod.
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10. A square-shaped copper coil has edges of length 50 cm and contains 50 turns. It
is placed perpendicular to an extended uniform magnetic field of strength 1.0 T
magnetic field. It is rotated in the magnetic field about one of its diagonal with
time period 0.25 s. Find the magnitude of the value of maximum emf induced in
the loop.
LEVEL II
## 1. A conducting bar of sufficient length is pulled with a constant
velocity in a conducting shaped rail as shown in the figure.
Inward magnetic field of induction B is present inside the X
area bounded by the bar & the rail. Find the external power v
delivered in moving the rail with constant velocity v at time t
(A = area of cross section of the bar, = resistivity of the bar)
## 2. An LR circuit having a time constant of 50 ms is connected with an ideal battery
of emf . Find the time elapsed before
(a) the current reaches half its maximum values,
(b) the magnetic field energy stored in the circuit reaches half its maximum value.
3. A long solenoid that has 800 turns per meter carries a current i = 3 sin (400t) A.
Find the electric field inside the solenoid at a distance 2 mm from the solenoid
axis. Consider only the field tangential to a circle having its center on the axis of
the solenoid.
## 4. A metallic rod of length & resistance R is free to
rotate about one of its ends over a smooth, rigid
circular metallic frame of radius in an inward O
magnetic field of induction B. What torque should be B
X
applied by an external agent to rotate the rod with
constant angular velocity ?
## 5. A sliding conducting bar of mass m, resistance R is B
x
released from rest. It starts sliding due to the current R
drawn from a battery of emf , in a steady inward magnetic m
field. Find the variation of its speed with time. Also find the
terminal speed of the bar.
## 6. An inductor of inductance 20 mH is connected across a charged capacitor of
capacitance 5 F & resulting L-C circuit is set oscillating at its natural frequency.
The maximum charge q is 200 C on the capacitor. Find the potential difference
across the inductor, when the charge is 100 C.
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## 7. The current in the inner coil is I = 2t2. b
Find the heat developed in the outer
coil between t = 0 and t seconds. The a
resistance of the inner coil is R and
take b >> a.
## 8. In the figure shown is a R-L circuit connected P Q
with a cell of emf through a key k. If key k k
is closed find the current drawn by the R
2R
battery
(a) just after the key k is closed
R
(b) long after the key k is closed L
9. A very small circular loop of area 5 10 -4 m2, resistance 2 ohm and negligible
self inductance initially coplaner and concentric with a much larger fixed circular
loop of radius 0.1 m. A constant current of 1.0 A is passed through the bigger
loop. The smaller loop is rotated with constant angular velocity rad/sec about
its diameter. Calculate the (a) induced emf and (b) the induced current through
the smaller loop as a function of time.
## 10. A wire in the form of a sector of radius and of x x x x x x x x
x x x x x x x x
angle ( = /4) having a resistance R is free to x x x x x x x x
rotate about an axis passing through point O x x x x x x x x
O
and perpendicular to horizontal plane. A vertical
magnetic field B = B0 k exists in the space. If
the sector rotates with constant angular velocity
so that Q Joules of heat is produced per
revolution, find the constant angular velocity.
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## 10. Assignment (Objective Problems)
LEVEL - I
1. An inductor coil of inductance L is divided into two equal parts and both parts are
connected in parallel. The net inductance is :
(A) L (B) 2L
(C) L/2 (D) L/4
## 2. An e.m.f. of 5 millivolt is induced in a coil when in a nearby placed another coil,
the current changes by 5 ampere in 0.1 second. The coefficient of mutual
induction between the two coils will be :
(A) 1 Henry (B) 0.1 Henry
(C) 0.1 millihenry (D) 0.001 millihenry
A
3. In figure when key is pressed the ammeter A reads i
ampere. The charge passing in the galvanometer circuit i
of total resistance R is Q. The mutual inductance of the
G
two coils is : C1
C2
(A) Q/R (B) QR
(C) QR/i (D) i/QR
2H
4. The equivalent inductance between points P and
Q in figure is : 2/3H
P
(A) 2 H (B) 6 H
(C) 8/3 H (D) 4/9 H
4H
## 5. A metal disc of radius R rotates with an angular velocity about an axis
perpendicular to its plane passing through its centre in a magnetic field of
induction B acting perpendicular to the plane of the disc. The induced e.m.f.
between the rim and axis of the disc is:
2B 2R 2
(A) B R2 (B)
BR 2
(C) B R2 (D)
2
## 6. In the circuit shown in the adjoining diagram R1 i1 i2 R3
S i3
E = 10 volts, R1 = 2 ohms, R2 = 3 ohms, R3 = 6 ohms
and L = 5 henry. The current i1 just after pressing the E R2 L
switch S is :
(A) 2.5 amp (B) 2 amp
(C) 5/6 amp (D) 5/3 amp
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## 7. A rectangular coil pqrs is moved away from an infinite,
straight wire carrying a current as shown in figure. Which of s r
the following statements is correct?
i
(A) There is no induced current in coil pqrs
p q
(B) The induced current in coil pqrs is in the clockwise sense
(C) The induced current in the coil pqrs is in anticlockwise
direction
(D) None of the above
## The current in the resistor at t = 0 and t = are
respectively.
(A) 0, 0 Amp. (B) 1, 0 Amp. C=2 F 10
(C) 0, 1 Amp. (D) 1, 1 Amp.
S 10 V
## 9. The two loops shown in the figure, have their planes
parallel to each other. A clockwise current flows in the
loop X as viewed from X towards Y. The two coils will
repel each other, if the current in the loop X is : X Y
## (A) increasing (B) decreasing
(C) constant (D) none of the above cases
10. A coil of area 500 cm2 having 1000 turns is placed such that the plane of the coil
is perpendicular to a magnetic field of magnitude 4 10 5 weber/m2. If it is
rotated by 180 about an axis passing through one of its diameter in 0.1 sec, find
the average induced emf.
(A) zero. (B) 30 mV
(C) 40 mV (D) 50 mV
## 11. For the L shaped conductor in a uniform magnetic field
B shown in figure, the emf across its ends when it x x
x
rotates with angular velocity ' ' about an axis through
one of its ends O and normal to its plane will be x x
O
(A) 2 B 2 (B) B 2 x
1 x
(C) B 2 (D) 4 B 2
2
## 12. A coil of inductance 8.4 mh and resistance 6 is connected to a 12 V battery.
The current in the coil is 1.0 A approximately after time
(A) 500 ms (B) 20 s
(C) 35 ms (D) 1 ms
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## 13. A uniform but time-varying magnetic field B(t) exists in a r
circular region of radius a and is directed into the plane of
x x
the paper, as shown. The magnitude of the induced
x ax P
electric field at point P at a distance r from the centre of x
the circular region is
## (A) is zero (B) proportional to r
(C) proportional to 1/ r (D) proportional to 1/r2
## 14. A conductor of length 5 cm, and A
resistance 2 is moving on
frictionless rails with a constant R
R=4
velocity of 5 cm/s in a magnetic R
v=5 cm/sec R
field of intensity 3 tesla as shown R
below. If conductor is connected B
to a circuit as shown, by two lead
wires of almost negligible
resistance, then current flowing
in it is
(A) 0.25 A (B) 2.5 Amp
(C) 2.5 mA (D) 0.25 10 4 amp
R
15. A wire cd of length , mass m, is sliding without friction on
a b
conducting rails ax and by as shown in figure. The vertical <
rails are connected to one another via an external <
resistance R. The entire circuit is placed in a region of space c d
## having a uniform magnetic field B. The field is to the plane
of circuit & directed outwards. The steady speed of rod cd is
y
(A) mg R/B (B) mg R/B2 2 x
2
(C) mg R/B (D) mg R/B2
16. A thin circular-conducting ring having N turns of radius R is
falling with its plane vertical in a horizontal magnetic field
B
B. At the position MNQ, the speed of ring is v, the induced
e.m.f. developed across the ring is N
v
(A) Zero
M Q
BV R 2N
(B) and M is at higher potential
2
(C) N BRv and Q is at higher potential
(D) 2RBvN and Q is at lower potential
## 17. A circular loop of radius 1m is kept in a A
magnetic field of strength 2 T directed
perpendicular to the plane of loop. Resistance 1 m/s
of the loop wire is 2/ /m. A conductor of
length 2 m is sliding with a speed 1 m/s as
shown in the figure. Find the instantaneous B
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## force acting on the rod [Assume that the rod
has negligible resistance]
(A) 8 N (B) 16 N
(C) 32 N (D) 64 N
18. Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A
causes a flux per turn of 10 3 Wb to link with A and a flux per turn of 0.8 10 3
Wb through B. The ratio of self-inductance of A and the mutual inductance of A
and B is :
(A) 5/4 (B) 1/1.6
(C) 1.6 (D) 1
19. A uniform conducting rod of mass M and length oscillates in a vertical plane
about a fixed horizontal axis passing through its one end with angular amplitude .
There exists a constant and uniform horizontal magnetic field of induction B
perpendicular to the plane of oscillation. The maximum e.m.f. induced in the rod is
B B
(A) 27 3 g 1 cos (B) 27 3 g 1 cos
8 8
3 3 g 1 cos 3 3 g 1 cos
(C) B (D) B
4 4
## 20. A copper rod moves with a constant angular velocity , about
a long straight wire carrying a current I. If the ends of the rod
from the wire are at distances a and b, then the e.m.f. induced
in the rod is I
i(wa) b i(wb) b a
(A) 0 ln (B) 0 ln
2 a 2 a b
0 iw(a b) b
(C) zero (D) ln
4 a
21. The time required for a current to attain the maximum value in a d.c. circuit
containing L and R, depends upon :
(A) R only (B) L only
(C) L/R (D) none of these
20
22. Consider the shown arrangement. When key k is
pressed, the steady value of current in 20 L=0.2H
resistance is :
(A) 0.1 A (B) 0.25 A
30
(C) 0.017 A (D) zero
V=5V K
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B
23. The resistances P, Q, R and S in the bridge shown are
adjusted such that the deflection in the galvanometer G P Q
## galvanometer will show a momentary deflection, if : A C
K2
(A) first K2 is inserted and then K1 S
## (B) first K1 is inserted and then K2 R
L
(C) K1 and K2 are both inserted but an additional D
resistance is put in the arm BD
(D) in all the above cases
K1
## 24. When a J shaped conducting rod is rotating in its
own plane with constant angular velocity about one l
L
of its ends P, in a uniform magnetic field B (directed
Q
normally into the plane of paper)then magnitude of P
emf induced across it will be
1
(A) B L2 l 2 (B) B L2
2
1 1
(C) B L2 l 2 (D) B l 2
2 2
## 25. The equivalent inductance between 2H 4H
points P and Q in the figure is 6H
8H
(A) 9 H P Q
2H 4H
(B) (24/13) H
(C) (12/13) H
(D) 12 H
LEVEL II
## 1. The SI unit of the inductance, the henry, can be written as
(A) weber / ampere (B) volt second / ampere
(C) joule / ampere 2 (D) ohm second
2. Two different coils have self-inductance L1 = 8 mH, L2 = 2mH. The current in one coil is
increased at a constant rate. The current in the second coil is also increased at the same
constant rate. At a certain instant of time, the power given to the two coils is the same.
At that time the current, the induced voltage and the energy stored in the first coils are i1
V1 and W 1 respectively. Corresponding values for the second coil at that instant rate i 2,
V2 and w2 respectively. Then :
i 1 i
(A) 1 (B) 1 4
i2 4 i2
w2 V 1
(C) 4 (D) 2
w1 V1 4
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3. The magnetic flux linked with a conducting coil depends on time as = 4tn + 6, where
n is positive constant. The induced emf in the coil is e
(A) If 0 < n < 1; e 0 and |e| decreases with time.
(B) If n = 1; e is constant.
(C) If n > 1; |e| increases with time.
(D) If n > 1; |e| decreases with time.
## 4. In figure, there is a conducting ring having resistance R A
placed in the plane of paper in a uniform magnetic field B0.
If the ring is rotating about in the plane of paper about an r
axis passing through point O and perpendicular to the C B
plane of paper with constant angular speed in clockwise
direction.
(A) Point A will be at higher potential than O.
(B) The potential of point B and C will be same. O
(C) The current in ring will be zero.
2B0 .r 2
(D) The current in the ring will be .
R
## magnetic field B = [(4 T/s) t + 0.3 T] in a cylindrical region + + +
+ +
of radius 4 m. An equilateral triangular conducting loop is + +
+ + O+
placed in the magnetic field with its centroide on the axis of + + ++
+
the field and its plane perpendicular to the field. B + + + C
(A) e.m.f. induced in any one rod is 16 V + +
## (B) e.m.f. induced in the complete ABC is 48 3 V
(C) e.m.f. induced in the complete ABC is 48 V
(D) e.m.f. induced in any one rod is 16 3 V
## 6. An electron moves in a uniform magnetic field and follows a sprial
path as shown in figure. Which of the following statements is/are
correct
## (A) Angular velocity of electron remains constant
(B) Magnitude of velocity of electron decreases continuously
(C) Net force on the particle is always perpendicular to its direction of motion
(D) Magnetic force on the electron is always perpendicular to direction of motion
decreases continuously
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7. A small magnet M is allowed to fall through a fixed horizontal conducting ring R. Let g be
the acceleration due to gravity. The acceleration of M will be
M
## (A) < g when it is above R and moving towards R
(B) > g when it is above R and moving towards R
(C) < g when it is below R and moving away from R
(D) > g when it is below R and moving away from R
8. The conductor AD moves to the right in a uniform magnetic field directed into the paper
D
v
B
A
(A) The free electrons in AD will move towards A
(B) D will acquire a positive potential with respect to A
(C) If D and A are joined by a conductor externally, a current will flow from A to D in AD
(D) The current in AD flows from lower to higher potential
9. The loop shown moves with a velocity v in a uniform magnetic field of magnitude B,
directed into the paper. The potential difference between P and Q is e
L
v P
L L/2
B Q
1
(A) e Blv (B) e = Blv
2
(C) P is positive with respect to Q (D) Q is positive with respect to P
10. The magnetic field perpendicular to the plane of a conducting ring of radius r changes at
the rate
dB/dt.
dB
(A) The emf induced in the ring is r 2
dt
dB
(B) The emf induced in the ring is 2 r
dt
(C) The potential difference between diametrically opposite points on the ring is half of
the induced emf
(D) All points on the ring are at the same potential
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COMPREHENSION
## I. In figure, a square loop consisting of an inductor of L
inductance L and resistor of resistance R is placed
between two long parallel wires. The two long straight i0cos t R i0cos t
wires have time varying current of magnitude I = I0 cos t
but the direction of current in them are opposite.
a a
3a
1. Total magnetic flux in this loop is
Ia 2 0Ia
(A) 0 ln 2 (B) ln 2
4 0Ia 0Ia
(C) ln2 ln 2 . (D)
2
2. Magnitude of emf in this circuit only due to flux change associated with two long straight
current carrying wires will be
aln 2 I0 2 0 aln 2 I0
(A) 0 sin t (B) sin t
0 aln 2 I0 0 aln 2 I0
(C) cos t (D)
cos t .
2
3. The instantaneous current in the circuit will be
2 0 aln2 I0 2 0aln2 I0
(A) sin( t ) (B) sin( t )
2 2 2
R L R2 2 2
L
0 aln 2 I0 0 aln 2 I0
(C) sin t (D) sin( t ).
R2 2 2
L R2 2 2
L
L
(where tan ).
R
II. A person wants to roll a solid non-conducting y
spherical ball of mass m and radius r on a surface
whose coefficient of static friction is . He placed the
ball on the surface wrapped with n turns of closely i
B
packed conducting coils of negligible mass at the
diameter. By some arrangement he is able to pass a
current i through the coils either in the clockwise
x
direction or in the anti-clockwise direction.
A constant horizontal magnetic field B is present throughout the space as shown in the
figure. (Assume is large enough to help rolling motion)
4. If current i is passed through the coils the maximum torque in the coil is
(A) nir2 B k (B) nir2B j
(C) nir2B j (D) nir2B k
5. Angular acceleration of the ball after it has rotated through an angle ( < 180o), is
5 niB 2 niB
(A) cos (B) cos
7 m 5 m
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7 niB 5 niB
(C) cos (D) cos
5 m 2 m
6. The minimum value of for which the rolling motion is possible, is
14 niB 5 niB
(A) r (B) r
5g m 7g m
7 niB
(C) 0 (D) r
5g m
## MATCH THE FOLLOWING
1. Match the Following :
Column I Column II
(A) LCR circuits (p) Resonant curve will be flattened
(B) Inductor (q) Sharpness indicates sensitivity
(C) More of friction or dampness (r) 1
Have resonant frequency with A
LC
Amax
(D) Radio Tuners characteristic curve (s) Mass
## 2. A conducting circular rigid loop near a long straight y-axis
current carrying wire as shown. Match the following
table :
i x-axis
Column I Column II
(A) If current is increased (p) Induced current in the loop is clockwise
(B) If current is decreased (q) Induced current in the loop is
anticlockwise
(C) If wire is moved away from the wire (r) Wire will attract the loop and there will be
maintaining constant current a torque about yaxis
(D) If wire is moved towards the wire (s) Wire will repel the loop and there will be
maintaining constant current no torque about yaxis
3. Match the following :
Column I Column II
(A) If an iron core is inserted in the current (p) Capacitor.
carrying solenoid, the quantities which
increases are
(B) The initial current in the circuit will be (q) Resistance.
zero when a resistance and battery in
series is connected to
(C) In L-C oscillation the maximum energy (r) Magnetic flux.
for an instant will be stored in
(D) A current carrying closed conducting (s) Inductance, inductor.
loop can have
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## 11. Hints (Subjective)
LEVEL- I
L
1. Time constant =
R
2. Maximum current = q0
Cha n g e of flux
3. Charge flown =
Re sis tan ce
4. First find current at the given time.
5. Find expression for induced emf across the inductor.
6. First find resistance.
di
7. EMF induced = L ; di is charge of current in dt second.
dt
8. Find current through the sliding wire first.
## 10. Find expression for emf induced at any instant t.
LEVEL- II
1. External power delivered = force on the rod due to magnetic field velocity of
the rod.
2. Use L-R circuit charging equation
r dB dB
3. Electric field = where r = 2mm, = rate of charge of magnetic field.
2 dt dt
4. Considering motion of the rod in horizontal plane only ; first find out current
through the rod, then force on a small element on the rod due to external
magnetic field.
5. Write equation of motion of the rod at any instant.
6. Use an L-C oscillating circuit relation.
7. Find flux enclosed by the smaller loop due to magnetic field of the larger loop.
8. Remember an inductor is an open circuit at t = 0 and shorting at t =
9. Find flux enclosed by the smaller loop due to magnetic field of the larger loop.
10. Induced current will flow while the loop enters in to field and again while it comes
out. There will be no current when the loop is completely in the magnetic field.
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LEVEL - I
## 1. 0.2 s. 2. C = 1.59 F, L = 15.9 mH
3. 2 Coulombs. 4. 0.97ms
B2 2 v
7. 0.4 H 8.
R
## 9. 2 B R2 10. 100 volts
LEVEL - II
2B 2 Av 3 tan ln2
1. t 2. (a) s (b) ( 50 10 3) ln(0.3) sec
20
B2 4
3. 1.2 10-3 cos 400t v/m 4.
4R
B2 2 t
E E
5. 1 e mR
; 6. 20 V
B B
2 2
4 a4 t3 3
7. 0
2
8. (a) (b)
bR 3 2R 5R
-9
9. (a) Induced emf = AB sin t = 3.14 10 sin t
9
(b) Induced current = 1.57 10 sin t
8QR
10.
B2 4
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LEVEL - I
1. (D) 2. (C)
3. (C) 4. (A)
5. (D) 6. (B)
7. (B) 8. (D)
9. (A) 10. (C)
11. (B) 12. (D)
13. (C) 14. (C)
15. (B) 16. (D)
17. (B) 18. (B)
19. (C) 20. (C)
21. (D) 22. (D)
23. (A) 24. (C)
25. (A)
LEVEL - II
## 1. (A), (B), (C), (D) 2. (A), (C), (D)
3. (A), (B), (C) 4. (A), (B), (C)
5. (B), (D) 6. (A), (B), (D)
7. (A), (C) 8. (A), (B), (C), (D)
9. (A), (C) 10. (A), (D)
COMPREHENSION
1. (A) 2. (A)
3. (D) 4. (A)
5. (A) 6. (B)
## MATCH THE FOLLOWING
1. (A) (q), (r); (B) (s); (C) (p); (D) (q)
2. (A) (r), (s); (B) (p), (r); (C) (p), (s); (D) (q), (s)
3. (A) (r), (s); (B) (s); (C) (p), (s); (D) (p), (q), (r), (s)
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ALTERNATING CURRENT
7. Assignment (Subjective Problems)
LEVEL - I
## 1. If the voltage in an ac circuit is represented by the equation, v=220 2 sin(314t )
V. Calculate (a) peak and rms value of the voltage (b) average voltage
(c) frequency of AC.
2. Find the maximum value of current when inductance of two henry is connected to
150V, 50 cycle supply?
## 3. A coil of resistance 300 and inductance 1.0 henry is connected across an
alternately voltage of frequency 300/2 Hz. Calculate the phase difference
between the voltage and current in the circuit.
## 4. A 0.21H inductor and a 12 resistance connected in series to a 220V, 50Hz ac
source. Calculate the current in the circuit and the phase angle between the
current and the source voltage.
## 5. A capacitor of 10 F and an inductor of 1H are joined in series. An ac of 50Hz is
applied to this combination. What is the impedance of the combination?
## 6. A coil has a inductance of 1 henry.
(a) At what frequency will it have a reactance of 3142 ?
(b) What should be the capacity of a condenser which has the same reactance at
that frequency?
## 7. A 100mH inductor, a 25 F capacitor and a 15 resistor are connected in series
to a 120V, 50Hz a.c. source. Calculate
(a) impedance of the circuit at resonance.
(b) current at resonance.
(c) Resonant frequency.
## 8. Find the value of an inductance which should be connected in series with a
capacitor of 5 F, a resistance of 10 and an ac source of 50Hz so that the
power factor of the circuit is unity.
## 9. A voltage of 10V and frequency 1000Hz is applied to a 0.1 F capacitor in series
with a resistor of 500 . Find the power factor of the circuit and the average
power dissipated.
10. The inductance of a choke-coil is 0.2 henry and its resistance is 0.50 . If a
current of 2.0 ampere (rms value) and frequency 50Hz be passed through it,
what will be the potential difference across its ends?
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LEVEL - II
1. When 100Volt D.C. is applied across a coil, a current of one ampere flows
through it, when 100V ac of 50Hz is applied to the same coil, only 0.5amp flows.
Calculate the resistance and inductance of the coil.
## 2. A 50W, 100V lamp is to be connected to an ac mains of 200V, 50Hz. What
capacitance is essential to be put in series with the lamp.
10 F
3. A 10 F capacitor, an inductor and a resistor of L
100 are connected to an alternating source of 100
emf 200 2 sin100t as shown in the figure. What
is the r.m.s current in the circuit and the value of
the inductance if the current and the source
voltage attain their maxima simultaneously?
What is the average power consumed in this 200 2 sin100t
case?
## 4. A 200Km long telegraph wire has capacity of 0.014 F /km. If it carries an
alternating current of 50KHz, what should be the value of an inductance required
to be connected in series so that impedance is minimum?
## 5. A coil of negligible resistance is connected in series with 90 resistor across a
120V - 60Hz line. A voltmeter reads 36V across the resistance. Find the voltage
across the coil and inductance of the coil.
## 6. An ac source of angular frequency is fed across resistor R and a capacitor C is
series. The current registered is I. If now the frequency of source is changed to
( but maintaining the same voltage), the current in the circuit is found to be
3
halved. Calculate the ratio of reactance to resistance at the original frequency .
## 7. An alternating current of 1.5mA and angular frequency = 300 rad/s flows
through 10K resistor and a 0.50 F capacitor in series. Find the r.m.s. voltage
across the capacitor and impedance of the circuit?
## 8. A 750 Hz, 20 V source is connected to a resistance of 100 , an inductance of
0.1803 H and a capacitance of 10 F all in series. Calculate the time in which the
resistance (thermal capacity 2 J/ C) will get heated by 10 C. (Ignore radiation)
## 9. In the circuit shown there is a box and capacitance C C
connected to alternating power source of angular
frequency of 2 rad/s. Box has power factor 1 2 and
circuit has overall power factor 1. Find the impedance of
the box. ~
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## 10. In the given circuit, power factor A B
between A and B is 0.5 and over all R L
power factor of the circuit is 1. Find V0sin10t
C = 2mF
the value of R and L.
***
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## 8. Assignment (Objective Problems)
LEVEL - I
1. The frequency for which a 5.0 F capacitor has a reactance of 1000 is given by
1000 100
(A) cycles / sec (B) cycles / sec
(C) 200 cycle /s (D) 5000 cycles /sec
2. In an a.c. circuit V and I are given by V = 50 sin50t volt and I = 100 sin(50t + /3)
mA. The power dissipated in the circuit
(A) 2.5 kW (B) 1.25 kW
(C) 5.0 kW (D) 500 watt
1
(A) 2
(B) 2Li2
2Li
Li2
(C) (D) zero.
4
## 4. Circuit as shown in figure below, choose R L C
the correct statement.
(A) current in resistance R and current in
inductor L will be in 90 phase difference. a.c. source
(B) potential drop across R and potential drop across L will be in same phase.
(C) current through C and current through L will be in 90 phase difference.
(D) current in R and current in L will be in same phase.
## 5. In a series L, R, C, circuit which is connected to a.c. source. When resonance is
obtained then net impedance Z will be
I
(A) Z = R (B) Z = L
C
1
(C) Z = L (D) Z =
C
6. An L,C, R series circuit is connected to a.c. source. At resonance, the applied
voltage and the current flowing through the circuit will have a phase difference of
(A) /4 (B) zero.
(C) (D) /2
## 7. The reciprocal of impedance is called
(C) inductance. (D) conductance.
8. The root-mean-square value of an alternating current of 50Hz frequency is 10
ampere. The time taken by the alternating current in reaching from zero to
maximum value and the peak value of current will be
(A) 2 10 2 sec and 14.14 amp. (B) 1 10 2 sec and 7.07 amp.
(C) 5 10 3 sec and 7.07 amp. (D) 5 10 3 sec and 14.14 amp.
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9. A coil of resistance 2000 and self-inductance 1.0 Henry has been connected to
an a.c. source of frequency 2000/2 Hz. The phase difference between voltage
and current is
(A) 30 (B) 60
(C) 45 (D) 75
10. In a series resonant circuit, the a.c. voltage across resistance R, inductance L
and capacitance C are 5V, 10V and 10V, respectively. The a.c. voltage applied to
the circuit will be
(A) 20V (B) 10V
(C) 5V (D) 25V
## 11. In the given figure, which voltmeter will read zero V1 V2 V3
(A) V1 (B) V2 R L C
(C) V3 (D) V4 V4
E=Eosin t
## 12. A resistance R is connected in series with capacitance C Farad value of
impedance of the circuit is 10 and R = 6 so, find the power factor of circuit.
(A) 0.4 (B) 0.6
(C) 0.67 (D) 0.9
## 13. In a R, L, C circuit, three elements is connected in series by an a.c. source. If
frequency is less than resonating frequency then net impedance of the circuit will be
(A) capacitive (B) inductive
(C) capacitive or inductive. (D) pure resistive.
14. Using an A.C. voltmeter, the potential difference in the electrical line in a house is
read to be 234 volts. If the line frequency is known to be 50 cycles per second,
the equation for the line voltage is
(A) V = 165 sin(100 t) (B) V = 331 sin(100 t)
(C) V = 234 sin(100 t) (D) V = 440 sin(100 t)
15. In an a.c. circuit, containing an inductance and a capacitor in series, the current
is found to be maximum when the value of inductance is 0.5henry and of
capacitance is 8 F. The angular frequency of the input A.C. Voltage must be
equal to
(A) 500 (B) 5 104
(C) 4000 (D) 5000
## 16. An alternating voltage E (in volts) = 200 2 sin (100t) is connected to a 1 F
capacitor through an a.c. ammeter. The reading of the ammeter shall be
(A) 10mA (B) 20mA
(C) 40mA (D) 80mA
17. In a series R, L, C circuit XL = 10 , XC = 4 and R = 6 . Find the power factor of
the circuit.
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1 3
(A) (B)
2 2
(C) 1/2 (D) none of the these.
18. In LCR circuit the capacitance is changed from C to 4C. For the same resonant
frequency, the inductance should be changed from L to
(A) 2L (B) L/2
(C) L/4 (D) 4L
## 19. A resistance (R) = 12 ; inductance (L) = 2 henry and capacitive reactance C = 5
mF are connected in series to an ac generator
(A) at resonance, the circuit impedance is zero.
(B) at resonance, the circuit impedance is 12 .
(C) the resonance frequency of the circuit is 1/2 .
(D) at resonance, the inductive reactance is less than the capacitive reactance.
20. In an A.C. circuit, the current is I = 5 sin(100 /2) amp and the A.C. potential is V
= 200 sin(100t) volt. Then the power consumption is
(A) 20 watts (B) 40 watts
(C) 1000Watts (D) 0 watts
LEVEL - II
## 1. An LC circuit has capacitance C1 = C and inductance L1 = L. A second circuit has
C2 = C/2 and L2 = 2L and third circuit has C3 = 2C and L3 = L/2. All the three capacitors
are charged to the same potential V and then made to oscillate. Then
(A) angular frequency of oscillation is same for al the three circuits
(B) maximum current is greatest in first circuit
(C) maximum current is greatest in second circuit
(D) maximum current is greatest in third circuit.
2. An LCR circuit with 100 resistance is connected to an ac source of 200 V and angular
frequency 300 rad/s. When only the capacitance is removed the current lags behind the
voltage by 60. When only the inductance is removed, the current leads the voltage by
60. Then in LCR circuit the current and power dissipated are :
(A) 2A (B) 1A
(C) 200W (D) 400W.
3. A current of 4A flows in a coil when connected to 12V dc source. If the same coil is
connected to a 12V, 50 rad/s source, a current of 2.4 A flows in the circuit. Then
(A) R = 4 (B) R = 3
(C) L = 4H (D) 0.08 H.
4. In the circuit shown in the figure R = 50 , E1 = 25 3 volt and R
1
E2 = 25 6 sin ( t) volt where = 100 s . The switch is
closed at t = 0 and remains closed for 14 minutes, then it is
opened ~ S
(A) The amount of heat produced in the resistor is 63000 J. 1
2
## (B) The amount of heat produced in the resistor is 7000 J.
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(C) If total amount of heat produced is used to heat 3 kg of water at 20C, the final
temperature will be 25C.
(D) The value of direct current that will produce same amount of heat in same time
through same resistor will be 1.5 A.
5. An alternating voltage (in volts) varies with time t (in seconds) as V = 200 sin (100 t)
(a) The peak value of the voltage is 200 V
(b) The rms value of the voltage is 220 V
(c) The rms value of the voltage is 100 2 V
(d) The frequency of the voltage is 50 Hz
## 6. A electric heater is connected to 100 V, 60 Hz ac supply.
(a) The peak value of the voltatge is 100 V
(b) The peak value of the current in the circuit is 2 2 A
(c) The rms value of the voltage is 100 V
(d) The rms value of the current is 2 A
## 7. L, C and R respectively represent inductance, capacitance and resistance. Which of the
following combinations have the dimensions of frequency ?
(a) R/L (b) 1/RC
(c) R / LC (d) 1/ LC
8. The network shown in figure is part of a circuit. The battery has negligible internal
resistance. At a certain instant the current I = 5 A and is decreasing at a rate of
103 As1. At that instant, the potential difference
E = 15 V L = 5mH
I
A R=1 B
(a) across L is 5 mV (b) across L is 5 V
(c) between points A and B is 15 V (d) between points A and B is 25 V
## 9. In a series LCR circuit
(a) the voltage VL across the inductance leads the current in the circuit by a
phase angle of / 2
(b) the voltage VC across the capacitance lags behind the current by a phase
angle of / 2
(c) the voltage VR across the resistance is in phase with the current
(d) the votage across the series combination of L, C and R is V = VL + VC + VR.
10. To convert mechanical energy into electrical energy, one can use
(a) DC dynamo (b) AC dynamo
(c) motor (d) transformer
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COMPREHENSION
## I. In the given circuit at t = 0, switch S is closed.
5mH 10 10V 20 F
S
2
1. The current through 10 resistor at any instant t; 0 < t < will be
1000 1000
1 t 5 t
(A) e 3 (B) e 3
6 6
1000 1000
1 3 t 6 3 t
(C) e (D) e .
6 5
95. The energy stored in the inductor at any instant t; 0 < t < will be
1000 1000
1 t 125 t
(A) (5 5e 3 )2 mJ (B) (1 e 3 )2 mJ
2 2
1000 1000
25 t 5 t
(C) (1 e 3 )2 mJ (D) (1 e 3 )2 mJ .
2 2
96. The energy stored in the capacitor and inductor respectively as t will be
(A) 1 mJ and 62.5 mJ (B) 62.5 mJ and 1 mJ
(C) 2 mJ and 62.5 mJ (D) 1 mJ and 60 mJ.
## II. The figure represents variation of peak current i0 with i0
applied frequency of the AC source of three different LCR
1
circuits having different resistances. The value of
inductance L and capacitance C are same for all the three
2
circuits.
3
## 97. If R1, R2 and R3 be the resistance of circuit 1, 2 and 3 respectively, then
(A) R1 > R2 > R3 (B) R1 < R2 < R3
(C) R1 > R2 = R3 (D) R1 = R2 = R3
900 40
98. If R1 = 1 , R2 = 5 , R3 = 10 and L = mH, C = F, then the value of 0 is
(A) 250 Hz (B) 125 Hz
250 250
(C) Hz (D) Hz
6 3
99. In the previous question, the frequency with which energy oscillates between
Electric Field Energy and Magnetic Field Energy, is
1 1
(A) (B)
4 LC LC
1
(C)
2 LC
(D) the energy in the electric field does not oscillate
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## 1. Match the Following :
Column I Column II
(A) For square wave having peak value v0 (p) v0 > vrms > vav
(B) For sinusoidal wave having peak value v0 (q) In a pure inductance.
(C) Current leads the voltage by /2 (r) vav = vrms = v0
(D) Wattless current (s) In a pure capacitance
## 2. Match the following :
Column I Column II
(A) In LR series circuit if switch is closed (p) Current at t = 0 is nonzero
at t = 0 (u sin of DC source)
(B) In LC series combination switch is (q) Nothing can be said about the
closed at t = 0 (if initially the capacitor is current
fully charged
(C) If voltage V=V0 sin t is applied to pure (r) Current in the circuit is zero at t = 0
inductor at t = 0
(D) If voltage V = V0 sin t is applied at t (s) Magnetic field energy in inductor is
= 0 to LCR series circuit zero at t = 0
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LEVEL - I
3
2. A 3. /4
2 2
## 4. 3.28A, 79.7 5. 4.47
6. (a) 500Hz. (b) 0.11 F
20
7. (a) 15 (b) 8A (c) 100.7 Hz 8. 2
LEVEL - II
## 1. 100 , ( 3/ ) Hz. 2. 9.2 F
3. 2A, 10H, 400W 4. 0.303 H
3
5. 114V, 0.76H 6.
5
7. 10V, 1.2 104 8. 5.8 min
1
9. 10. 100 , 5H
C 2
***
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LEVEL - I
1. (B) 2. (B)
3. (D) 4. (A)
5. (A) 6. (B)
7. (D) 8. (D)
9. (C) 10. (C)
11. (D) 12. (B)
13. (A) 14. (B)
15. (A) 16. (B)
17. (A) 18. (C)
19. (B) 20. (D)
LEVEL - II
## 1. (A), (D) 2. (A), (D)
3. (B), (D) 4. (A), (C), (D)
5. (A), (C), (D) 6. (B), (C), (D)
7. (A), (B), (C) 8. (B), (C)
9. (A), (B), (C) 10. (A), (B)
COMPREHENSION
4. (B) 5. (B)
6. (A) 7. (B)
8. (D) 9. (B)
## 1. (A) (r); (B) (p); (C) (s); (D) (q), (s)
2. (A) (r), (s); (B) (r), (s); (C) (p); (D) (q)
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OPTICS
8. Assignment (Subjective Problems)
LEVEL I
## 1. A thin converging lens forms a magnified image (magnification :p) of an object.
The magnification factor becomes q when the lens is moved a distance a
towards the object. Find the focal length of the lens.
2. A parallel beam of light is incident normally onto a solid glass sphere of radius R
( = 1.5). Find the distance of the image from the outer edge of the glass sphere.
3. A point object is placed in front of a silvered plano-convex lens of
refractive index n, radius of curvature R, so that its image is O
## formed on itself. Calculate the object distance.
4. A convex lens focuses a distant object on a screen placed 10 cm away from it. A
glass plate (n = 1.5) of thickness 1.5 is inserted between the lens and the screen.
Where should the object be placed so that its image is again focused on the
screen?
5. A parallel beam of light tavelling in water (refractive index = 4/3) is refracted by a
spherical air bubble of radius 2 mm situated in water. Assuming the light rays to
be paraxial (i) find the position of image due to refraction at first surface and
position of final image. (ii) draw a ray diagram showing the position of both
images.
6. Find the focal length of the lens shown in the 1 < 2 < 3
figure. The radii of curvature of both the surfaces 3
1 2
are equal to R.
R R
## 7. A converging lens which has a focal length of 20 cm is placed 60 cm to the left of
a concave mirror of focal length 30 cm. An object is placed 40 cm to the left of
lens. Find the position, nature and magnification of the final image.
8. A cylindrical glass rod has its two coaxial ends of
spherical form bulging outward. The front end has a
radius of curvature 5 cm and the back end which is O P
silvered has a radius of curvature 8 cm. The
50 cm
thickness of the rod along the axis is 10 cm. R = 5 cm R = 8 cm 1 2
## Calculate the position of the image of a point object
at the axis 50 cm from front face (ang = 1.5)
9. A thin bi-convex lens of refractive index 3/2 and radius of curvature 50cm is placed
on a reflecting convex surface of radius of curvature 100cm. A point object is placed
on the principal axis of the system such that its final image coincides with itself. Now
few drops of a transparent liquid is placed between the mirror and lens such that final
image of the object is at infinity. Find refractive index of the liquid used. And also find
position of the object.
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10. An object of height 2.5 cm is placed at a 1.5 f from a concave mirror where f is
the magnitude of the focal length of the mirror. The object is placed perpendicular
to the principal axis. Find the height of the image. Is the image erect or inverted ?
11. In Young's double slit experiment the fringe width obtained is 0.6 cm, when light
of wavelength 4800 A0 is used. If the distance between the screen and the slit is
reduced to half, what should be the wavelength of light used to obtain fringes
0.0045-m width?
12. In a Young's double slit experiment, the slits are 1.5 mm apart. When the slits
are illuminated by a monochromatic light source and the screen is kept 1 m apart
from the slits, width of 10 fringes is measured as 3.93 mm. Calculate the
wavelength of light used. What will be the width of 10 fringes when the distance
between the slits and the screen is increased by 0.5 m. The source of light used
remains the same.
## 13. A beam of light consisting of two wavelengths 6500Aand 5200A is used to
obtain interference fringes in a Youngs double slit experiment. Find the distance
of the third fringe on the screen from the central maximum for the wavelength
6500A.
## 14. In a two-slit experiment with monochromatic light, fringes are obtained on a
screen placed at some distance from the slits. If the screen is moved by 5 10-2
m towards the slits, the change in fringe width is 3 10-5 m. If the distance
between the slits is 10-3 m, calculate the wave length of the light used.
15. At a certain point on a screen the path difference for the two interfering rays is
(1/8)th of a wavelength. Find the ratio of the intensity at this point to that at the
centre of a bright fringe.
## (a) Show that in this case d = 2 D/3. d
A
(b) Show that the intensity at P0 is three times the P0
individually.
## 600 nm are placed at a distance of 1.0 10 2 cm. A
detector can be moved on the line S1P which is d
perpendicular to S1S2. Find out the position of first P
minimum detected. S1 D
18. White light may be considered to have from 4000 A0 to7500 A0. If an oil film
has thickness 10-6 m, deduce the wavelengths in the visible region for which the
reflection among the normal direction will be (i) weak, (ii) strong. Take of the
oil as 1.40.
19. Find the maximum intensity in case of interference of n identical waves each of
intensity I0 if the interference is (a) coherent (b) incoherent
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## 20. A monochromatic light of = 5000 A0 is incident on two identical slits separated
by a distance of 5 10 4 m. The interference pattern is seen on a screen placed
at a distance of 1m from the plane of slits. A thin glass plate of thickness
1.5 10 6 m and refractive index = 1.5 is placed between one of the slits and
screen. Find the intensity at the centre of the screen if the intensity there is I0 in
the absence of the plate. Also find the lateral shift of the central maxima.
LEVEL II
## the figure is formed at the bottom of the /4
O
tank filled with water. Using the values
given in the figure, calculate the value 36 cm 1m
85 cm
of h, i.e. the water level in the tank.
h
2. (a) The refracting angle of a prism is equal to /2. It is given that is the angle of
minimum deviation and is the deviation of the ray at grazing incidence.
Prove that sin = sin2 and cos = cos
(b) A ray of light passes through a prism, deviation equal to the angle of
incidence which, again, is equal to 2 . It is given that is the angle of prism.
1
Show that cos2 = 2
, where is the refractive index of the material
8 1
of prism.
## a plate of refractive index n1 (n1 < n) cemented to its diagonal
n1
face. The assembly is in air. A ray is incident on AB.
(i) Calculate the angle of incidence at AB for which the ray n
## strikes the diagonal face at the critical angle. B C
(ii) Assuming n = 1.352, calculate the angle of incidence at AB
for which the refracted ray passes through the diagonal
face undeviated.
4. Three thin equi-convex lenses each of focal length f are separated by distance
f apart. A point object is placed at a distance of 3f in front of the first lens. Find
the position of the final image.
5. A cylindrical vessel of radius R and height 3a is completely filled
with three different immiscible liquids each of height a and n1 a
having refractive indices n1, n 2 and n3 (where n1>n2>n3)>1. A a
n2
point object is placed at the centre of the bottom of the vessel.
n3 a
The rays just suffer total internal reflection at the edge of the
vessels mouth. Find the radius of curvature of the vessel.
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## 6. A point object is placed at a distance of 0.3 m L1
from a convex lens (focal length 0.2 m) cut into
two halves each of which is displaced by 5 S 0.0005 m
10-4m as shown in the figure. Find the position
of the image. If more than one image is
formed, find their number and the distance L2
## distance 5 mm from the centre and the
two parts are placed at a separation of 5 5 mm 5 mm
x
mm as shown. A concave lens is also (0, 0) 5 mm 5 mm
## divided into two parts but in the opposite
sense that of convex lens. The focal
lengths of convex and concave lenses are 90 cm 20 cm
## 30 cm and -50 cm respectively. Find the
co-ordinate(s) of real images when an
object is placed at a distance of 90 cm
from the plane of the convex lens.
## 8. Two thin convex lenses of focal lengths f1 and f2 Y
L2
are separated by a horizontal distance d (where L1
## d < f1, d < f2) and their centres are displaced by a
vertical separation as shown in figure. Taking
O X
the origin of coordinates O, as the centre of first
lens, what would be the x and y coordinates of
the focal point of this lens system, for a parallel d
beam of rays coming from the left ?
## 9. A convex lens of focal length 15 cm and a concave A
mirror of focal length 30cm are kept with their optic B
P
axes PQ and RS parallel but separated in vertical R
0.6
S
Q
## direction by 0.6 cm as shown. The distance between
the lens and the mirror is 30 cm. An upright object
AB of height 1.2 cm is placed on the optic axis PQ of 30 cm 20 cm
## the lens at a distance of 20 cm from the lens. If A B
is the image after refraction from the lens and
reflection from the mirror, find the distance of A B
from the pole of the mirror and obtain its
magnification . Also locate positions of A and B with
respect to the optic axis RS.
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## 10. A transparent solid sphere of radius 2 cm and density
10 cm
floats in a transparent liquid of density 2 kept in a beaker. 2 cm
The bottom of the beaker is spherical in shape with its radius
of curvature 8 cm and is silvered to make it a concave mirror h
as shown in the figure. When an object is placed at a
distance of 10 cm directly above the centre of the sphere its
final image coincides with it. Find h (as shown in the figure),
the height of the liquid surface in the beaker from the apex of
the bottom. Consider paraxial rays only. The refractive index
of the sphere is (3/2) and that of the liquid is (4/3).
11. Sodium light has two wavelengths 1 = 589 nm and 2 = 589.6 nm. As the path
difference increases, when is the visibility of the fringes minimum ?
## 12. In Youngs double slit experimental setup as TA = T0
shown in the figure, the two glass plates A and S1
B each of thickness TA = T0 and TB = T0 + t2
(where is constant, t is time in sec) are placed d
S O
in the paths of rays of light coming from S1 and TB = T0+ t2
Screen
S2
S2. Find out the minimum time after which the
central maxima position O will again appear D
## bright in the presence of light of wavelength .
[Take = refractive index of glass]
## 13. Two transparent slabs having equal thickness 0.45 mm S
and refractive indices 1.40 & 1.42 are pasted on the
S1
two slits of a double slit aparatus. The separation of O
slits equals 1 mm. Wavelength of light used equals S2
600 nm. The screen S is placed at a distance 1 m from
the plane of the slits. Find the position(s) of first
maxima from the centre O of the screen.
## 14. A vessel ABCD of 10cm width has two small
A D
slits S1 and S 2 sealed with identical glass
plates of equal thickness. The distance S1
P
between the slits is 0.8 mm. POQ is the line O
Q
perpendicular to the plane AB and passing S 2
40cm
through O, the middle point of S1 and S2. A
monochromatic light source is kept at S, 40cm 2m 10 cm
S C
below P and 2m from the vessel, to illuminate B
## the slits as shown in the figure below.
Calculate the position of the central bright fringe on the other wall CD with respect to
the line OQ. Now, a liquid is poured into the vessel and filled up to OQ. The central
bright fringe is found to be at Q. Calculate the refractive index of the liquid.
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## 15. Consider the situation of the interference experiment P Q
P
set up as shown in the figure. The S1S2 part of the set
air
up is put in a medium whose refractive index varies rn
as = 0(1 + x) where x is the displacement from 0
0
## the line PP . [Take (D>>d, D>>r) and =constant].
S O
(a) Find the nature of fringes obtained on the screen. S1
II
(b) Find the distance of nth bright fringe from the I
mediu
III
m
central fringe on the screen. [take to be small]. d Air Screen
## Taking 0 as refractive index in medium I and III
P Q
and also of S1 and S2 sources.
## 16. In the Youngs Double Slit
experiment the point source S S 1
P
## central axis as shown in the S
O
2
figure. 1m 2m
(a) Find the nature and
order of the interference
at the point P.
(b) Find the nature and order of the interference at O.
(c) Where should we place a film of refractive index = 1.5 and what should be
its thickness so that a maxima of zero order is placed at O?
## 17. In a modified Youngs double-slit experiment, a monochromatic, uniform and
parallel beam of light of wavelength 6000 and intensity (10/ ) W-m-2 is incident
normally on two circular apertures A and B of radii 0.001 m and 0.002 m
respectively. A perfect transparent film of thickness 2000 and refractive index
1.5 for the wavelength of 6000 is placed in front of aperture A (figure).
Calculate the power (in watt) received at the focal spot F of the lens. The lens is
symmetrically placed with respect to the apertures. Assume that 10% of the
power received by each aperture goes in the original direction and is brought to
the focal spot.
## 18. In the Youngs double slit experiment the space
between the light source and the two slits is filled = 4/3 S1
Air
with a liquid of refractive index 4/3. Whereas the S
medium between the slits and screen is air. Find Screen
the position of the first bright fringe from the central S2
maxima?
D = 2 m and d = 0.25 mm and medium = 5000A .
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## 19. The Youngs double slit experiment is done in a medium of Y
refractive index 4/3. A light of 600 nm wavelength is falling
on the slits having 0.45 mm separation. The lower slit S2 is S1
covered by a thin glass sheet of thickness 10.4 m and O
refractive index 1.5. the interference pattern is observed S2
on a screen placed 1.5 m from the slits as shown in figure.
(a) Find the location of the central maximum (bright fringe with zero path difference) on
the y-axis.
(b) Find the light intensity at point O relative to the maximum fringe intensity.
(c) Now, if 600 nm light is replaced by white light of range 400 to 700 nm, find the
wavelengths of the light that from maxima exactly at point O.
[All wavelengths in this problem are for the given medium of refractive index 4/3.
Ignore dispersion.]
## (emitting monochromatic light of
wavelength ) and screen are placed as
shown in the figure. Now whole of the d
setup is kept in a liquid of medium = S
Wood plate
## 4/3. If the angle is very small, find out P
O
the position of third minima on the screen
in terms of , r, R and .
=4/3
S2 R
r
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LEVEL I
## 1. If the behavior of light rays through a convex lens is as
shown in the adjoining figure, then;
(A) = 2 (B) < 2
(C) > 2 (D) 2
## angle i, it emerges finally parallel to the surface of water, W = 4/3
r
then the value of g would be Water r
(A) (4/3)sin(i) (B) [1/sin(i)]
(C) 4/3 (D) 1 Glass i
3. A converging lens is used to form an image on a screen. When the upper half of
the lens is covered by an opaque screen :
(A) half of the image will disappear (B) image will not form on the screen.
(C) intensity of image will increase (D) intensity of image will decrease
4. A spherical convex surface separates object and image space of refractive index
1 and 4/3 respectively. If radius of curvature of the surface is 0.1 m, its power is :
(A) 2.5 D (B) 2.5 D
(C) 3.3 D (D) 3.3 D
5. A ray of light passes through an equilateral prism such that the angle of
incidence is equal to the angle of emergence and latter is equal to 3/4th the angle
of prism. The angle of deviation is :
(A) 45o (B) 39o
o
(C) 20 (D) 30o
## 6. A liquid is placed in a hollow prism of angle 60o. If angle of the minimum
deviation is 30o, what is the refractive index of the liquid?
(A) 1.41 (B) 1.50
(C) 1.65 (D) 1.95
7. A prism can produce a minimum deviation in a light beam. If three such prisms
are combined, the minimum deviation that can be produced in this beam is:
(A) 0 (B)
(C) 2 (D) 3
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## ray is incident on face PQ at an angle of 45o as shown
in figure. The refracted ray undergoes reflection on
face PR and retraces its path. The refractive index of 45
0
the prism is :
(A) 2 (B) 3/ 2
Q R
(C) 1.5 (D) 1.33
9. A particle moves towards a concave mirror of focal length 30 cm along its axis
and with a constant speed of 4 cm/ sec. What is the speed of its image when the
particle is at 90 cm from the mirror?
(A) 2 cm/ sec. (B) 8 cm/sec.
(C) 1 cm/sec. (D) 4 cm/sec.
10. A thin prism of glass is placed in air and water successively. If a g = 3/2 and
a w = 4/3, then the ratio of deviations produced by the prism for a small angle of
incidence when placed in air and water is :
(A) 9 : 8 (B) 4 : 3
(C) 3 : 4 (D) 4 : 1
11. A thin prism P1 with angle 4o and made from glass of refractive index 1.54 is
combined with another thin prism P2 made from glass of refractive index 1.72 to
produce dispersion without deviation. The angle of the prism P2 is :
(A) 5.33o (B) 4o
(C) 3o (D) 2.6o
12. Focal lengths of two lenses are f and f and dispersive powers of their materials
are and . To form achromatic combination from these, which relation is
correct?
(A) = 0, = 2 0, f = 2f (B) = 0, = 2 0, f = 2f
(C) = 0, = 2 0, f =f/2 (D) = 0, = 2 0, f = f/2.
13. A lens of refractive index is put in a liquid of refractive index . If the focal
length of the lens in air is f, its focal length in liquid will be
f 1 f
(A) (B)
1
1 f
(C) (D)
f
14. A convex lens, a glass slab, a glass prism and a spherical solid ball have been
prepared from the same optically transparent material. Dispersive power will be
possessed by:
(A) the prism only (B) the convex lens and the prism
(C) all except glass slab. (D) all the four
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15. A beam of white light is incident on a hollow prism of glass as shown in the
figure. Then
(A) The light emerging from prism gives no spectrum
(B) The light emerging from prism gives spectrum but the bending i
of all colours is away from base. White
(C) The light emerging from prism gives spectrum, all the colours light
bend towards base, the violet most and red the least.
(D) The light emerging from prism gives spectrum, all the colours
bend towards base, the violet the least and red the most.
## 16. A beam of light consisting of red, green and blue colours
is incident on a right-angled prism. The refractive indices
of the material of prism for the above red, green and blue
wavelengths are 1.39, 1.44 and 1.47 respectively. The
prism will: 45
o
## (A) separate part of the red colour from the green
and blue colours
(B) separate part of the blue colour from the red and green colours
(C) separate all the three colours from one another
(D) not separate even partially any colour from the other two colours.
17. A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm
are kept along the same axis with a distance d between them. If a parallel beam
of light falling on A leaves B as a parallel beam, then the distance d in cm will be:
(A) 25 (B) 15
(C) 10 (D) 30
18. When the distance between the object and the screen is more than 4f, we can
obtain the image of the object on the screen for the two different positions of a
convex lens of focal length f. If I1 and I2 be the sizes of the two images, then the
size of the object is:
(A) (I1 + I2)/2 (B) I1 I2
(C) (I1 I2) (D) (I1/I2)
## 19. A layered lens as shown in the figure is made of two
types of transparent materials indicated by different
shades. A point object is placed on its axis. The object
will form:
(A) 1 image (B) 2 images
(C) 3 images (D) 7 images
20. In the displacement method, a convex lens is placed in between an object and a
screen. If the magnification in the two positions be m1 and m2 and the
displacement of the lens between the two positions is X, then the focal length of
the lens is :
(A) X/(m1 m2) (B) X/|m1 m2 |
(C) X/|m1 + m2 | (D) X/(m1 m2)2
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21. Two coherent monochromatic light beams of intensities I and 4I are superposed.
The maximum and minimum intensities in the resulting beam are
(A) 5I and I (B) 5I and 3I
(C) 9I and I (D) 9I and 3I
22. In Young's double slit experiment, the fringe width is . If the entire arrangement
is now placed inside a liquid of refractive index , the fringe width will become
(A) (B) /
(C) (D)
1 1
23. In a Young's double slit experiment, let S1 and S2 be the two slits, and C be the
centre of the screen. If S1CS2= and is the wavelength, the fringe width will be
(A) (B)
(C) 2 / (D) /2
24. The speed of light in air is 3 10 8 m/s. If the refractive index of glass is 1.5, find
the time taken by light to travel a distance 50 cm in glass.
(A) 2.5 10 9 sec. (B) 0.5 10 9 sec.
(C) 0.16 10 9 sec. (D) 3 10 9 sec.
25. In the Youngs double slit experiment, films of thickness tA and tB and refractive
indices A and B are placed in front of A and B respectively. If AtA = BtB , the
central maximum will
(A) not shift
(B) shift towards A
(C) shift towards B
(D) option (B), if tB > tA and option (C) if tB < tA
26. In the Youngs double slit experiment both the slits are similar. If the length of
one of the slits is halved, which of the following is true?
(A) Bright fringes becomes narrower. (B) Bright fringes become wider.
(C) Dark fringes become darker. (D) Dark fringes become brighter.
27. Waves from two different sources overlap near a particular point. The amplitude
and the frequency of the two waves are same. The ratio of the intensity when the
two waves arrive in phase to that when they arrive 900 out phase is
(A) 1 : 1 (B) 2 : 1
(C) 2 : 1 (D) 4 : 1
28. Instead of using two slits as in Young's experiment, if we use two separate but
identical sodium lamps, which of the following occur ?
(A) general illumination (B) widely separate interference
(C) very bright maximum (D) very dark minimum
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29. For best contrast between maxima and minima in the interference pattern of
Young's double slit experiment, the intensity of light emerging out of the two slits
should be
(A) equal (B) double
(C) small (D) large
30. The path difference between two interfering waves at a point on a screen is 11.5
times the wavelength. The point is
(A) dark (B) bright
(C) neither dark nor bright (D) data is inadequate
31. In an interference pattern produced by two identical slits, the intensity at the site
of maxima is I. When one of the slit is closed, the intensity at the same spot is I0.
What is the relation between I and I0
(A) I = 2I0 (B) I = 4I0
(C) I = 16I0 (D) I = I0
32. In a Youngs double slit experiment, the position of first bright fringe coincides
with S1 and S2 respectively on the either side of central maxima. What is the
wavelength of the light used? [Take D = 1m and d = 1.2 mm]
(A) 3600A (B) 5400A
(C) 7200A (D) none of these.
33. In a Young's double slit experiment, if the slits are of unequal width,
(A) fringes will not be formed
(B) the positions of minimum intensity will not be completely dark.
(C) bright fringe will not be formed at the centre of the screen
(D) distance between two consecutive bright fringes will not be equal to the
distance between two consecurive dark fringes.
## 34. Two identical coherent sources of light S1 and S2 separated by
a distance 'a' produce an interference pattern on the screen. S1
The wave length of the monochromatic light emitted by the
sources is . The maximum number of interference fringes
S2
that can be observed on the screen is nearly equal to
2a a D
(A) +1 (B)
a
(C) (D) 1
a
35. In Young's double slit experiment, we get 60 fringes in the field of view of
monochromatic light of wavelength 4000 A0. If we use monochromatic light of
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wavelength 6000 A0, then the number of fringes obtained in the same field of
view is
(A) 60 (B) 90
(C) 40 (D) 1.5
36. In Young's double slit experiment, the 7th maximum with wavelength 1 is at a
distance d1 and that with wavelength 2 is at a distance d2 . Then d1/d2 is
(A) 1/ 2 (B) 2 / 1
2 2 2 2
(C) 1/ 2 (D) 2 / 1
37. In a two slit experiment with white light, a white fringe is observed on a screen
kept behind the slits. When the screen is moved away by 0.05 m, this white
fringe
(A) does not move at all
(B) gets displaced from its earlier position
(C) becomes coloured
(D) disappears
38. A source emits electromagnetic waves of wavelength 3m. One beam reaches
the observer directly and other after reflection from a water surface, travelling
1.5m extra distance and with intensity reduced to 1/4 as compared to intensity
due to the direct beam alone. The resultant intensity will be
(A) (1/4) fold (B) (3/4) fold
(C) (5/4) fold (D) (9/4) fold
39. Ratio of intensities of two waves are given by 4 :1. Then the ratio of the
amplitudes of the two waves is
(A) 2 :1 (B) 1 : 2
(C) 4 : 1 (D) 1 : 4
40. In the Young's experiment with sodium light, the slits are 0.589 m apart. What is
the angular width of the fourth maximum ? Given that = 589 nm.
(A) sin-1 (3 10-6) (B) sin -1 (3 10-8)
(C) sin-1 (0.33 10-6) (D) sin-1 (0.33 10-8)
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LEVEL II
1. An image of a bright square is obtained on a screen with the aid of a convergent lens.
The distance between the square and the lens is 40 cm. The area of the image is nine
time larger than that of the square. Select the correct statement(s) :
(A) Image is formed at a distance 120 cm from lens.
(B) Image is formed at a distance 360 cm from lens.
(C) Focal length of lens is 30 cm.
(D) Focal length of lens is 36 cm.
2. In a prism of angle A and refractive index , the maximum deviation occurs when
(A) the angle of incidence is 90
(B) the angle of incidence may be is sin 1 2
1sin A cos A
1
(C) the angle of emergence is sin sin A C
## (D) the angle of emergence is equal to the angle of incidence
3. A lens of focal length f is placed in between an object and screen at a distance D. The
lens forms two real images of object on the screen for two of its different positions, a
distance x apart. The two real images have magnifications m1 and m2 respectively (m1 >
m2).
x
(A) f (B) m1m2 = 1
m1 m 2
D2 x 2
(C) f (D) D 4f.
4D
4. An interference pattern is formed on the screen, when light from two different
monochromatic sources are allowed to interfere. Then, it is true that,
(A) frequencies of light from the two sources are equal to each other
(B) the sources are coherent
(C) the sources should be located in the same medium
(D) the path difference should either be an even or, an odd multiple of , where is the
2
wavelength of light
5. A thin paper of thickness 0.02 mm having refractive index 1.45 is pasted across one of
the slit in a Youngs double slit experiment. The paper transmits 4/9 of light falling on it.
( light = 600 nm).
(A) Amplitude of light wave transmitted through the paper will be 2/3 time of incident
wave.
(B) The ratio of maximum and minimum intensity in the fringe pattern will be 25.
(C) The total number of fringe crossing the centre if an identical paper is pasted on the
other slit is 15.
(D) The ratio of maximum and minimum intensity in the pattern will be 5.
6. For refraction through a small angled prism, the angle of minimum deviation :
(A) increases with the increases in R.I. of the prism
(B) will be 2D for a ray of R.I. 2.4, if it is D for a ray of R.I. 1.2
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## (C) is directly proportional to the angle of the prism
(D) will decrease with the increase in R.I. of the prism
7. The radius of curvature of the left and right surface of the concave lens are 10 cm and
15 cm respectively. The radius of curvature of the mirror is 15 cm :
Water
Air
(n=4/3)
Glass
(n=3/2)
(A) equivalent focal length of the combination is 18 cm
(B) equivalent focal length of the combination is + 36 cm
(C) the system behaves like a concave mirror
(D) the system behaves like a convex mirror
3
8. A point object is placed at 30 cm from a convex glass lens g of focal length 20
2
cm. The final image of object will be formed at infinity if :
(A) another concave lens of focal length 60 cm is placed in contact with the previous
lens
(B) another convex lens of focal length 60 cm is placed at a distance of 30 cm from the
first lens
(C) the whole system is immersed in a liquid of refractive index 4/3
(D) the whole system is immersed in a liquid of refractive index 9/8
9. The upper portion of lens is painted black in situation as shown in figure. Which of the
following statement(s) is/are correct :
## (A) the intensity of image will reduce by a factor of 2
(B) the distribution of brightness of image will not be symmetric
(C) the lower half of image will be brighter than upper half
(D) the upper half of image will be brighter than lower half
## 10. In a Youngs double-slit interference experiment the fringe pattern is observed on a
screen placed at a distance D. The slits are separated by d and are illuminated by light
of wavelength . The distance from the central point where the intensity falls to half the
maximum is :
D D
(A) (B)
3d 2d
D D
(C) (D)
d 4d
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COMPREHENSION
## I. A ray of light enters a spherical drop of water of
refractive index as shown in the figure.
## 1. Select the correct statement :
(A) Incident rays are partially reflected at point A.
(B) Incident rays are totally reflected at point A.
(C) Incident rays are totally transmitted through A.
(D) None of these.
## 2. An expression of the angle between incidence ray and
emergent ray (angle of deviation) as shown in the figure is
(A) 0 (B)
(C) (D) 4 + 2 . A
3. Consider the figure of question 8, the angle for which minimum deviation is produced
will be given by
2 2
1 1
(A) cos2 (B) cos2
3 3
2 2
1 1
(C) sin2 (D) sin2 .
3 3
## II. A thin biconvex lens of refractive index 3/2 is placed on a horizontal
plane mirror as shown in the figure. The space between lens and the
mirror is then filled with water of refractive index 4/3.
It is found that when a point object is placed 15 cm above the lens
on the principal axis the object coincides with its own image.
4. At what distance object should be placed before water is filled so that image coincides
with object if R is radius of curvature of lens
(A) 1.5 R (B) R
(C) 2R (D) R/2
5. In the above experiment when water is present, and parallel rays are incident then it will
converge at a distance
(A) 2.25 cm (B) 15 cm
(C) 10 cm (D) 7.5 cm
6. On repeating the above experiment in which water is replaced by a liquid of refractive
index image again coincide at a distance 25 cm from the lens then refraction index of
liquid is
(A) 1.5 (B) 1.4
(C) 1.8 (D) 1.6
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## MATCH THE FOLLOWING
1. An object located between the focus and the pole of a concave mirror moves towards
the pole with a constant velocity along its principal axis. Consider the image formed by
paraxial rays. Let 0 and I represent the magnitudes (absolute values) of the angles
subtended by the object and its image at the pole of the mirror respectively; and let m be
I
defined as . Use the New Cartesan Sign Convention.
0
Column I Column II
(A) Velocity of image (p) Positive.
(B) Acceleration of image (q) Negative.
(C) d 0 (r) Zero.
, i.e., the rate at which 0
dt
changes with time
(D) dm (s) Changes from positive to negative.
dt
## angle . 1 and 2 are refractive indices of media with ( 2 > 1).
Column I Column II
(A) (p)
1 1 1 1
When sin then deviation in sin
2 2 2
the path of ray is
(B) Maximum deviation in the path of ray (q)
1 1
for refraction at boundary 2 sin
2
## (C) Maximum deviation in the path of ray (r) Zero
for reflection at the boundary
(D) Deviation in the path at grazing angle of (s)
1 1
incidence sin sin
2
3. A plane mirror is tied to the free end of an ideal spring. The Mirror
V
other end of the spring is attached to a wall. The spring with
mirror is held vertically to the floor, can slide along it smoothly.
When the spring is at its natural length, the mirror is found to 2V
be moving at a speed of V with respect to ground frame. An
object is moving towards the mirror with speed 2V with respect
to ground frame. Then, Match the following :
Column I Column II
(A) Speed of image with respect to ground (p) V
frame when spring is at natural length
(B) Speed of image with respect to mirror (q) O
when spring is at natural length
(C) Speed of image with respect to object (r) 2V
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## when spring is at natural length
(D) Speed of image with respect to ground (s) 3V
frame when spring is at maximum
compressed state
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## 10. Answers to the Subjective Assignment
LEVEL I
apq R ( 2 n)
1. 2.
q p 2 (n 1)
3. R/n 4. 190cm, right of the lens
3R
5. 5 mm from left of 2nd surface 6.
3 1
## 7. 60 cm behind the mirror, virtual & inverted, 3
8. 9.365cm 9. 7/6 and 100 cm
10. 5 cm and inverted 11. 72 10-7 m
17. 1.7 cm
## For strong reflection 6222, 5091, 4308 A0
19. n2I0 , nI0 20. Zero, 1.5 mm
LEVEL II
1. h = 20 cm.
3. (i) sin 1 {n sin (450 n1/n)} (ii) i = 72.90
4. -2f from third lens
1 1 1
5. R =a
n12 1 n22 1 n231 1
## 6. 0.6m from lens, 0.003m
7. (160, -0.5), (135, -0.75), (160, -0.15)
f1f2 d(f1 d) (f1 d)
8. ,
f1 f2 d f1 f2 d
9. 15cm, -1.5, 1.5cm below RS, 0.3cm above RS
10. 15 cm
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1
0 d
2
## 19. (a) y0 = 4.33 10 3m (below X axis) (b) 0.75 Imax
(c) 650 nm, 433.34 nm
15 2r R
20.
32 r
## 11. Answers to the Objective Assignment
LEVEL I
1. (B) 2. (B)
3. (D) 4. (A)
5. (D) 6. (A)
7. (B) 8. (A)
9. (C) 10. (D)
11. (C) 12. (B)
13. (A) 14. (C)
15. (A) 16. (A)
17. (B) 18. (C)
19. (B) 20. (B)
21. (C) 22. (B)
23. (A) 24. (A)
25. (D) 26. (D)
27. (C) 28. (A)
29. (A) 30. (A)
31. (B) 32. (C)
33. (B) 34. (A)
35. (C) 36. (A)
37. (A) 38. (D)
39. (A) 40. (A)
LEVEL II
1. (A), (C) 2. (A), (B), (C)
3. (A), (B), (C), (D) 4. (A), (B)
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## 5. (A), (B), (C) 6. (A), (C)
7. (D) 8. (A), (D)
9. (A), (B), (C) 10. (D)
COMPREHENSION
1. (A) 2. (D)
3. (B) 4. (B)
5. (D) 6. (D)
## 3. (A) (q); (B) (p); (C) (r); (D) (r)
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MODERN PHYSICS
9. Assignment (Subjective Problems)
LEVEL - I
## 1. Frequency of a photon emitted due to transition of electron of a certain element
from L to K shell is found to be 4.2 1018 Hz. Using Moseleys law, find the
atomic number of the element, given that the Rydbergs constant R=1.1 107 m 1.
2. A radioactive sample emits n -particles in 2 sec. In next 2 sec it emits 0.75 n
-particle, what is the mean life of the sample?
3. The energy of a K-electron in tungsten is 20 KeV and of an L electron is
2 KeV. Find the wavelength of X-rays emitted when there is electron jump from
L to K shell.
4. One milliwatt of light of wavelength 4560 is incident on a cesium surface.
Calculate the photoelectric current produced, if the efficiency of the surface for
photoelectric emission is only 0.5%.
5. If the wavelength of the light falling on a surface is increased from 3000 to
3040 , then what will be the corresponding change in the stopping potential?
(Given that hc = 12.4 103 eV )
6. In an experiment on photoelectric emission, following observations were made:
(i) Wavelength of the incident light = 1.98 10 7 m, (ii) stopping potential = 2.5
volt. Find threshold frequency, work-function and energy of photoelectrons with
maximum speed. (Given e = 1.6 10 19 C, h = 6.6 10 34 J-s., c = 3 108 m/s)
7. Light of wavelength 180 nm ejects photo-electrons from a plate of metal whose
work-function is 2 eV. If a uniform magnetic field of 5 10 5 Tesla be applied
parallel to the plate, what would be the radius of the path followed by electrons
ejected normally from the plates with maximum energy (h = 6.62 10 34 J-s,
m = 9.1 10 31 kg and e = 1.6 10 19 coulomb).
8. Photoelectric threshold wavelength of metallic silver is = 3800 . Ultra-violet
light of = 2600 is incident on silver surface. Calculate
(a) the value of work function in joule and eV,
(b) maximum Kinetic energy of the emitted photo electrons,
(c) the maximum velocity of the photo electrons.
(Mass of the electron=9.11 10 31 kg).
9. Consider the fusion reaction 1H2 + 1H3 2He4. If 20 MeV of energy is released
per fusion reaction, mass of 1H2 consumed per day is 0.1 gm, what is the Power
of the reactor?
10. The radiation emitted due to de-excitation of electron from n=2 to n=1 in H2 atom
falls on a metal to produce photo electrons. The electrons from the metal surface
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with maximum kinetic energy are made to move perpendicular to a magnetic field
1
of T in a radius 10-3m. Find the threshold wavelength for the metal.
160
LEVEL II
1. A hydrogen atom moving with a velocity 6.24 104 m/s makes a perfectly
inelastic head on collision with another stationary hydrogen atom. Both atoms are
in ground state before collision. Up to what state either one atom may be excited.
2. An X-ray tube with a copper target is found to emit lines other than those due to
copper. The K line of copper is known to have a wavelength 1.5405 A 0 and the
other two K lines observed have wavelengths 0.7092 A0 and 1.6578 A0. Identify
the impurities (find the value of Z, atomic number). What is the minimum voltage
at which the X-ray tube should be operated?
3. Radiation falls on a target kept within a solenoid with 20 turns per cm, carrying a
current 2.5 A. Electrons emitted move in a circle with a maximum radius of 1 cm.
Find the wavelength of radiation, given that the work function of the target is 0.5
volts, e = 1.6 10 19 coulomb, h = 6.625 10 34 J s, m = 9.1 10 31 Kg.
4. Electrons in a hydrogen like atom (Z = 3) make transitions from 5th to 4th orbit
and from the 4th to the 3rd orbit. The resulting radiation is incident normally on a
metal plate and the photo- electrons are ejected. The stopping potential for the
photoelectrons ejected by light of shorter wavelength is 3.95V. Calculate the
work function of the metal and the stopping potential the photo electrons for the
longer wavelength.
## 5. When the voltage applied to an X-ray tube increased from V1 = 10 KV to
V2=20 KV, the wavelength interval between the K -line and the short wavelength
cut-off of the continuous X-ray spectrum increases by a factor of 3. Find the
atomic number of the element of the target.
## 6. The maximum kinetic energy of photoelectrons emitted from a metallic surface is
30 eV when monochromatic radiation of wavelength falls on it. When the same
surface is illuminated with light of wavelength 2 , the maximum kinetic energy
photo electrons is observed to be 10 eV. Calculate the wavelength and
determine the maximum wavelength of incident radiation for which
photoelectrons can be emitted by this surface.
(h = 6.62 10 34 J-S = 4.14 10 15 eV-s, c = 3 108 m/s)
## 7. A monochromatic beam of light ( = 4900 ) incident normally upon a surface
produces a pressure of 5 x 10-7 N/m2 on it. Assuming that 25% of the light
incident is reflected and the rest absorbed, find the number of photons falling per
second on a unit area of thin surface.
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## 8. A nuclear explosion is designed to deliver 1 MW of heat. How many fission
events must be required in a second to attain this power level? Assume that this
explosion is designed with nuclear fuel consisting of uranium -235. Calculate the
amount of fuel needed to run a reactor at this power level for one year. You can
assume that the amount of energy released per fission event is 200 MeV.
## 9. An electron in the ground state of the hydrogen atom is n B
300
revolving in the anti-clockwise direction in a circular orbit of
(a) Obtain an expression for the orbital magnetic dipole moment
of the electron.
(b) The atom is placed in a uniform magnetic induction B such
that the plane normal to the electron orbit makes an angle
30o with the magnetic induction. Find the torque experienced
by the orbiting electron.
## 10. A radionuclide A1 with decay constant 1 transforms into a radionuclide A2 with
decay constant 2. Assuming that at the initial moment the preparation contained
(a) the equation describing accumulation of the radionuclide A2.
(b) the time interval after which the activity of radionuclide A2 reaches the
maximum value. Assume concentration of A1 at t = 0 to be No.
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## 10. Assignment (Objective Problems)
LEVEL I
1. The total energy of the electron in the hydrogen atom in the ground state is 13.6 eV.
Which of the following is its kinetic energy in the first excited state?
(A) 13.6 eV (B) 6.8 eV
(C) 3.4 eV (D) 1.825 eV
2. A freshly prepared radioactive source of half life 2 hrs emits radiation of intensity
which is 32 times the permissible safe value of intensity. Which of the following is the
minimum time after which it would be possible to work safely with this source?
(A) 16 hrs (B) 5 hrs
(C) 10 hrs (D) 32 hrs
3. The ionisation potential of a hydrogen atom is 13.6 volt. The energy required to
remove an electron from the second orbit of hydrogen is:
(A) 3.4 eV (B) 6.8 eV
(C) 13.6 eV (D) 27.2 eV
4. The radius of the first Bohr orbit is a0. The nth orbit has a radius:
(A) na0 (B) a0/n
2
(C) n a0 (D) a0/n2
## 5. The ionisation energy of the ionised sodium atom Na+10 is :
(A) 13.6 eV (B) 13.6 11 eV
(C) (13.6/11) eV (D) 13.6 (112) eV
## 6. Radius of the second Bohr orbit of a singly ionised helium atom is
(A) 0.53 A0 (B) 1.06 A0
0
(C) 0.265 A (D) 0.132 A0
## 7. The potential difference applied to an X-ray tube is increased. As a result, in the
(A) the maximum wavelength increases
(B) the minimum wave length increases
(C) the minimum wavelength remains unchanged
(D) the minimum wave length decreases
## 8. A beam of electrons accelerated by a large potential difference V is made to
strike a metal target to produce X-rays. For which of the following values of V, the
resulting X-rays have the lowest minimum wave length:
(A) 10 KV (B) 20 KV
(C) 30 KV (D) 40 KV
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## 9. The X-ray beam emerging from an X-ray tube
(A) is monochromatic
(B) contains all wavelengths smaller than a certain maximum wavelength
(C) contains all wave lengths larger than a certain minimum wavelength
(D) contains all wave lengths lying between a minimum and a maximum wavelength.
10. The relation between half-life T of a radioactive sample and its mean life is:
(A) T = 0.693 (B) = 0.693 T
(C) = T (D) = 2.718 T
11. The stopping potential for the photo electrons emitted from a metal surface of work
function 1.7 eV is 10.4 V. Identify the energy levels corresponding to the transitions
in hydrogen atom which will result in emission of wavelength equal to that of incident
radiation for the above photoelectric effect
(A) n = 3 to 1 (B) n = 3 to 2
(C) n = 2 to 1 (D) n = 4 to 1
12. An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom
gets excited and the colliding electron loses all its kinetic energy. Consequently the
hydrogen atom may emit a photon corresponding to the largest wavelength of the
Balmer series. The K.E. of colliding electron will be
(A) 10.2 eV (B) 1.9 eV
(C) 12.1 eV (D) 13.6 eV
13. When a radioactive isotope 88Ra228 decays in series by the emission of three -
particles and a particle the isotope finally formed is :
(A) 84X220 (B) 86X222
216
(C) 83X (D) 83X215
14. Photo electric effect supports the quantum nature of light because:
(A) there is a minimum frequency of light below which no photo electrons are
emitted.
(B) the maximum K.E. of photoelectrons depends only on the frequency of light
and not on its intensity.
(C) even when the metal surface is faintly illuminated by light of the approximate
wavelength, the photo electrons leave the surface immediately.
(D) electric charge of photoelectrons is quantized.
15. If the electron in the hydrogen atoms is excited to n = 5 state, the number of
frequencies present in the radiation emitted is :
(A) 4 (B) 5
(C) 8 (D) 10
16. The ratio of magnetic dipole moment of an electron of charge e and mass m in the
Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:
(A) e/m (B) e/2m
(C) m/e (D) 2m/e
17. The wave length of K -ray line of an anticathode element of atomic number Z is
nearly proportional to:
(A) Z2 (B) (Z 1)2
1 1
(C) (D)
( Z 1) (Z 1)2
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18. If 1 and 2 are the wavelengths of characteristic X-ray and gamma rays
respectively, then the relation between them is :
(A) 1 2 (B) 1 = 2
(C) 1 > 2 (D) 1 < 2
## 19. In the nuclear reaction given by 2He4 + 7N14 1
1H + X then the nucleus X is :
(A) Nitrogen of mass 16 (B) Nitrogen of mass 17
(C) Oxygen of mass 16 (D) Oxygen of mass 17
20. If 10% of a radioactive material decays in 5 days, then the amount of the original
material left after 20 days is approximately:
(A) 60% (B) 65%
(C) 70% (D) 75%
LEVEL II
1. Suppose the potential energy between electron and proton at a distance r is given by
Ke 2
. Application of Bohrs theory of hydrogen atom in this case shows that :
3r 3
(A) energy in the nth orbit is proportional to n6
(B) energy is proportional to m 3 (m : mass of electron)
(C) energy in the nth orbit is proportional to n 2
(D) energy is proportional to m3 (m = mass of electron)
2. X ray from a tube with a target A of atomic number Z shows strong K lines for target A
and weak K lines for impurities. The wavelength of K lines is z for target A and 1 and
2 for two impurities.
z 1
4 and z .
1 2 4
Screening constant of K lines to be unity. Select the correct statement(s)
(A) The atomic number of first impurity is 2z 1.
(B) The atomic number of first impurity is 2z + 1.
(z 1)
(C) The atomic number of second impurity is .
2
z
(D) The atomic number of second impurity is 1.
2
3. Energy liberated in the deexcitation of hydrogen atom from 3rd level to 1st level falls on
a photocathode. Later when the same photocathode is exposed to a spectrum of
some unknown hydrogen like gas, excited to 2 nd energy level, it is found that the de
Broglie wavelength of the fastest photoelectrons, now ejected has decreased by a factor
of 3. For this new gas, difference of energies of 2 nd Lyman line and 1st Balmer line is
found to be 3 times the ionization potential of the hydrogen atom. Select the correct
statement(s) :
(A) The gas is lithium.
(B) The gas is helium.
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## (C) The work function of photocathode is 8.5 eV.
(D) The work function of photocathode is 5.5 eV.
## 4. Hydrogen atoms absorbs radiation of wavelength 0 and consequently emit radiations of
6 different wavelengths of which two wavelengths are shorter than 0.
(A) The final excited state of the atoms is n = 4
(B) The initial state of the atoms may be n = 2
(C) The initial state of the atoms may be n = 3
(D) There are three transitions belonging to Lyman series.
5. For a certain radioactive substance, it is observed that after 4 hours, only 6.25% of the
original sample is left undecayed. If follows that
(A) the half life of the sample is 1 hour
(B) the mean life of the sample is 1 hour
ln2
(C) the decay constant of the sample is ln 2 hour-1
(D) after a further 4 hours, the amount of the substance left over would by only 0.39% of
the original amount
6. Let 1 be the frequency of the series limit of the Lyman series, 2 be the frequency of the
first line of the Lyman series, and 3 be the frequency of the series limit of the Balmer
series
(A) 1 2 3 (B) 2 1 3
1
(C) 3 ( 1 2) (D) 1 2 3
2
2. An electron in a hydrogen atom makes a transition from n = n1 to n = n2. The time period
of the electron in the initial state is eight times that in the final state. The possible values
of n1 and n2 are
(A) n1 = 4, n2 = 2 (B) n1 = 8, n 2 = 2
(C) n1 = 8, n2= 1 (D) n1 = 6, n 2 = 3
## 3. Whenever a hydrogen atom emits a photon in the Balmer series,
(A) it may emit another photon in Balmer series
(B) it must emit another photon in Lyman series
(C) the second photon, if emitted, will have a wavelength of about 122 nm
(D) it may emit a second photon, but the wavelength of this photon cannot be predicted
4. When an electron moving at a high speed strikes a metal surface, which of the following
are possible ?
(A) the entire energy of the electron may be converted into an X-ray photon
(B) any fraction of the energy of the electron may be converted into an X-ray photon
(C) the entire energy of the electron may get converted to heat
(D) the electron may under go elastic collision with the metal surface
5. There are two radioactive nuclei A and B. A is an alpha emitter and B is a beta emitter.
Their disintegration constants are in ratio of 1 : 2. What should be the ratio of number of
atoms of A and B at any time t so that probabilities of getting alpha and beta particle are
same at that instant
(A) 2 : 1 (B) 1 : 2
(C) 3 (D) e-1
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6. An electron in hydrogen atom first jumps from second excited state to first excited state
and then from first excited state to ground state. Let the ratio of wavelength, momentum
and energy of photons emitted in these two cases be a, b and c respectively. Then
1 9
(A) c (B) a
a 4
5 5
(C) b (D) c
27 27
7. The wavelengths and frequencies of photons in transitions 1, 2 and 1
3 for hydrogen like atom are 1, 2, 3, 1, 2 and 3 respectively.
Then 3
2
(A) 3 1 2 (B) 3
1 2
1 2
(C) 3 1 2 (D) 3
1 2
1 2
## 8. When the intensity of a light source is increased,
(A) the number of photons emitted by the source in unit time increases
(B) the total energy of the photons emitted per unit time increases
(C) more energetic photons are emitted
(D) faster photons are emitted
## 9. If the wavelength of light in an experiment on photoelectric effect is doubled,
(A) the photoelectric emission will not take place
(B) the photoelectric emission may or may not take place
(C) the stopping potential will increase
(D) the stopping potential will decrease
10. The collector plate in an experiment on photoelectric effect is kept vertically above the
emitter plate. Light source is put on and a saturation photocurrent is recorded. An
electric field is switched on which has a vertically downward direction.
(A) the photocurrent will increase
(B) the kinetic energy of the electrons will increase
(C) the stopping potential will decrease
(D) the threshold wavelength will increase
COMPREHENSION
I : [Question No. 1 to 3]
Many unstable nuclei can decay spontaneously to a nucleus of lower mass but different
combination of nucleons. The process of spontaneous emission of radiation is called
Radioactive decay is a statistical process. Radioactivity is independent of all external conditions.
The number of decay per unit time or decay rate is called activity. Activity exponentially
decreases with time.
## Mean life time is always greater than half lift time.
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(B) Radioactivity is independent of high temperature and high pressure
(C) When a nucleus undergoes or decay, its atomic number changes
(D) All of these
2. If the decay constants of a radioactive element for and decay are 1 and 2
respectively. The total decay constant ( ) is :
(A) 1 2
(B) 1 2
1 2 1 2
(C) 1 2 (D) 1 2
3. The activity of radioactive substance is R1 at time t1 and R2 at time t2 ( > t1) the decay
constant is
(A) R1t1 R 2 t 2 (B) R2 R1e ( t2 t1 )
R R2
(C) R2 R1e ( t1 t 2 ) (D) 1 constant
t 2 t1
II : [Question No. 4 to 6]
All nuclei consist of two type of particles protons and neutrons. Nuclear force is the strongest
force. Stability of nucleus is determined by the neutron proton ratio or mass defect or Binding
energy per nucleons or packing fraction. Shape of nucleus calculated by quadrupole moment.
Spin of nucleus depends on even or odd mass number. Volume of nucleus depends on the
mass number. Whole mass of atom (nearly 99%) is centred at the nucleus. Magnetic moment of
nucleus measured by the nuclear magnetons.
## 4. The correct statement(s) about nuclear force is/are
(A) Charge independent (B) Short ranges forces
(C) Non conservative force (D) All option are correct
## 5. Volume (V) of the nucleus is related with mass number (A) as
(A) V A2 (B) V A1/3
2/3
(C) V A (D) V A
6. The mass defect in a particular nuclear reaction is 0.5 gram. The amount of heat energy
liberated in Joule is
(A) 4.5 10 13 Joule (B) 45 1016 Joule
15
(C) 45 10 Joule (D) 0.5 931 Joule
## 1. Match the following :
Column I Column II
(A) Particle behaviour of light (p) Reflection
(B) Electron microscope (q) Refraction
(C) Xray photon (r) Interference
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## (D) Spectrum (s) Photoelectric effect
2. Some quantities related to the photoelectric effect are mentioned under Column I and
Column II. Match each quantity in Column I with the corresponding quantities in Column
II on which it depends.
Column I Column II
(A) Saturation current (p) Frequency of light
(B) Stopping potential (q) Work function
(C) de-Broglie wavelength of photoelectron (r) Area of photosensitive plate
(D) Force due to radiation falling on the (s) Intensity of light (at constant
photoplate. frequency)
## 3. Match the following :
Column I Column II
(A) decay (p) For atoms of high atomic number
(B) Fusion (q) Mass energy equivalence
(C) Fission (r) For atoms of low atomic number
(D) Exothermic nuclear reaction (s) Involves weak forces
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LEVEL I
1 2
1. 42 2. sec
ln(4 / 3)
5. -0.055 V
## 6. 9.1 1014 per sec, 6.0 10-19 J, 4.0 10-19 J
7. 0.148 m.
8. (a) 5.23 10-19 J, 3.27 eV (b) 1.5 eV (c) 0.7289 106 m/s
LEVEL II
## 7. 3 1020 m-2 sec-1 8. 3.125 1016, 356.3 gm
eh ehB 1 1t 2t
9. ; 10. N2 = N0 e e
4 m 8 m 2 1
## 12. Answers to the Objective Assignment
LEVEL I
1. C 2. C
3. A 4. C
5. D 6. B
7. D 8. D
9. C 10. A
11. A 12. C
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13. C 14. C
15. D 16. B
17. D 18. C
19. D 20. B
LEVEL II
## 1. (A), (B) 2. (A), (C)
3. (B), (C) 4. (A), (B), (D)
5. (A), (B), (C), (D) 6. (A)
7. (A), (D) 8. (B), (C)
9. (A), (B), (C) 10. (A)
11. (A), (C), (D) 12. (A), (D)
13. (A), (B) 14. (B), (D)
15. (B)
COMPREHENSION
1. (D) 2. (C)
3. (B) 4. (D)
5. (D) 6. (A)
## MATCH THE FOLLOWING
1. (A) (p), (q), (s); (B) (r); (C) (p), (q), (r), (s); (D) (q)
2. (A) (s); (B) (p), (q); (C) (p), (q); (D) (p), (r), (s)
3. (A) (q), (s); (B) (q), (r); (C) (p), (q); (D) (q)
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Under ideal conditions, a service bay at a Fast Lube
Not my Question
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# Question : Under ideal conditions, a service bay at a Fast Lube : 6605
Under ideal conditions, a service bay at a Fast Lube can serve 7 cars per hour. The effective capacity of a Fast Lube service bay is 5.5 cars per hour, with efficiency known to be 0.85. The minimum number of service bays Fast Lube needs to achieve an anticipated production of 250 cars per 8-hour day . ??? service bays (enter your response rounded up to the next whole number).
## Solution 4 (1 Ratings )
Solved
Mathematics 1 Year Ago 15 Views
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### try, throw, catch exceptions
Hi all, ive been writing this same code for the last few days. I'm trying to utilize the try, throw, and catch exceptions. So my code compiles as I posted below. But I know for a fact I'm not using the exceptions correctly. I've only gone over exceptions the last two days, so I really still only understand a little bit of them. How would I get the exceptions to run my code but check for any unexpected errors?
``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127`` ``````#include #include //Exception handling. using namespace std; int main() { int bArray[16] = { 0 }; int k = 0; int oArray[8] = { 0 }; int i = 0; int hArray[8] = { 0 }; int j = 0; int decimal = 0; int number; cout << "Enter 'B' to convert a decimal number to binary, 'O' for an octal conversion,and'H' for a hexadecimal conversion. " << "Hit space to quit program." << endl; try{ while ((number = cin.get()) != ' ') { cout << "\nEnter a decimal number to be converted (No more than 65,535): " << endl; switch (number) { case 'B': case 'b': cin >> decimal; while ((decimal != 0) & (k <= 15)) { bArray[k] = decimal % 2; decimal = decimal / 2; k++; } cout << "Here is the binary conversion: "; for (k = 15; k >= 0; k--) { cout << bArray[k]; if ((k % 4) == 0) cout << " "; } cout << endl; break; case 'O': case 'o': cin >> decimal; while ((decimal != 0) & (i <= 7)) { oArray[i] = decimal % 8; decimal = decimal / 8; i++; } cout << "Here is the octal conversion: "; for (i = 7; i >= 0; i--) { cout << oArray[i]; if ((i % 4) == 0) cout << " "; } cout << endl; break; case 'H': case 'h': cin >> decimal; for (int j = decimal; j > 0; j--){ hArray[7] = hArray[7] + 1; if (hArray[7] > 15){ for (int n = 6; n > -1; n--){ hArray[n + 1] = 0; hArray[n] = hArray[n] + 1; if (hArray[n] < 16){ n = -1; } } } } cout << "\n\nConverted to Hexidecimal is: "; for (int j = 0; j < 8; j++) { if (hArray[j] == 10){ cout << 'A'; } else if (hArray[j] == 11){ cout << 'B'; } else if (hArray[j] == 12){ cout << 'C'; } else if (hArray[j] == 13){ cout << 'D'; } else if (hArray[j] == 14){ cout << 'E'; } else if (hArray[j] == 15){ cout << 'F'; } else { cout << hArray[j]; } } break; case '\n': case '\t': case ' ': break; default: cout << "Wrong decimal input." << "Enter a new number." << endl; break; } } throw 99; } catch (int x) { cout << "Error: " << x << endl; } cout << "Thank you." << endl;//end main }``````
Just wanted to update my situation. I solved a bit of it on my own. However my code is "throwing" the wrong function I guess. Originally my exception handling is suppose to deal with the input for B, O, or H. If someone put in like K I'll display the error with the exception. Instead my exception and error display shows up when the user is entering a decimal after they've chosen what they want converted (B, H, or O). Here's my updated code:
``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142`` `````` #include #include //Exception handling. using namespace std; int main() { int bArray[16] = { 0 }; int k = 0; int oArray[8] = { 0 }; int i = 0; int hArray[8] = { 0 }; int j = 0; int decimal = 0; int number; cout << "Enter 'B' to convert a decimal number to binary, 'O' for an octal conversion,and'H' for a hexadecimal conversion. " << "Hit space to quit program." << endl; try{ while ((number = cin.get()) != ' ') { cout << "\nEnter a decimal number to be converted (No more than 65,535): " << endl; switch (number) { case 'B': case 'b': cin >> decimal; if ((decimal != 'B') && (decimal)){ throw 99; } while ((decimal != 0) & (k <= 15)) { bArray[k] = decimal % 2; decimal = decimal / 2; k++; } cout << "Here is the binary conversion: "; for (k = 15; k >= 0; k--) { cout << bArray[k]; if ((k % 4) == 0) cout << " "; } cout << endl; break; case 'O': case 'o': cin >> decimal; if ((decimal != 'O') && (decimal != 'o')){ throw 99; } while ((decimal != 0) & (i <= 7)) { oArray[i] = decimal % 8; decimal = decimal / 8; i++; } cout << "Here is the octal conversion: "; for (i = 7; i >= 0; i--) { cout << oArray[i]; if ((i % 4) == 0) cout << " "; } cout << endl; break; case 'H': case 'h': cin >> decimal; if ((decimal != 'H') && (decimal != 'h')){ throw 99; } for (int j = decimal; j > 0; j--) { hArray[7] = hArray[7] + 1; if (hArray[7] > 15) { for (int n = 6; n > -1; n--) { hArray[n + 1] = 0; hArray[n] = hArray[n] + 1; if (hArray[n] < 16) { n = -1; } } } } cout << "\n\nConverted to Hexidecimal is: "; for (int j = 0; j < 8; j++) { if (hArray[j] == 10){ cout << 'A'; } else if (hArray[j] == 11){ cout << 'B'; } else if (hArray[j] == 12){ cout << 'C'; } else if (hArray[j] == 13){ cout << 'D'; } else if (hArray[j] == 14){ cout << 'E'; } else if (hArray[j] == 15){ cout << 'F'; } else { cout << hArray[j]; } } break; case '\n': case '\t': case ' ': break; default: cout << "Wrong decimal input." << "Enter a new number." << endl; break; } } } catch (int x) { cerr << "Error: please run program again." << x << endl; } cout << "Thank you." << endl;//end main } ``````
I appreciate all the help I've been getting with this code.
You are trying to enter a character ('H') into an integer (decimal). You are also trying to compare an integer (number) with a character (' ', or 32).
I have a few recommendations:
1) replace decimal with the datatype of char
2) Change lines 38 and 61 to use a logical AND rather than binary AND
3) Remove your throw statements from within the cases and instead throw from inside the `default` case
4) Put in a check for if the decimal is larger than 65536, and if so, throw
5) Rather than throwing error codes (what does "99" mean?) throw meaningful data, such as std::runtime_error or std::range_error(see http://www.cplusplus.com/reference/stdexcept/ )
6) Have a catch block that catches arbritatry exceptions, and another to catch everything (just in case) (this isn't really necessary, but I like to know that my program crashed due to exception rather than something like segfault). Here is an example:
``12345678910`` ``````try { //... main function } // specialized catch blocks for error handling, and then: catch (std::exception& e) { std::cerr << "Unhandled exception caught: " << e.what() << "\n"; } catch (...) { std::cerr << "Unrecognized expection caught.\n"; }``````
Thanks for the reply. I've changed everything you've recommended. I can't believe I've been using the datatype int instead of char for my switch statements. Anyway I really don't know how use the throw in the default case. So far I have:
``123`` ``````default: throw invalid_argument ("Wrong input"); break;``````
First, looking over my post before, I seem to have made a mistake (I do that a lot). I meant to say for point one, to make "number" a `char` (get the type of the number), not make "decimal" a` char`.
What you have there is pretty much what you want. However, rather than just saying "Wrong Input", which we have gathered from it being an "invalid argument"-type error, you could give details on the error (such as saying what character was recieved and what were required, like this:
``123`` ``````default: throw std::invalid_argument ('"' + number + "\" recieved; B O or H expected"); // break not required due to the throw terminating program run ``````
Then you could have a catch block to catch instances of std::invalid_argument, and display an error saying something like:
``123`` ``````catch (const std::invalid_argument& e) { std::cerr << "Error: Invalid Argument: " << e.what() << "\n"; }``````
In this case, it would say if I input L:
`Error: Invalid Argument: "L" recieved, B O or H expected`
However, it seems that you get the idea.
Topic archived. No new replies allowed.
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# If y is the smallest positive integer such that 3,150
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Director
Joined: 29 Aug 2005
Posts: 808
If y is the smallest positive integer such that 3,150 [#permalink]
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23 Nov 2008, 08:21
If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then
y must be
A. 2
B. 5
C. 6
D. 7
E. 14
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Director
Joined: 29 Aug 2005
Posts: 808
Re: GMAT Set 27 - 30 [#permalink]
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23 Nov 2008, 10:33
DavidArchuleta wrote:
3150= 25 . 9. 14
David, how did you arrive to the answer? To be honest, i dont get the question .
Manager
Joined: 05 Jul 2008
Posts: 131
GMAT 2: 740 Q51 V38
Re: GMAT Set 27 - 30 [#permalink]
### Show Tags
24 Nov 2008, 10:07
1
botirvoy wrote:
DavidArchuleta wrote:
3150= 25 . 9. 14
David, how did you arrive to the answer? To be honest, i dont get the question .
Oh my gosh, ROFL..
I really love that.
You call me David haha. I'm a girl, David Archuleta is my fav singer.
Anyway, the question ask you the smallest number that if you multiple that number with 3150, the result is is the square of an integer.
3150= 5^2. 3^2. 14=15^2 . 14
So you must multiple 14 to get:
14. y = 15^2 . 14^2=(15.14)^2
Director
Joined: 29 Aug 2005
Posts: 808
Re: GMAT Set 27 - 30 [#permalink]
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30 Nov 2008, 11:30
DavidArchuleta wrote:
botirvoy wrote:
DavidArchuleta wrote:
3150= 25 . 9. 14
David, how did you arrive to the answer? To be honest, i dont get the question .
Oh my gosh, ROFL..
I really love that.
You call me David haha. I'm a girl, David Archuleta is my fav singer.
Anyway, the question ask you the smallest number that if you multiple that number with 3150, the result is is the square of an integer.
3150= 5^2. 3^2. 14=15^2 . 14
So you must multiple 14 to get:
14. y = 15^2 . 14^2=(15.14)^2
In this side of the ocean we (act, I) have got no clue about David Archuleta! Sings well though.
Thanks for explaining the q.
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Re: GMAT Set 27 - 30 &nbs [#permalink] 30 Nov 2008, 11:30
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View Full Version : Another course 2 question
bg23516
04-21-2002, 04:15 PM
Company A invests \$10,000 in machinery that is expected to yield net cash flows of \$7000 in each of the next 2 years. Company A intends to record depreciation costs of \$5000 in each of the next 2 years.
Find the Economic Rate of Return and Book Rate of Return on Investment if cost of capital is 10%.
---
I understand how to find the the economic rate of return....
Economic Income = CF1 + PV1 - PV0 = 1215
EROR = EI/PV0 = 10%
My problem is with finding book ROI...
BROI = Book Income / BV0 where:
Book Income = CF1 + BV1 - BV0
According to the solutions, BV0 = 10,000 (2*5000) and BV1 = 5000. Why is book value equal to the depreciation, and why is it just added, despite the different years?
Gandalf
04-21-2002, 04:49 PM
Why am I trying to discuss course 2 having none of the study material?
That said, I suspect BV0 = 10,000 because that's what you paid for it. It's somewhat of a coincidence that it equals 2 * 5,000, but not completely coincidence. It is true that BV0 = sum of depreciation in all future periods + any residual value after the depreciation, and that in many situations such as this one residual value (at least residual book value) is 0.
Normally BV1 does not equal depreciation. BV1 = BV0 - depreciation during 1. Here, BV1 = 10,000 - depreciation just happens to equal depreciation.
Those are my guesses, anyway.
ASA_Woman
04-21-2002, 06:39 PM
Gandalf is correct. BV0 is the initial value of the machinery, ie 10,000. BV1 is the value at the end of 1 year, ie after one year's worth of depreciation. So BV1 = BV0 - 5,000 = 10,000 - 5,000 = 5,000.
The question states that the depreciation is 5,000 each year. It just works out this time that BV0 is 2 times the depreciation. That will not always be the case.
GuyInWestGrove
04-12-2004, 03:35 PM
Sorry to resurrect this thread two years later, but ..
This appears to be problem 47 from the sample exam (Adobe PDF at http://casact.org/admissions/studytools/exam2/sampleExam2.pdf
The original post seems to indicate that the economic income calculation in the solution was in error. The solution shows CF = 7, PV at beginning of year = 10, PV at EOF = 7/1.1 = 6.36. Economic income = 7 - (10-6.36) = 3.36. Return = 3.36 / 10.
Calculating the PV's from the discounted future cash flows as bg23516 has jives with what Braeley and Meyers seem to do.
A couple of questions --
1. Is the solution in the PDF correct?
2. How would you know to look at the return in only the first year, and not over say both years, or even just the second year?
3. What is the origin of the sample exam? (i.e. Did these questions appear on real exams in the past? They seem harder than the other published exams.)
Svak
04-13-2004, 12:58 AM
B&M are referring to year when calculating Rate of Return either Economic Rate of Return or Book Rate of Return.
If we assume question is on return during the "first year", then,
Economic Rate of Return
Change in PV (from year 0 to year 1) /PV at year 0
[CF<sub>1</sub>+ PV<sub>1</sub>- PV<sub>0</sub>]÷PV<sub>0</sub>
[7000+6363.63-12147.70]÷12147.70
=10%
Solution is different from this that it is using 10000 as PV<sub>0</sub>
It makes sense to take 10000 in the denominator, since this is what we are investing. But is it the Present Value at time zero?
Book Rate of Return
Book Rate Return, I think is straight forward. There is no confusion.
[CF<sub>1</sub>+BV<sub>1</sub>-BV<sub>0</sub>]÷BV<sub>0</sub>
[7000+5000-10000]÷10000
Investing 10000 and receiving 2000 net income (7000 cash flow and losing 5000 as depreciation).
Gandalf
04-13-2004, 08:47 AM
B&M is referring to year when calculating Rate of Return either Economic Rate of Return or Book Rate of Return.
If we assume question is on return during the "first year", then,
Economic Rate of Return
Change in PV (from year 0 to year 1) /PV at year 0
[CF<sub>1</sub>+ PV<sub>1</sub>- PV<sub>0</sub>]÷PV<sub>0</sub>
[7000+6363.63-12147.70]÷12147.70
=10%
Solution is different from this that it is using 10000 as PV<sub>0</sub>
It makes sense to take 10000 in the denominator, since this is what we are investing. But is it the Present Value at time zero?
You really need to see if the text specifies what to do. One thing that seems clear to me is that you must use the same value for PV0 in both numerator and denominator.
Which to use depends on what is meant by "first year" (and the problem itself didn't even specify "first year" :swear: )
If the first year is from the time you had 10,000 and ends 1 year later, then PV0 should be 10,000. You started with 10,000, and added economic value during the period by making the investment.
If the first year is one year starting immediately after the moment the investment was made, then the economic gain at the moment of investment is before the first year, and PV0 should be 12,148.
Becoming Actuary
04-14-2004, 10:20 AM
In book rate of return -
The initial value BV0 is always equal to initial investment (10,000). and book value at time 1, BV1 is BV0-depr (which in this case is 5000. sometimes they say its a st. line depr. for lets just say 5 years then you calculate the depr = initial invest./#of yrs.=10,000/5 = 2000).
san
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