Question:
A plane electromagnetic wave of wavelength $\\lambda\$ has an
intensity I. It is propagating along the positive Y-direction. The
allowed expressions for the electric and magnetic fields are given by

$\overset{\rightarrow}{E} = \sqrt{\frac{2I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\;\ \overset{\rightarrow}{B} = + \frac{1}{c}E\overset{\hat{}}{i}$

$\overset{\rightarrow}{E} = \sqrt{\frac{I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\overset{\hat{}}{k}\;\ \overset{\rightarrow}{B} = + \frac{1}{c}E\overset{\hat{}}{i}$

$\overset{\rightarrow}{E} = \sqrt{\frac{2I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\overset{\hat{}}{k}\;\ \overset{\rightarrow}{B} = - \frac{1}{c}E\overset{\hat{}}{i}$

$\overset{\rightarrow}{E} = \sqrt{\frac{I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\;\ \overset{\rightarrow}{B} = \frac{1}{c}E$

Solution:
If E_0 is magnitude of electric field then
$\frac{1}{2}\varepsilon_{0}E^{2} \times C = I$
$E_{0} = \sqrt{\frac{2I}{C\varepsilon_{0}}}$
$B_{0} = \frac{E_{0}}{C}$
Direction of $\overset{\rightarrow}{E} \times \overset{\rightarrow}{B}$
will be along$+ \overset{\hat{}}{j}$.

Physics


Question:
A block of mass m is at rest on an incline having an angle of
inclination θ with the horizontal. The incline is now being rotated with
constant angular velocity ω about an axis passing through the
perpendicular side. If the surface of the incline is rough having
coefficient of friction as µ, then find the distance along the
incline so that block just remains stationary with respect to incline
and is on the verge of slipping.
 from the top of the incline as shown in the figure.]

Solution:
N + mrω2 sin θ = mg cos θ
mg sin θ + mrω2 cos θ = Fdown
µ (mg cos θ - mrω2 sin θ) = Fup
Fdown = Fup
r =[IMAGE] x cos θ = r
x =[IMAGE] =[IMAGE] =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
In new system of unit 1 unit mass is 5 kg 1 unit length is 20
meter and 1 unit time is 2 second then 1 unit force in new system is
how much newton ?

Solution:
1 unit of mass = 5kg
1 unit of length = 20 m
1 unit of time = 2 sec
1 unit of F = [IMAGE] = [IMAGE] = 25 N
TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
A reflecting surface is represented by the equation
[IMAGE] A ray travelling horizontally becomes vertical after reflection. The
co-ordinates of points where this ray incident is

Solution: The slope of curve at such points is

Physics


Question:
In the potentiometer circuit shown, AB is a $10\Omega$ uniform
potentiometer wire of length 50 cm, E_1 is 2V accumulator of negligible
internal resistance. R_1 and R_2 are $15\Omega$ and 5$\Omega$
respectively. When K_1 and K_2 are both open, the galvanometer shows
no deflection when AJ is 31.25 cm. When K_1 and K_2 both are closed
the balance length AJ =5 cm. The emf of cell E_2is ____

Solution:
$V_{AB} = i10 = \frac{20}{25}$
$E_{2} = \frac{V_{AB}}{50} \times 31.25$
$= \frac{20}{25} \times \frac{31.25}{50}$
$= 0.5V$

Physics


Question:
One end of a straight uniform [IMAGE] long bar
is pivoted on horizontal table. It is released from rest when itmakes an
angle [IMAGE] from the horizontal (see figure). Its
angular speed when it hits the table is given as$\sqrt{3n}s^{- 1},$where
n is an integer. The value of n is.......

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
In the shown figure three slits, separated by
[IMAGE] and illuminated by monochromatic parallel beam of
light of wavelength[IMAGE]. P is a point on the line
perpendicular to the plane of slits through[IMAGE]. If
[IMAGE] is the intensity of the wave from each slit, the
intensity at p is

Solution:
Phase difference between
[IMAGE] &[IMAGE] Phase difference between [IMAGE] &[IMAGE] Now using phase diagram

Physics


Question:
An electron gun is placed inside a long solenoid of radius R on
its axis. The solenoid has n turns/length and carries a current I. The
electron gun shoots an electron along the radius of the solenoid with
speed
v. If the electron does not hit the surface of the solenoid, maximum
possible value of v is (all symbols have their standard meaning)

Solution:
$R_{\max} = \frac{R}{2} = \frac{mv_{\max}}{e\mu_{0}in}$
$V_{\max} = \frac{Re\mu_{0}in}{2m}$

Physics


Question:
A Carnot engine having an efficiency of$\frac{1}{10}$is being
used as a refrigerator. If the work done on therefrigerator is 10J, the
amount of heat absorbed from the reservoir at lower temperature is 10x.
Find x.,d dkuksZa bUtu $\frac{1}{10}$ bl s,d jsfÝtjsVj ds:i
 tk jgk
10J gks rks fuEurki okys rkidq.M tkus okyh

Solution:
For Carnot engine using as refrigerator
dkuksZV btau dks jsfÝtjsVj dh Hkkfr fy,
$W = Q_{2}\left( \frac{T_{1}}{T_{2}} - 1 \right)$
$\eta = 1 - \frac{T_{2}}{T_{1}}$ $\Rightarrow$
$\frac{T_{2}}{T_{1}} = \frac{9}{10}$
So, vr Q_2 = 90 J (as W = 10 J)

Physics


Question:
The loss of energy in KWH in a specimen of Iron is
$\frac{n}{25}$. The hysteresis loop of which is equivalent in area equal
to 250 Jm^-3 cycle^-1and frequency 50 min^-1 and specific gravity
of iron is 7.5 and mass of iron is 3.6kg. Find n.

Solution:
Hysteresis loss = 250 cycle
No. of cycle per hour = 50 ×60 = 3000
Hence total loss per hour = 250 × (volume) × (cycles)
$= 250 \times \frac{3.6}{7.5 \times 1000} \times 3000$
= 360 WH = 0.36 kWH

Physics


Question:
A plane electromagnetic wave is propagating along the
direction$\frac{\overset{\hat{}}{\mathbf{i}} + \overset{\hat{}}{\mathbf{j}}}{\sqrt{2}}$,
with its polarization along the direction$\widehat{k}$. The correct form
of the magnetic field of the wave would be (here B0 is an
appropriateconstant)

$B_{0}\frac{\overset{\hat{}}{i} + \overset{\hat{}}{j}}{\sqrt{2}}cos\left( \omega t - k\frac{\overset{\hat{}}{i} + \overset{\hat{}}{j}}{\sqrt{2}} \right)$

$B_{0}\frac{\overset{\hat{}}{i} - \overset{\hat{}}{j}}{\sqrt{2}}cos\left( \omega t + k\frac{\overset{\hat{}}{i} + \overset{\hat{}}{j}}{\sqrt{2}} \right)$

$B_{0}\frac{\overset{\hat{}}{i} - \overset{\hat{}}{j}}{\sqrt{2}}cos\left( \omega t - k\frac{\overset{\hat{}}{i} + \overset{\hat{}}{j}}{\sqrt{2}} \right)$

$B_{0}\overset{\hat{}}{k}cos\left( \omega t - k\frac{\overset{\hat{}}{i} + \overset{\hat{}}{j}}{\sqrt{2}} \right)$

Solution:
EM wave is in
direction$\ \rightarrow \frac{\overset{\hat{}}{i} + \overset{\hat{}}{j}}{\sqrt{2}}$.
Electric field is in direction$\ \rightarrow \overset{\hat{}}{k}$.
$\overset{\rightarrow}{E} \times \overset{\rightarrow}{B} \rightarrow$
Direction of propagation of EM wave
Therefore, B is along
$\frac{\overset{\hat{}}{i} - \overset{\hat{}}{j}}{\sqrt{2}}$.

Physics


Question:
A uniform thick rope of length 5m is kept on frictionless
surface and a force of 5N is applied to one of its end. Find tension in
the rope at 1 m from this end

Solution:
[IMAGE] Equation of motion
$$F - T = \times a\ldots.(i)$$
$$T = \times a\ldots(ii)$$
Solving (i) and (ii)
T = 4 N
TOPIC: Newton's second law
SUB TOPIC: Tension in a rope

Physics


Question:
Nuclei of radioactive element A are produced at rate
at any time t. The element A has decay constant. Let N be the number of
nuclei of element A at any time t. At time [IMAGE] is
minimum. Then the number of nuclei of element A at time

Solution: $\frac{dN}{dt} = t^{2} - \lambda N$

Physics


Question:
Two semicircular rings of linear mass densities
each are joined to form a complete ring. The distance of the center of
the mass of complete ring from its geometrical centre is

Solution:
Let the two half rings be placed in left and right of y-axis
with centre as shown in figure.
[IMAGE] Then the coordinate of the centre of mass of left and right half rings
X-coordinates of the centre of mass of complete ring is

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
The figure shows the P-V plot of an ideal gas taken through a
cycle DCBAD then
dks pØ DCBAD

maximum temperature will occur during the process C → B → A.

negative work is done by the gas in the cycle DCBAD.

heat flows out of the gas during C → B → A.

Solution:
Since temperature at C and A are same so maximum temperature
occurs at a state between them during process CBA.
Since cycle in PV graph is anticlockwise so work done by gas in
negative.
During CBA work done by gas is negative and [IMAGE] so
[IMAGE]. So heat is released by gas.
D;ksafdC rFkk A ij rkieku leku gS rks eku bu izØe CBA
ds nkSjku gksxkA D;ksfd PV vkjs[k okekorZ

Physics


Question:
A capacitor is made of two square plates each of side
making a very small angle $\alpha$ between them, as shown in figure. The
capacitance will be close to

$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

$\frac{\varepsilon_{0}a^{2}}{d}(1 + \frac{\alpha a}{2d})$

$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{4d})$

Solution:
$\Rightarrow c = \frac{\varepsilon_{0}a}{\alpha}\lbrack ln(d + \alpha x)\rbrack_{0}^{a}$
$= \frac{\varepsilon_{0}a}{\alpha}In(1 + \frac{\alpha a}{d}) \approx \frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

Physics


Question:
Starting from rest, a body slides down a $45^{\circ}$ 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

Solution:
$\mu = tan\theta\left( 1 - \frac{1}{n^{2}} \right)$
$\theta = 45^{\circ}$ and $n = 2$ (Given)
$\therefore\mu = tan45^{\circ}\left( 1 - \frac{1}{2^{2}} \right) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$

Physics


Question:
A system undergoes three quasi-static processes sequentially as
indicated in the given figure. $1 \rightarrow 2$is an isobaric process,
$2 \rightarrow 3$ is a polytrophic process with
$\gamma = \frac{4}{3}$and $3 \rightarrow 1$ is a process in which PV=
constant.
$P_{2} = P_{1} = 4 \times 10^{5}Nm^{- 2},P_{3} = 1 \times 10^{5}Nm^{- 2}$and$\ V_{1} = 1m^{3}$.
The heat transfer of the cycle is $\Delta Q,$ change in internal
energy is $\Delta U$ and the work done is $\Delta W$. for polytropic
process $PV^{\gamma} =$ constant.
(Take $\mathcal{l}$n2 = 0.693). Then

Solution:
For the process $3 \rightarrow 1$
$P_{1}V_{1} = P_{3}V_{3} \Rightarrow V_{3} = \frac{P_{1}V_{1}}{P_{3}} = \frac{4 \times 10^{5} \times 1}{1 \times 10^{5}} = 4m^{3}$
For the process $2 \rightarrow 3$
$P_{2}V_{2}^{\gamma} = P_{3}V_{3}^{\gamma} \Rightarrow V_{2} = \left( \frac{P_{3}}{P_{2}} \right)^{\frac{1}{\gamma}}V_{3} = 4\left( \frac{1}{4} \right)^{\frac{3}{4}} = 4^{\frac{1}{4}} = \sqrt{2}m^{3}$
For the process $1 \rightarrow 2$
$W_{12} = P_{1}\left( V_{2} - V_{1} \right) = \left( \sqrt{2} - 1 \right)4 \times 10^{5}J$
For the process $2 \rightarrow 3$
$W_{23} = \frac{\left( P_{2}V_{2} - P_{3}V_{3} \right)}{(\gamma - 1)} = 12(\sqrt{2} - 1) \times 10^{5}J$
For the process $3 \rightarrow 1$,
$w_{31} = - P_{1}V_{1}ln\frac{V_{3}}{V_{1}} - 4 \times 10^{5} \times 1 \times ln(4)$
For cyclic process, $\Delta\ U = 0$
$\therefore\Delta Q = \Delta U + \Delta W = \Delta W = 4 \times 10^{5}(\sqrt{2} - 1)$
$1 \rightarrow 2 \rightarrow 3 \rightarrow 1$
$\Delta Q = + 12 \times 10^{5}(\sqrt{2} - 1) - 4 \times 10^{5} \times 2 \times 0.693$
$= 1.08 \times 10^{5}J$

Physics


Question:
Rain is falling with constant and uniform velocity v at angle
37º vertical. An observer starts from rest and moves with uniform
acceleration $2m/s^{2}$ in horizontal direction (with the rain). After 6
sec. velocity of rain drop again appear at angle 37º from vertical.
Velocity of rain fall with respect to ground
is:$\left( tan37^{\circ} \right.\ = 3/4)$

Solution:
w.r.t observer
let final velocity of rain be v'
initial velocity of
rain$= \frac{3v}{5}\overset{\hat{}}{i} - \frac{4v}{5}\overset{\hat{}}{j}$
[IMAGE] w.r.t observer.
acceleration of rain$= - 2\overset{\hat{}}{j}m/s^{2}$
final velocity of
rain$= - \frac{3v^{'}}{5}\overset{\hat{}}{i} - \frac{4v^{'}}{5}\overset{\hat{}}{j}$
in y- direction
$V_{y\ final\ } = V_{y\ initial\ }$
$v^{'} = v$

Physics


Question:
An electric field
$\overset{\rightarrow}{E} = 4x\overset{\hat{}}{i} - (y^{2} + 1)jN/C$
passes through the box shown in figure. The flux of the electric field
through surface ABCDand BCGF and marked as $\phi_{1}$ and $\phi_{2}$
respectively. The difference between
$(\phi_{2} - \phi_{1})is\ (in{Nm}^{2}/C)$
[IMAGE] fudyrk gSA;fn cD ABCD rFkk BCGF leryksa

Solution:
Flux via ABCD ls ¶yDl
$\phi_{1} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
[IMAGE] Flux via BCEF ls ¶yDl
$= 16x,x = 3$
$\phi_{2} = 48\frac{N - m^{2}}{C};\phi_{2} - \phi_{1} = 48\frac{N - m^{2}}{C}$

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A bar of mass M & length L is in pure translatory motion with
its centre of mass velocity V. It collides with and sticks to a second
identical bar which is initially at rest. (Assume that it becomes one
composite bar of length 2 L). The angular velocity of the composite bar
will be

Solution:
Cons. of ang. momentum about P gives
$MV\frac{L}{2} = \frac{(2M)(2L)^{2}}{12}w$
$\Rightarrow \frac{V}{2} = \frac{2Lw}{3}$
$w = \frac{3V}{4L}$

Physics


Question:
In the arrangement shown in figure, wavelength of light used is
$\lambda$. The distance between slits S_1 and S_2 is $d(d < < D)$. The
distance between S_3 and S_4 is $u = \frac{\lambda D}{3d}$. If the
ratio of maximum to minimum intensity received on screen P is K. Then 10
K is

Solution:
If intensity of light incident on slits S_1 and S_2 is
$I_{0}$
The intensity through S_4 will be $I_{1} = 4I_{0}$
The intensity through S_3 will be
$I_{2} = 4I_{0}\cos^{2}(\frac{\phi}{2}) = I_{0}$
$K = \frac{I_{\max}}{I_{\min}} = \frac{{(\sqrt{4I_{0}} + \sqrt{I_{0}})}^{2}}{{(\sqrt{4I_{0}} - \sqrt{I_{0}})}^{2}} = {(\frac{3}{1})}^{2} = \frac{9}{1}$

Physics


Question:
Three metallic plates are given and middle plate is given charge
Q, as shown in the figure. The outer plates can be earthed with switches
S1 and S2. The area of plates is same.The charge that will flow to earth
when only switch S1 is closed and right most plate is uncharged is

Solution:
At point D,[IMAGE] must be zero because potential
of left most plate is zero.
[IMAGE] So, vr x = 0
[IMAGE] (conductor pkyd)
x = y + Q
y = -Q

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A uniform semi circular ring of radius R = 0.4m and mass m is
free to rotate in vertical plane about fixed horizontal axis passing
through A as shown in the figure. The angular acceleration (in
rad/sec^2) of ring immediately after it is released from rest from the
position

Solution:
${mgR}\left( 1 - \frac{2}{\pi} \right) = I\alpha$
$I = {mR}^{2} - m\left( \frac{2R}{\pi} \right)^{2} + m\left( R - \frac{2R}{\pi} \right)^{2}$
$I = 2{mR}^{2}\left( 1 - \frac{2}{\pi} \right)$
${mgR}\left( 1 - \frac{2}{\pi} \right) = 2\ m\ R^{2}\left( 1 - \frac{2}{\pi} \right)\alpha$
$\Rightarrow \alpha = \frac{g}{2R} = \frac{10}{2 \times 0.4} = 12.50rad/\sec^{2}$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
The magnifying power of a telescope with tube length 60 cm is5.
The focal length in cm of its eye piece is 2f. Find f?

Physics


Question:
A particle of mass m is attached to an end of a uniform rod of
mass [IMAGE] and length [IMAGE] which is
suspended through its mid point by an inextensible string as shown.
Initially the rod is in horizontal position and at rest. The system is
released from this position. Just after the release

The angular acceleration of the system is [IMAGE]

Solution:
$\frac{mgl}{2} = \left( \frac{2ml^{2}}{12} + \frac{{ml}^{2}}{4} \right)\alpha$
$\alpha = \frac{6\text{g}}{5l}$
$\text{a}_{\text{cm}} = \frac{\alpha l}{6} = \frac{\text{g}}{5}$

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
In a single slit diffraction of different light of wavelength
$\lambda$ by a slit of with t_1, the size of the central maximum on a
screen at a distance t_2 is

Solution:
$t_{1}sin\theta = \lambda$
$\theta = \lambda/t_{1}$
Width of central Max. $y = 2t_{2}\theta + t_{1}$
$y = \frac{2t_{2}\lambda}{t_{1}} + t_{1}$

Physics


Question:
A 60 pF capacitor is fully charged by a 20V supply. It is then
disconnected from the supply and isconnected to another uncharged 60 pF
capacitor in parallel. The electrostatic energy that is lost in
thisprocess by the time the charge is redistributed between them is (in
nJ)________.
 A rRi'pkr~ bls lzksr ls gVkdj 60 pF ds,d nwljs =k (parallel connection) esa tksM+k tc
 esa forfjr gks tk;s rks bl
izfØ fLFkj oS|qr

Solution:
[IMAGE] V_0 = 20 V
Heat loss gkfu = U_i - U_f
$= \frac{1}{2}CV_{0}^{2} - 2\left\lbrack \frac{1}{2}C\left( \frac{V_{0}}{2} \right)^{2} \right\rbrack$
$= \frac{CV_{0}^{2}}{4}$
$= \frac{\left( 60 \times 10^{- 12} \right)(20)^{2}}{4}J$
= 6 × 10^-9 J = 6 nJ

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
The charge appearing on outer surface of external plates when
S_1 and S_2 are open and external plates are uncharged.

Solution:
${\overrightarrow{E}}_{A} = 0$ (Conductor)
-x = x + Q + y - y
2x = - Q
$x = \frac{- Q}{2}$
$- x = \frac{Q}{2}$

Physics


Question:
Consider a circular coil of wire carrying constant current I,
forming a magnetic dipole. The magnetic flux through an infinite plane
that contains the circular coil and excluding the circular coil area is
given by $\phi_{i}$.The magnetic flux through the area of the circular
coil area is given by $\phi_{0}$. Which of the following optionis
correct?

Solution:
As magnetic field lines always form a closed loop, hence every
magnetic field line creating magnetic flux in the inner region must be
passing through the outer region. Since flux in two regions are in
opposite direction,
$\therefore$ $\phi_{i} = - \phi_{0}$

Physics


Question:
In the shown figure a ray of light enters the face XY of a right
angered prism at grazing incident. It emerges from the adjacent face. YZ
at an angle$\ \theta$. If x is the critical angle of the prism held in
water of refractive index $\frac{4}{3}$, then the angle$\ \theta$ with
normal is

$\theta = \sin^{- 1}\lbrack tanx\rbrack$

$\theta = \sin^{- 1}\lbrack\frac{4}{3}tanx\rbrack$

$\theta = \sin^{- 1}\lbrack\frac{3}{4}cotx\rbrack$

$\theta = \sin^{- 1}\lbrack cotx\rbrack$

Solution:
[IMAGE] As at XY
$\frac{4}{3}sin90 = \mu sinx$
At YZ
$\mu cosx = \frac{4}{3}sin\theta$
On solving
$\theta = \sin^{- 1}(cotx)$

Physics


Question:
A mixture of two moles of hydrogen and one mole of argon gas is
taken in a closed container at room temperature. Consider the following
two statements
(i) The average kinetic energy of each molecule of H2 and Ar are the
same.
(ii) The partial pressure due to argon gas is more than that due to
hydrogen gas.

Statement (i) is correct while statement (ii) is incorrect

Both statement (i) and (ii) are incorrect

Statement (i) is incorrect while statement (ii) is correct

Solution:
Kav = [IMAGE] kT
Kav = [IMAGE] kT
p1 = [IMAGE] × p
n1 > n2 ⇒ p1 > p2

Physics


Question:
A charge q is projected from origin with a velocity along
positive x-axis in a region having uniform magnetic field directed
towards negative y-axis. If T is the time period of circular motion then
the velocity vector of charge q at some instant t where
[IMAGE] may be

Solution:
[IMAGE] The charge executes circular motion in x-z plane with its center lying
on the negative z-axis. At time t where[IMAGE] the charge
has completed [IMAGE] th of the circle and so its velocity
vector is pointing towards first quadrant in x-z plane.
⇒ V_x is positive and V_z is positive and V_y is zero.

Physics


Question:
Both the diodes used in the circuit shown are assumed to be
ideal and have negligible resistance when these are forward biased.
Built in potential in each diode is[IMAGE]. For the
input voltages shown in the figure, the voltage (in volts) at point A is
6x. Find x.

Solution: Let [IMAGE] Right diode is reversed biased and left diode is forward biased

Physics


Question:
In the arrangement shown in the figure there is friction only
between the light string and the sharp spike at the point P. When the
blocks are in motion, the tensions in the vertical and horizontal
sections of the string are 4 N & 3 N respectively. The co-efficient of
friction between the string and the spike is equal to

Solution:
[IMAGE] N = 5
on string
$4 - 3 - \mu N = 0$
$\mu N = 1$
$\mu = 0.2$

Physics


Question:
A ball is given velocity [IMAGE] as shown. If the
ratio of centripetal acceleration totangential acceleration at the point
where the ball leaves the circular path is [IMAGE] then n
is

Solution:
and [IMAGE].....(2)
[IMAGE] Now TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate

Physics


Question:
A block is moved very slowly from A to D with the help of a
variable force on block which is always tangential to the surface of the
fixed curve path. Coefficient of friction between block and surface is
µ. (The block always remain in contact with plane). Work done by
friction in this process is

Solution:
Work done by
friction$= - \int\mu mgcos\theta d\mathcal{l = - \int}\mu mgdx = - 3\mu mg\mathcal{l}$

Physics


Question:
A conducting circular loop of copper is placed as shown in
figure. Cross section area of part abc is A and part adc is A/3 then
magnetic field at point is

Solution:
$R = \frac{\rho l}{A} \Rightarrow \frac{R_{abc}}{R_{adc}} = \left( \frac{\mathcal{l}}{A} \right)_{abc} \times \left( \frac{A}{\mathcal{l}} \right)_{adc}$
$= \frac{3/4\mathcal{l}}{A} \times \frac{A/3}{\mathcal{l/}4} = \frac{1}{1}$
$\frac{R_{abc}}{R_{adc}} = \frac{1}{1} \Rightarrow I_{abc} = I_{adc} = \frac{I}{2}$
$B_{0} = \frac{\mu_{0}I_{abc}\alpha_{1}}{4\pi R} - \frac{\mu_{0}I_{adc}\alpha_{2}}{4\pi R}$
$= \frac{\mu_{0}\left( \frac{I}{2} \right)}{4\pi R}\left( \alpha_{1} - \alpha_{2} \right) = \frac{\mu_{0}\left( \frac{I}{2} \right)}{4\pi R}\left( \frac{3\pi}{2} - \frac{\pi}{2} \right)$
$= \frac{\mu_{0}I}{8R} \otimes$

Physics

Magnetic field due to current carrying conductor


Question:
A vessel of depth 2h is half filled with a liquid of refractive
index[IMAGE] and the upper half with another liquid of
refractive index[IMAGE]. The liquids are immiscible. The
apparent depth of the inner surface of the bottom of vessel will be

Physics


Question:
One mole of an ideal gas is kept enclosed under a light piston
(area=[IMAGE] ) connected by acompressed spring (spring
constant 100 N / m). The volume of gas is [IMAGE] and
its temperature is 100K. The gas is heated so that it compresses the
spring further by 0.1 m. Then the work done (in Joule)by the gas in the
process is 1.5n. Find n: (Take R =8.3 J / K -mole and suppose there is
no atmosphere).,d 100 N / m½

Solution:
Before heating let the pressure of gas be
[IMAGE] from the equilibrium piston.
 -
[IMAGE] Since during heating process, pwafd xeZ izfØ nkSjku,
The spring is compressed further by 0.1 m fLizax 0.1 m

Physics


Question:
Power radiated by a black body is P_0 and the wavelength
corresponding to the maximum energyis around$\lambda_{0}$. On changing
the temperature of the black body, it is observed that the power
radiatedis increased to$\frac{256}{81}P_{0}$. The shift in the
wavelength
corresponding to the maximum energy will be

Solution:
$P = \sigma{AT}^{4}$
$\lambda_{m}T = b$
$T = \frac{b}{\lambda_{m}} \Rightarrow P = \frac{\sigma{Ab}^{4}}{\lambda_{m}^{4}}$
$\Rightarrow P \propto \frac{1}{\lambda_{m}^{4}} \Rightarrow \frac{P_{i}}{P_{2}} = {(\frac{\lambda_{m_{2}}}{\lambda_{m_{4}}})}^{4}$
$\frac{P_{0}}{\frac{256}{81}P_{0}} = {(\frac{\lambda_{m_{2}}}{\lambda_{0}})}^{4} \Rightarrow \lambda_{m_{2}} = \frac{3}{4}\lambda_{0}$
$\Delta\lambda = \frac{3}{4}\lambda_{0} - \lambda_{0} = - \frac{\lambda_{0}}{4}$

Physics


Question:
A solid sphere of mass m and radius r is gently placed on a
conveyer belt moving with constant velocityV. If the coefficient of
friction between the belt and sphere is$\frac{2}{7}$, the distance
travelled by the centre of the sphere before it starts pure rolling is.

Solution:
Initial situation w.r.t. the conveyer belt:
$\mu mgR = \frac{2}{5}{mR}^{2} \cdot \alpha$ $\Rightarrow$
$\alpha = \frac{5}{7} \cdot \frac{g}{R}$
At pure rolling:
$v^{'} = v - (\mu g)t$and $\omega^{'} = 0 + \left( \frac{5}{7} \cdot \frac{g}{R} \right)t$
(Since at pure rolling)
$\Rightarrow$ $v - \mu gt = \frac{5}{7} \cdot g \cdot t$
$\Rightarrow$ $t = \frac{v}{g}\left( \because\mu = \frac{2}{7} \right)$
For centre of the sphere:$S = v\left( \frac{v}{g} \right) - \frac{1}{2} \cdot \left( \frac{2}{7} \cdot g \right)\left( \frac{v^{2}}{g^{2}} \right)$
$S = \frac{v^{2}}{g} - \frac{v^{2}}{7g}$ (w.r.t. belt)
In this time, the belt with speed will move a distance
of$s_{b} = vt = v\left( \frac{v}{g} \right) = \frac{v^{2}}{g}$
[IMAGE] Distance travelled w.r.t. ground will be:$S_{g} = \left| S - S_{b} \right| = \left| \left( \frac{v^{2}}{g} - \frac{v^{2}}{7g} \right) - \left( \frac{v^{2}}{g} \right) \right| = \frac{v^{2}}{7g}$

Physics


Question:
Initially two particles A and B are present at (0,0) and (d, 0)
respectively. They start moving with speed
${\overset{\rightarrow}{v}}_{A} = v\overset{\hat{}}{i} + v\overset{\hat{}}{j}$and$V_{B} = - v\overset{\hat{}}{j}$.
If R is the separation between them and $T_{0}$ be the time when
separationbetween them is minimum, then

Solution:
$V_{BA} = - v\overset{\hat{}}{i} - 2v\overset{\hat{}}{j}$
$tan\theta = 2$
[IMAGE] at any time t
$x = d - vt$
$y = - 2vt$
$\therefore R^{2} = (d - vt)^{2} + 4v^{2}t^{2}$
$R = \sqrt{5v^{2}t^{2} - 2vdt + d^{2}}$
It is neither st. line nor circle
$R_{\min} = dsin\theta = \frac{2d}{\sqrt{5}}$
Distance $BC = dcos\theta$
$= \frac{d}{\sqrt{5}}$
time, $T_{0} = \frac{d}{\sqrt{5}v_{BA}}$
$T_{0} = \frac{d}{\sqrt{5} \times \sqrt{5v}};\ T_{0} = \frac{d}{5v}$

Physics


Question:
In the figure shown cross-section area of holes are same and
very small in comparison to area of cross-section of water tank:
(tanks are the open from the top)

Solution:
For case -1
$\rho gh = 1/2\rho v_{1}^{2}$
$V_{1} = \sqrt{2gh}$
$t_{1} = \sqrt{(10h/g)}$
For case -2:
$\rho g2h + 2\rho gh = \frac{1}{2}2\rho v_{2}^{2}$
$V_{2} = \sqrt{4gh}$
$t_{2} = \sqrt{\frac{6h}{g}}$
For case-3:
$\rho g2h + 2\rho gh + 3\rho gh = \frac{1}{2}3\rho v_{3}^{2}$
$V_{3} = \sqrt{6gh}$
$t_{3} = \sqrt{\frac{2h}{g}}$
$x_{1} = v_{1}t_{1} = \sqrt{20}$ $x_{2} = \sqrt{24}$ $x_{3} = \sqrt{12}$
$x_{1}:x_{2}:x_{3} = \sqrt{20}:\sqrt{24}:\sqrt{12}$
$x_{1}:x_{2}:x_{3} = \sqrt{10}:\sqrt{12}:\sqrt{6}$
$x_{1}:x_{2}:x_{3} = \sqrt{5}:\sqrt{6}:\sqrt{3}$

Physics


Question:
Radiation, with wavelength 656.1 nm falls on a metal surface to
produce photoelectrons. The electrons are made to enter a uniform
magnetic field of[IMAGE]. If the radius of the largest
circular path followed by the electrons is[IMAGE], the
work function of the metal is close to

Physics


Question:
In a single slit diffraction of different light of wavelength
[IMAGE] by a slit of with t, the size of the central
maximum on a screen at a distance[IMAGE] is

Solution: [IMAGE] Width of central Max. [IMAGE]

Physics


Question:
Two immiscible liquids are poured in a U-tube having densities
[IMAGE]. If the ratio of heights (of the liquids
above their interface)
[IMAGE] is x, then find the value of x.
U-ufy nks vfeJ.kh; nzo ftuds ?kuRoρ_1= 1.0 × 10^3
kg/m^3rFkkρ_2= 3.0 × 10^3 kg/m^3gSA;fn (nksuks nzoks ds vUr i`"B ls) nzoksa dh vuqikr $\frac{h_{1}}{h_{2}}$dk eku
x

Physics


Question:
A wire has a mass $0.3\ \pm \ 0.003$ g, radius
$0.5\ 0 \pm 0.005$cm and length $6 \pm 0.06$cm. What is the maximum
percentage error in the measurement of density?

Solution:
$\frac{d\rho}{A} = \frac{dm}{m} + 2\frac{dr}{r} + \frac{dl}{l}$
$\%\frac{d\rho}{A} = \left( \frac{0.003}{0.3} + 2*\frac{0.005}{0.5} + \frac{0.06}{6} \right)*100 = 4\%$

Physics


Question:
A non-viscous fluid of density [IMAGE] is
flowing in a tube as shown in figure. Area of section

Work done by gravitational force per unit volume from
section

Work done by elastic forces (pressure) per unit volume from
section

Solution:
Applying Bernoulli is equation from section-(1) and (2)
[IMAGE] and $P_{1} - P_{2} = \rho g(2h)$
Solving we get, [IMAGE] (3) Work done by gravitation force per unit volume = negative of the
increase in gravitational potential energy per unit volume =
$- (\rho gh)$
(4) Work done by elastic force per unit volume = decrease in
pressure = [IMAGE] = $2\rho gh$
TOPIC: Fluid dynamics
SUB TOPIC: Bernoulli's theorem

Physics


Question:
Two particles of equal mass m have respective initial
velocities[IMAGE] and [IMAGE]. They collide
completely inelastically. The energy lost in the process is

Solution:
Conserving momentum
[IMAGE] On solving

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
The charge that will flow to earth when only switch S1 is closed
and right most plate is uncharged is

Solution:
At point D,$\overrightarrow{E}$must be zero because potential
of left most plate is zero.
So, x = 0
${\overrightarrow{E}}_{A} = 0$ (Conductor)
x = y + Q
y = - Q

Physics


Question:
If finding the electric field using gauss law the
formula$|\overset{\rightarrow}{E}| = \frac{q_{enc}}{\varepsilon_{0}|A|}$is
applicable. In the formula $\varepsilon_{0}$ is permittivity of free
space, A is the area of Gaussian surface and q_enc is charge enclosed
by the Gaussiansurface. This equation can be used in which of the
following situation?

Only when the Gaussian surface is an equipotential surface.

Only when $|\overset{\rightarrow}{E}| =$ constant on the
surface.

Only when the Gaussian surface is an equipotential
surface and $\left| \overset{\rightarrow}{E} \right|$is constant of
the surface.

Solution:
Magnitude of electric field is constant & the surface is
equipotential

Physics


Question:
In YDSE arrangement, white light is used to illuminate the slits.
At point on the screen directly in front of slits, which of these
wavelengths is missing?
([IMAGE] = wavelength of light used)
(d = distance between slits)
(D = distance between slits and screen)
(Given d<<D)

Solution: Wavelength missing is condition for destructive superposition

Physics


Question:
Figure shows the variation of internal energy "U" with the
density "$\rho$ " of one mole of ideal diatomic gas. Process BA is a
part of rectangular hyperbola. If the work done by gas in the process BA
is 7n joules. Find n ?
 &ijek.kqd xSl ds,d fLFkfrt
 ?kuRo [IMAGE]

Solution: XY = constant constant (isobaric process) fu;rkad¼lenkch

Physics


Question:
A vessel of depth 2h is half filled with a liquid of refractive
index[IMAGE] and the upper half with another liquid of
refractive index[IMAGE]. The liquids are immiscible. The
apparent depth of the inner surface of thebottom of vessel will be

Solution: $d = \frac{h}{\sqrt{2}} + \frac{h}{2\sqrt{2}}$

Physics


Question:
A plane electromagnetic wave of frequency 25 GHz is propagating
in vacuum along the z-direction. At aparticular point in space and time,
the magnetic field is given
by$\overset{\rightarrow}{B} = 5 \times 10^{- 8}\overset{\hat{}}{j}T$.The
corresponding electric field E is (Speed of light
c = 3 × 10^8ms^-1)

Solution:
$\frac{E}{B} = c$
$\overset{\rightarrow}{E}$ = $\overset{\rightarrow}{B}$ ×
$\overset{\rightarrow}{c}$
= $15\overset{\hat{}}{i} \vee /m$

Physics


Question:
Diagrams show readings of a micrometer screw gauge. Figure (i)
shows the zero error reading when the screw gauge is closed, figure (ii)
the reading when the screw gauge is being used to measure the diameter
of a ball-bearing. What is the diameter of the ball-bearing ? There are
50 divisions on circular scale.

Solution:
Zero error is 3 division
LC = [IMAGE] = 0.01 mm
zero error = 0.03 mm
Reading = 5.0 + 29 × 0.01
d = 5.29 - 0.03 = 5.26 mm
SECTION - B [INTEGER ANSWER section contains

Physics


Question:
The acceleration-displacement graph of a particle moving in a
straight line is as shown in figure. Initial velocity of particle is
zero. What is the velocity (in m /s) of the particle when displacement
of the particle is s=16 m.

Solution:
$v\ dv = a\ ds$
$\therefore\int_{0}^{v}vdv = \int_{0}^{12m}ads$
$\therefore\frac{v^{2}}{2} = area\ under\ a - s\ graph\ froms = 0\ to\ s = 16m$
$= 2 + 12 + 6 + 12$
$= 32m/s$
or$v = \sqrt{64}m/s = 8m/s$

Physics


Question:
The activity of a radioactive sample falls from
[IMAGE] to [IMAGE] in 30 minutes. Its half
life is close to

Solution: $\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$

Physics


Question:
A physical quantity A is related to fourobservation a, b, c and
d as follows$A = \frac{a^{2}b^{3}}{c\sqrt{d}}$
The percentage errors of measurement in a, b, c and d are 1, 3, 2 and 2 respectively. Whatis the percentage error in the quantity A ?

Solution:
$\frac{dA}{A} = 2\frac{da}{a} + 3\frac{db}{b} + \frac{dc}{c} + \frac{1}{2}\frac{dd}{d}$
$\%\frac{dA}{A} = 2*1 + 3*3 + 1*2 + \frac{1}{2}*2 = 14\%.$

Physics


Question: The internal resistance of cell E2

Solution:
[IMAGE] V_AB = 2
$i^{'} = \frac{0.5}{r + 5}$
$0.5 - i^{'}r$
$= \frac{V_{AB}}{50} \times 5$
$0.5 - \frac{0.5}{r + 5}r = \frac{2}{10}$
$0.5 - 0.2 = \frac{0.5r}{r + 5}$
0.3 r +1.5=0.5 r
0.2r = 1.5
r = 7.5 $\Omega$

Physics


Question:
100 m
 XY.: (Y) )

Solution:
Range=[IMAGE] Rmax =[IMAGE] The speed becomes minimum at the highest point.
H =[IMAGE] = 100[IMAGE] =25 m.
Hence, ([IMAGE] ) are the coordinates.
TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
Three metal rods of the same material an identicalin all
respects are joined as shown in the figure. The temperatures at the ends
are maintained asindicated. Assuming no loss of heat from thecurved
surfaces of the rods, the temperature at thejunction X inºCis 10n. Find
n.

Solution:
Let T be the temperature at the junction.
Let L and A be the length and area of cross-sectionof each rod
respectively.
[IMAGE] $\therefore$ Heat current from Y to X is
$H_{1} = \frac{KA\left( 90^{\circ}C - T \right)}{L}$
$H_{2} = \frac{KA\left( 90^{\circ}C - T \right)}{L}$
Heat current from X to W is
$H_{3} = \frac{KA\left( T - 0^{\circ}C \right)}{L}$
 At the junction X
$H_{1} + H_{2} = H_{3}$
$\therefore 90^{\circ}C - T + 90^{\circ}C - T = T$
or $3T = 180^{\circ}C\ or\ T = 60^{\circ}C$

Physics


Question:
A point moves in a straight line under the retardation av^2,
where is a positive constant and v is speed. If the initial speed
is u, the distance covered in seconds is

Solution:
The retardation is given by
$\frac{dv}{dt} = - av^{2}$
integrating between proper limits
$\Rightarrow - \int_{u}^{v}\mspace{2mu}\frac{dv}{v^{2}} = \int_{0}^{t}\mspace{2mu} adt$or$\frac{1}{v} = at + \frac{1}{u}$
$\Rightarrow \frac{dt}{dx} = at + \frac{1}{u} \Rightarrow dx = \frac{udt}{1 + aut}$
integrating between proper
limits$\Rightarrow \int_{0}^{s}\mspace{2mu} dx = \int_{0}^{t}\mspace{2mu}\frac{udt}{1 + aut} \Rightarrow S = \frac{1}{a}\ ln\ (1 + \ aut\ )$

Physics


Question:
A uniform elastic rod of cross-section area A, natural length L
and Young's modulus Y is placed on asmooth horizontal surface. Now two
horizontal forces (of magnitude F and 3F) directed along the lengthof
rod and in opposite direction act at two of its ends as shown. After the
rod has acquired steady state,the extension of the rod will be.

Solution:
Tension in rod at a distance x from right edge is
$T = F\left( 3 - 2\frac{x}{L} \right)$
[IMAGE] $\therefore$net extension in
rod $= \int_{0}^{L}\mspace{2mu}\frac{T}{YA}dx = \frac{2F}{YA}L$

Physics

Elongation in rod due to unbalanced force


Question:
The aperture diameter of a telescope is 5 m. The separation
between the moon and the earth is [IMAGE]. With light of
wavelength of[IMAGE], the minimum separation between
objects on the surface ofmoon, so that they are just resolved, is close
to

Solution:
[IMAGE] Distance [IMAGE] Distance [IMAGE] Answer from options =60 m
(Minimum distance)

Physics


Question:
A circular loop is placed with its plane parallel to a long
current carrying wire as shown. Consider the following statements.
[IMAGE] (i) Circular loop is rotated about y-axis
(ii) Circular loop is rotated about x-axis
(iii) Circular loop is rotated about z-axis
(iv) Circular loop is rotated about the current carrying wire.
Then emf induced in the loop is zero in

Solution:
[IMAGE] Due to current carrying wire [IMAGE] is towards negative
z-axis. Also the area vector of the square loop is along positive
z-axis.
∴ If square loop is rotated about x-axis or y-axis the angle
⇒ Flux through the square loop changes
⇒ emf is induced in the loop in case (i) & (ii)

Physics


Question: In the figure, potential difference between A and B is

Solution:
Diode is in forward bias, so it will behave as simple wire so,
[IMAGE] So, $V_{ab} = \frac{30}{5 + 10} \times 5 = 10V$

Physics


Question:
The sinusoidal voltage of an a.c. source is given as
V = 100 sin (100πt) cos (100πt). Then

Peak value of voltage is 100 V and frequency is 100 Hz.

Peak value of voltage is 50 V and frequency is 50 Hz.

Peak value of voltage is 100 V and frequency is 200 Hz.

Peak value of voltage is 50 V and frequency is 100 Hz.

Solution:
The above relation can be written as
V = 50 (2 sin (100 πt) cos (100 πt))
or V = 50 sin (200πt)
∴ V_o = 50 V
f = 100 Hz

Physics


Question:
A vertical capillary tube with inner radius 0.5 mm, is submerged
into water so that the length of its part above the water surface is h =
25 mm. Radius of curvature of meniscuses formed is
$\frac{\alpha}{10}mm$. Find $\alpha$. (surface tension of water = 0.075
N/m. Density of water 1000kgm^-3, angle of contact = 0°)

Solution:
$h_{0} = \frac{2{Scos}\theta}{r_{0}\rho g} = \frac{2 \times 7.5 \times 10^{- 2}}{5 \times 10^{- 4} \times 10^{3} \times 10}$
$= \frac{15 \times 10^{- 2}}{5} = 3 \times 10^{- 2}m = 30mm$
$h_{0}r_{0} = hr$
$\Rightarrow r = \frac{h_{0}}{h}\ r_{0} = \left( \frac{30}{25} \right)(0.5) = 0.6mm$

Physics


Question:
A weight W is to be moved from the bottom to the top of an
inclined plane of inclination $\theta$ to the horizontal. If a smaller
force is to be applied to drag it along the plane in comparison to lift
it verticallyup, the coefficient of friction should be such that

Solution:
Force applied to drag$F_{1} = Wsin\theta + \mu Wcos\theta$
Force applied to lift $F_{2} = W$
As, $F_{1} < F_{2} \Rightarrow Wsin\theta + \mu Wcos\theta < W$
$\Rightarrow \mu < \frac{1 - sin\theta}{cos\theta} = \frac{\sin^{2}\frac{\theta}{2} + \cos^{2}\frac{\theta}{2} - 2sin\frac{\theta}{2}cos\frac{\theta}{2}}{\cos^{2}\frac{\theta}{2} - \sin^{2}\frac{\theta}{2}}$
$= \frac{{(cos\frac{\theta}{2} - sin\frac{\theta}{2})}^{2}}{(\cos^{2}\frac{\theta}{2} - \sin^{2}\frac{\theta}{2})} = \frac{cos\frac{\theta}{2} - sin\frac{\theta}{2}}{cos\frac{\theta}{2} + sin\frac{\theta}{2}} = \frac{1 - tan\frac{\theta}{2}}{1 + tan\frac{\theta}{2}}$
$\Rightarrow \mu < tan(\frac{\pi}{4} - \frac{\theta}{2})$

Physics


Question:
A convex lens is placed between a fixed object and a fixed
screen. The distance between the object and the screen is 150 cm. The
real images of the object are formed on the screen for two successive
positions of the lens separated by a distance of 50 cm. The focal length
of the lens is

Solution:
B
According to displacement method
[IMAGE] = 33.3 cm
∴ (B)

Physics


Question:
A uniform sphere of mass 20 kg and radius 10 cm is placed on a
rough horizontal surface. The coefficient of friction between the sphere
and the surface is 0.5. A horizontal force of magnitude 14 N is applied
on the sphere as shown. If the friction force (in N) acting on the
sphere is F the value of F is (g = 10 ms^-2)

Solution:
$f_{e} = \mu mg = 100N$
$F - f = ma$
$fR = I\alpha \Rightarrow f = \frac{2}{5}ma$
$\therefore f = \frac{2}{5} \times 20 \times 0.5 = 4N$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
A very long wire ABDMNDC is shown in figure carrying current I.
AB and BC parts are straight, long and at right angle. At D wire forms a
circular turn DMND of radius R. AB, BC parts are tangential to circular
turn at N and D. Magnetic field at the centre of circle is

Solution:
$\frac{\mu_{0}i}{4\pi R}\lbrack sin90^{\circ} - sin45^{\circ}\rbrack \otimes + \frac{\mu_{0}I}{2R} \odot + \frac{\mu_{0}i}{4\pi R}(sin45^{\circ} + sin90^{\circ}) \odot$
$= \frac{- \mu_{0}i}{4\pi R}\lbrack 1 - \frac{1}{\sqrt{2}}\rbrack + \frac{\mu_{0}i}{2R} + \frac{\mu_{0}i}{4\pi R}\lbrack\frac{1}{\sqrt{2}} + 1\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack - 1 + \frac{1}{\sqrt{2}} + 2\pi + \frac{1}{\sqrt{2}} + 1\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack\sqrt{2} + 2\pi\rbrack \odot$
$= \frac{\mu_{0}i}{2\pi R}\lbrack\frac{1}{\sqrt{2}} + \pi\rbrack \odot$

Physics

Magnetic field due to current carrying conductors


Question:
A uniform cylinder floats with its axis vertical with
[IMAGE] parts of its length in the upper liquid and
[IMAGE] part of its length in lower liquid. If the
densities of upper and lower liquids are [IMAGE] then the
density of the material of the cylinder is

Solution:
Let [IMAGE] be the be the length of the cylinder then
 is the density of the cylinder.
TOPIC: Hydrostatics
SUB TOPIC:Bouyancy
LEVEL: Moderate

Physics


Question:
A 20 gm particle is subjected to two simple harmonic motions
x_1 = 2 sin 10 t,
$x_{2} = 4sin\left( 10t + \frac{\pi}{3} \right)$where x_1& x_2 are in
metre& t is in sec.

The displacement of the particle at t = 0 will be $2\sqrt{3}m$

Maximum speed of the particle will be m/s. $20\sqrt{7}m$

Magnitude of maximum acceleration of the particle will be

Solution:
At t = 0
Displacement $x = x_{1} + x_{2} = 4sin\frac{\pi}{2} = 2\sqrt{3}m$
Resulting Amplitude
$A = \sqrt{2^{2} + 4^{2} + 2(2)(4)cos\pi/3} = \sqrt{4 + 16 + 8} = \sqrt{28} = 2\sqrt{7}m$
Maximum speed $= A\omega = 20\sqrt{7}m/s$
Maximum acceleration $= A\omega^{2} = 200\sqrt{7}m/s^{2}$

Physics


Question:
Kinetic energy of a particle of mass 2 kg moving along a circle
varies with time t as $K = 3t^{2}$, where t in second and K is in
Joules. Then rate at which work is being done on particle at t=1 s in
watt is

Solution:
$K = \frac{1}{2}{mv}^{2} = 3t^{2}$
$v = \sqrt{3}t$
$a_{t} = dv/dt = \sqrt{3}$
$= ma_{t}v$
$= 2.\sqrt{3} \cdot \sqrt{3} \times 1 = 6watt$

Physics


Question:
A system undergoes three quasi-static processes sequentially as
indicated in the given figure. $1 \rightarrow 2$is an isobaric process,
$2 \rightarrow 3$ is a polytrophic process with
$\gamma = \frac{4}{3}$and $3 \rightarrow 1$ is a process in which PV=
constant.
$P_{2} = P_{1} = 4 \times 10^{5}Nm^{- 2},P_{3} = 1 \times 10^{5}Nm^{- 2}$and$\ V_{1} = 1m^{3}$.
The heat transfer of the cycle is $\Delta Q,$ change in internal
energy is $\Delta U$ and the work done is $\Delta W$. for polytropic
process $PV^{\gamma} =$ constant.
(Take $\mathcal{l}$n2 = 0.693). Then,d s fp=kuqlkj rhu vnZ~/k&fLFkj çØeks ls xqtkjk A
 vkarfjd

Solution:
For the process izØe ds fy, 3 → 1
[IMAGE] For the processizØe ds fy, 2→ 3
[IMAGE] For the processizØe ds fy, 1 → 2
[IMAGE] For the processizØe ds fy, 2 → 3
[IMAGE] For the process 3 → 1
[IMAGE] For cyclic process pØh
$= 1.08 \times 10^{5}J$

Physics


Question:
A rod AB is free to rotate on a smooth horizontal plane about one
of its end A which is hinged. Initially the rod is at rest and D and E
are fixed points located in horizontal plane in such a way that DA is
perpendicular to rod and AE is along the length of the rod. A particle P
hits the rod perpendicularly at point C some distance from point A and
gets imbedded in it, then

Angular momentum of rod plus particle system about point C must
increase after collision

Angular momentum of rod plus particle
system about point D is conserved during collision.

Angular momentum of rod plus particle system about point D is not
conserved during collision.

Angular momentum of rod plus particle system about point E is
conserved

Solution:
The force applied by hinge on rod may be towards point D or
away from D or zero depending on distance of point C from A. Reaction at
hinge has torque about E therefore, angular momentum is not conserved.

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
Three charged particles A, B and C with charges -4q, 2q and
-2q are present on the circumference of acircle of radius d. The
charged particles A, C and centre O of the circle formed an equilateral
triangle asshown in figure. Electric field at O along x-direction is

Solution: $= \frac{q\sqrt{3}}{\pi\varepsilon_{0}d^{2}}$

Physics


Question:
In a certain series RLC circuit it is found that at resonance,
the inductive reactance is equal to the resistance. It is also found
that rms current in this circuit at resonance was
[IMAGE]. If now angular frequency of source is doubled
without changing any other parameter, what will be new rms current in
ampere in circuit?

Solution:
At resonance:
[IMAGE] Atangular frequency of source is double of resonance frequency:
[IMAGE] TOPIC:Alternating current
SUB TOPIC:RLC circuit

Physics


Question:
An ideal gas whose internal energy U is at absolute zero
temperature is equal to zero undergoes areversible adiabatic
compression. If U, P, V, T represents the internal energy, pressure,
volume andabsolute temperature respectively for ideal gas then,d ije ij mRØe.kh;
fy, U, P, V, T Øe'k vkUrfjd rFkk ije rki

Solution: [IMAGE] For adiabatic process [IMAGE] constant

Physics


Question:
Two forces of 6 N and 3 N are acting on the two blocks of 2 kg
and 1 kg kept on frictionless floor. The force exerted on 2 kg block by
1 kg block is 2 n then find the value of n.

Solution:
[IMAGE] Both blocks are constrained to move with same acceleration.
$6 - N = 2a\ \ \ \ \ \ \ $[Newtons II law for 2kg block]
$N - 3 = 1a$ [Newton's II law for 1 kg block]
This gives a= 1m/s^2
Therefore, N =4 N

Physics


Question:
A force F= 4t acts on a particle of mass 1kg. If the particle
starts from rest the work done by the force during the first 2 second
will be 8n J, find n.

Solution:
$\frac{dv}{dt} = 4t$ $\int_{0}^{v}{dv} = 4\int_{0}^{t}{t\ dt}$
$V = 2t^{2}$
$ds = 2t^{2}dt$
$W = \int_{}^{}{Fds\ cos0^{\circ}}$
$= \int_{0}^{2}\mspace{2mu} 4t \cdot 2t^{2}dt$
$= 8\left\lbrack \frac{t^{4}}{4} \right\rbrack_{0}^{2}$ $= 32J$

Physics


Question:
An object is moving along x-axis. It is definitely moving
towards origin if

Solution:
When x > 0, the particle will be moving towards origin if
[IMAGE] ⇒ [IMAGE] When x < 0, the particle will be moving towards origin if
[IMAGE] ⇒ [IMAGE] Hence in both cases [IMAGE]

Physics


Question:
Four point mass, each of mass m are connected at a corner of a
square of side by massless rods as shown in the figure. x and y
axis are in the plane of the system and z axis is perpendicular to the
plane and passing through the centre of the square.

Moment of inertia of the system about x axis is

Moment of inertia of the system about y axis

Moment of inertia of the system about the diagonal axis

Moment of inertia of the system about z axis is

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
The activity of a radioactive sample falls from
[IMAGE] to [IMAGE] in 30 minutes. Its
half life is close to

Solution: $\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$

Physics


Question:
Consider a mixture of n moles of helium gas and 2n moles of
oxygen gas (molecules taken to be rigid)as an ideal gas. Its CP/CV value
will be:
ds 2n eksYl dh feJ.k xSl ekus rks bl feJ.k

Solution:
$\gamma_{mix} = \frac{n_{1}C_{p_{1}} + n_{2}C_{p_{2}}}{n_{1}C_{v_{1}} + n_{2}c_{v_{2}}}$
$= \frac{n\left( \frac{5}{2}R \right) + 2n\left( \frac{7}{2}R \right)}{n\left( \frac{3}{2}R \right) + 2n\left( \frac{5}{2}R \right)}$
$= \frac{5 + 14}{3 + 10} = \frac{19}{13}$

Physics


Question:
The temperature needed to excite the hydrogen atoms to first
excited level is:
(Boltzman constant = 1.381 × 10^-23 J/K)

Solution:
According to kinetic interpretation of temperature, the average
kinetic energy per atom
[IMAGE] Energy required to excite the atoms to first excited state is
[IMAGE] 10.2 × 1.6 × 10^-19
T = 7.88 ´ 10^4 K

Physics


Question:
Four identical plates (equally spaced) and a battery are
connected as shown. If the capacitancebetween two consecutive plates is
C then choose the correct statement
pkj,d leku IysVs( leku nwjh ij),oa,d cSVjh ls fp=kuqlkj tqM+h gSA;fn nks Øekxr IysVksa C

Energy supplied by the battery isCV^2.
cSVjh }kjk lIykbZ CV^2

Energy linked in the space between plates 1 and 2 is$\frac{1}{6}$
CV^2

Potential difference between plates 2 and 4 is V/2
IysV 2 rFkk 4

The surface charge density on plate 3 on its right side is equal to
that on its left side.
IysV 3 ij i`" ?kuRo ml rFkk cka;h lrg ij leku gSA

Solution:
Equivalent Circuit rqY; ifjiFk
[IMAGE] Combination of Capacitors (1) and (2) are short circuited.
energy supplied by battery cSVjh }kjk lIykbZ
= QV
= CV.V
= CV^2
Potential difference between 1 and 3 = 0
1 rFkk 3 foHkokUrj= 0
Hence, potential difference between 2 and 4 = V
vr IysV 2 rFkk 4 foHkokUrj= V
Charge is only on right side of plate

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
Under the action of a force, a 2 kg body moves such that its
position x as a function of time is given by $x = t^{3}/3$,where xis in
meters and t is in seconds. The work done by the force in the first two
seconds starting form rest is

Solution:
$x = \frac{t^{3}}{3}$
$\frac{dx}{dt} = \frac{3t^{2}}{3}$
$V = t^{2}$
$V(2) = 4$
$V(0) = 0$
$\Rightarrow \Delta K = \frac{1}{2}m\left( 4^{2} \right) - \frac{1}{2}m\left( 0^{2} \right)$
$\Delta K = 16\ J$
From work energy theorem
Work done by force = $\Delta K = 16\ J$

Physics


Question:
If a particle of mass m start from A along curved circular path
shown in figure with constant tangential acceleration a, then net force
at B on the particle has the magnitude

Solution:
Fnet = matotal = [IMAGE] at = a
ac = [IMAGE] v2 = 0 + 2a[IMAGE] ⇒ ac =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: KIMATIC

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
In a uniform gravitational field g, which is acting vertically
downward, a ball is thrown at an angle with horizontal such that the
initial (immediately after projection) radius of curvature is 8 times
the minimum radius of curvature. If angle of projection is θ°, then find
the value of θ.
 g,
 ( )
 8 θ°, θ

Solution:
TOPIC: CIRCULAR MOTION
SUB TOPIC: RADIUS OF CURVATURE

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
A sphere of mass is given some angular velocity about a
horizontal axis through its centre and gently placed on a plank of mass
 The co-efficient of friction between sphere and plank
is[IMAGE]. The plank rests on a smooth horizontal
surface. The initial acceleration of the centre of sphere relative to
the plank will be

Solution:
FBD for sphere & block
$a_{2} = \frac{f_{r}}{m} = \frac{\mu mg}{m}$
${\overrightarrow{a}}_{1} = \mu g\overset{\hat{}}{i}$
${\overrightarrow{a}}_{2} = - \mu g\overset{\hat{}}{i}$
${\overset{\rightarrow}{a}}_{rel} = {\overset{\rightarrow}{a}}_{1} - {\overset{\rightarrow}{a}}_{2} = 2\mu g\overset{\hat{}}{i}$
$a_{rel} = 2\mu g$.

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
A container filled with viscous liquid is moving vertically
downwards with constant speed
[IMAGE]. At the instant shown, a sphere of radius
r is moving vertically downwards (in liquid) has speed
[IMAGE]. The coefficient of viscosity is η. There is no relative motion between
the liquid and the container. Then at the shown instant, the magnitude
of viscous force acting on sphere is[IMAGE]. Find x.,d ik=k tks fd,d pky 3v_0ls
f= xksyk uhps dh rjQ (nzo esa) tk jgk gS mlds xksys dh
ugh gSA rks bl.k ij xksys ij cy

Solution:
Relative to liquid, the velocity of sphere is
[IMAGE] viscous force on sphere
$= 12\pi\eta\text{r}\text{v}_{0}$downward
nzo xksys dk osx dh rjQ gSA
$\therefore$xksys ij nzo = 6 πη r2v_0uhps

Physics


Question:
Three particles each of mass m can slide on fixed frictionless
horizontal circular tracks in the same horizontal plane as shown in the
figure. The coefficient of restitution being e=0.5. Assuming that
m_2and m_3 are at rest initially and lie along a radial line before
impact and the spring is initially unstretched, then maximum extension
in spring in subsequent motion.

Solution:
Velocity just after collision $V_{0} = V_{1} + V_{2}$.....(1)
$\Rightarrow \frac{1}{2} = \frac{- V_{1} + V_{2}}{V_{0}}$
$- V_{1} + V_{2} = \frac{V_{0}}{2}$.....(2)
From (1) & (2)
$\Rightarrow V_{1} = \frac{V_{0}}{4}\& V_{2} = \frac{3V_{0}}{4}$
When maximum extension occurs then angular speed with rest to centre
for mass m_2& m_3 are same. Using angular momentum conservation
about centre of circle.
$m\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\omega + m(2R)^{2}\omega$
$\Rightarrow \omega = \frac{3}{10}\frac{V_{0}}{R}$
So velocity of$m_{2} = \frac{3}{5}V_{0}$
& velocity of$m_{3} = \frac{3}{10}V_{0}$
When extension is maximum using energy conservation.
$\frac{1}{2}m{(\frac{3}{4}V_{0})}^{2} = \frac{1}{2}m{(\frac{3}{5}V_{0})}^{2} + \frac{1}{2}m{(\frac{3}{10}V_{0})}^{2} + \frac{1}{2}kx_{max.}^{2} \Rightarrow x_{\max} = \frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$

Physics


Question:
32 g of [IMAGE] is contained in a cubical
container of side 1 m and maintained at a temperature of
[IMAGE]. The isothermal bulk modulus of elasticity of the
gas is [IMAGE]. Then the value of K is:[IMAGE] 32 s 1m Hkqtk ds ?kukdkj crZu esa 127°C A rks K

Solution: Bulk modulus of elasticity at constant temperature ij

Physics


Question:
The activity of a radioactive sample falls from
[IMAGE] to [IMAGE] in 30 minutes. Its half
life is close to

Solution:
[IMAGE] From (i)(ii)
$\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$

Physics


Question:
The time period of revolution of electron in its ground state
orbit in a hydrogen atom is[IMAGE]. The frequency of
revolution of the electron in its first excited state (in
[IMAGE] ) is

Physics


Question:
A convex lens forms inverted image of a real object on a fixed
screen. The size of image is 12 cm. When lens is displaced 20 cm along
principle axis it again forms a real image of size 3 cm on the screen.
Focal length (in cm) of the lens is 40/n. Find n. (Assume image
formation only by paraxial rays)

Solution:
since [IMAGE] For first position of the lens
[IMAGE] From displacement method
Using
$f = \frac{D^{2} - \Delta d^{2}}{4D} = \frac{60^{2} - 20^{2}}{4 \times 60} = \frac{40}{3}\text{cm}$

Physics


Question:
The dependence of potential with distance r from the centre of a
negatively charged non-conducting solid sphere is given by the following
curve

Solution: [IMAGE] Where, r → distance from circle

Physics

Electric potential due to non-conducting solid sphere


Question:
A small spherical droplet of density d is floating exactly half
immerse in a liquid of density [IMAGE] and surface
tension T. The radius of the droplet is (take note that the surface
tension applies an upward force on the droplet):
?kuRo d dh,d NksVh xksykdkj cwan ?kuRo [IMAGE] rFkk
i`"B ruko T ds æo esa Bhd vk
dk eku gS nsa fd i`"B ruko cwan ij

Physics


Question:
A street car moves rectilinearly from station A (here car stops)
to the next station B (here also car stops) with an acceleration varying
according to the law f = a - bx, where a and b are positive constants
and x is the distance from station A. The distance between the two
stations & the maximum velocity are
$x = \frac{na}{b};v_{\max} = \frac{na}{2\sqrt{b}}$. Find n

Solution:
f = a - bx means particle will do SHM.
At mean position; f = 0
[IMAGE] In the figure shown, 'C' is the mean position and A & B are extreme
positions

Physics


Question:
A uniform elastic rod of cross-section area A, natural length L
and Young's modulus Y is placed on asmooth horizontal surface. Now two
horizontal forces (of magnitude F and 3F) directed along the lengthof
rod and in opposite direction act at two of its ends as shown. After the
rod has acquired steady state,the extension of the rod will be.

Solution:
Tension in rod at a distance x from right edge is
$T = F\left( 3 - 2\frac{x}{L} \right)$
[IMAGE] $\therefore$net extension in
rod $= \int_{0}^{L}\mspace{2mu}\frac{T}{YA}dx = \frac{2F}{YA}L$

Physics

Elongation in rod due to unbalanced force


Question:
All surfaces are smooth. Find the acceleration (in$m/s^{2}$) of
sphere in the given figure. $(g = 10m/s^{2})$

Solution:
Suppose of acceleration of wedge a_1, towards right and
acceleration of sphere a_2 downward
[IMAGE] $mg - Ncos\theta = {ma}_{2}$.....(1)
$Nsin\theta = {ma}_{1}$.....(2)
[IMAGE] From wedge constraint
$a_{2}cos\theta = a_{1}sin\theta$
From (i),(ii) & (iii)
$a_{2} = g\sin^{2}\theta$
$= 5m/s^{2}$

Physics


Question:
In the given circuit diagram, a wire is joining points B and D.
The current in this wire is

Solution:
$i = \frac{20}{2} = 10A$
$I = \frac{4i}{5} - \frac{3i}{5} = + \frac{i}{5} = 2A$

Physics


Question:
A U tube filled with a liquid is accelerating horizontally with
an acceleration a. The acceleration (in m/s^2) of the tube is:
æo ls Hkjh gqbZ,d U uyha Roj.k Rofjr gSA ufydk

Solution: For free surface eqDr lrg

Physics


Question:
When a gas in a vessel expands, its internal energy decreases.
The process involved is

Solution:
Internal energy decreases $\Rightarrow$ temperature drops.
Expansion $\Rightarrow$ not isochoric
 Cooling $\Rightarrow$ not isothermal
In isobaric process, $V \propto T \Rightarrow \ $expansion raisestemp.

Physics


Question:
A metal sample carrying a current along x-axis with density Jx
is subjected to a magnetic field Bz (along z-axis). The electric field
Ey (Hall field) developed along y-axis is directly proportional to Jx as
well as Bz. The constant of proportionality (Hall coefficient) has SI
unit

Solution:
Jx = Jx [IMAGE] I = JxA
evdBz = eEy
Ey = vdBz
neAvd = JxA
Ey = [IMAGE] Constant of proportionality = [IMAGE] =

Physics


Question:
An electric dipole of moment
$\overset{\rightarrow}{p} = ( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k}) \times 10^{- 29}$
cm. is at the origin (0,0,0). The electric fielddue to this dipole at
$\overset{\rightarrow}{r} = ( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} + 5\overset{\hat{}}{k})$
(note that$\overset{\rightarrow}{r}.\overset{\rightarrow}{p} = 0$) is
parallel to,d fo|qr f}/kzqo ftldk vk?kw.kZ ¼moment) [IMAGE] C.m.
[IMAGE] ij cuus okys fo|qr

$( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

$( - \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

$( + \overset{\hat{}}{i} - 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

$( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k})$

Solution:
Since [IMAGE] must be antiparallel to [IMAGE], izfrlekUrj gksxk
So,[IMAGE] Where [IMAGE] is a arbitrary positive constant
Let
$\overset{\rightarrow}{A} = a\overset{\hat{}}{i} + b\overset{\hat{}}{j} + c\overset{\hat{}}{k}$
is parallel to $\overset{\rightarrow}{E}$
So, blfy, [IMAGE]

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
An unknown quantity is expressed as
α =[IMAGE], where m = mass,
a = acceleration, = length. The unit of α should be

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
A light ray parallel to x-axis is incident on a parabolic
concave mirror in XY plane. The focus of this parabola is at (1,0). The
unit vector along the reflected ray will be

Solution:
After reflection reflected ray passes through focus
So vector along reflected ray[IMAGE] TOPIC:Ray optics
SUB TOPIC:Vectors

Physics


Question:
A capacitor is made of two square plates each of side
making a very small angle $\alpha$ between them, as shown in figure. The
capacitance will be close to

$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

$\frac{\varepsilon_{0}a^{2}}{d}(1 + \frac{\alpha a}{2d})$

$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{4d})$

Solution:
$\Rightarrow c = \frac{\varepsilon_{0}a}{\alpha}\lbrack ln(d + \alpha x)\rbrack_{0}^{a}$
$= \frac{\varepsilon_{0}a}{\alpha}In(1 + \frac{\alpha a}{d}) \approx \frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

Physics


Question:
A uniform elastic rod of cross-section area A, natural length L
and Young's modulus Y is placed on a smooth horizontal surface. Now two
horizontal forces (of magnitude F and 3F) directed along the length of
rod and in opposite direction act at two of its ends as shown. After the
rod has acquired steady state, the extension of the rod will be,d,dleku izR;kLFk NM+ ftl V {ks=k AgS rFkk izkd`frd yEckbZ LrFkk
 xq.kkad YgS bldks,d fpdus {kSfrt lrg ij j
vc nks {kSfrt cy ¼ftuds ifj.kke Fo 3FgS½ dks NM+ ds yEckbZ
fp=kkuqlkj,d nwljs ds foijhr fljksa ij fp=kkuqlkj A tc NM+ LFkk;h voLFkk izkIr dj ysrh

Solution:
Tension in rod at a distance x from right edge is
nka;s fljs ls NM+ ij xnwjh ij ruko
$T = F\left( 3 - 2\frac{x}{L} \right)$
[IMAGE] net extension in
rod$= \int_{0}^{L}\mspace{2mu}\frac{T}{YA}dx = \frac{2F}{YA}L$
NM+

Physics

Elongation in rod due to unbalanced force


Question:
A particle is dropped from a certain height onto a fixed surface
at which it collides inelastically. Which of the following represents
true phase-space (momentum-position) graph? Taking upward direction as
positive and x=0 at the surface

Solution:
Initial velocity is zero and x=+H
Just before collision velocity hence momentum is negative and position
is zero. Just after collision, particle will change its direction. So,
option (C) is correct.

Physics


Question:
A point object moves on a circular path such that distance covered
by it is given by function [IMAGE] meter (t in second).
The ratio of the magnitude of tangential and radial acceleration at t =
2 sec. is 1: 2 then radius of the circle is ( in meter )

[IMAGE] (t ) t = 2 sec.
 1:2

Solution:
S = [IMAGE] + 2t
v = t + 2
[IMAGE] R = 8 m
TOPIC: CIRCULAR MOTION
SUB TOPIC: KINEMATIC

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
Figure shows a capillary tube C dipped in a liquid that wets it.
The liquid rises to a point A. If we blow airthrough the horizontal tube
H, what will happen to the liquid column in the capillary tube?
iznf'kZr fp=k esa C æo ds vUnj Mwch gqbZ,d dSiyjh V~ A æo
fcUnq A rd mBrk gSA;fn ge kjk gok izokfgr djsa
rks dSiyjh V~;wc

Level will remain at A Lrj A ij gh jgsxk

Solution:
Level rise because to compensate for the reduction in pressure
above A, due to high velocity (fromBernoulli's principle pressure
decreases if velocity increases.)
mPp osx j.k nkc A ds ?kVsxk
¼cjukWyh

Physics


Question:
[IMAGE] As shown in the figure, a battery of emf $\varepsilon$ is connected to
an inductor L and resistance R in series. The switch is closed at t=0.
The total charge that flows from the battery, between t=0 and
$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is

Solution:
$q = \int_{0}^{T_{C}}idt$
$= \frac{\varepsilon}{R}{\lbrack t - \frac{e^{- t/T_{C}}}{\frac{- 1}{T_{C}}}\rbrack}_{0}^{T_{C}}$;
$= \frac{\varepsilon}{R}\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\rbrack$
$= \frac{\varepsilon}{R} \times \frac{1}{e} \times \frac{L}{R}$
$= \frac{\varepsilon L}{R^{2}e}$

Physics


Question:
An ideal gas obeys the law $PV^{x} =$ constant. x is value for
which it has zero molar specific heat at normal temperature. Then find 5
x. (Cv for the gas is 2.5 R)

Solution:
For a gas at normal temperature,$\ c_{v} = \frac{5}{2}R$
Specific heat for a polytropicprocess
$C = C_{v} + \frac{R}{1 - x} = \frac{5}{2}R + \frac{R}{1 - x}$
According to question C = 0
$\frac{5(1 - x) + 2}{2(1 - x)} = 0\ or\ \frac{7 - 5x}{1 - x} = 0 \Rightarrow x = 1.4$

Physics


Question:
[IMAGE] m =
 a = = 

Solution:
α] =[IMAGE] TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
Mass of plank shown is 102.5 kg. If plank remains horizontal and
slipping does not take place at any contact, Radius of uniform
cylindrical rollers are 10 cm and 5 cm and masses 40 kg and 20 kg
respectively, then choose the correct options. Given that during the
first second starting from rest plank gets displaced by 10 cm.

Ratio of magnitude of acceleration of centre of mass of bigger and
smaller rollers is 1

Magnitude of friction between bigger roller and ground is more than
that between smaller roller and ground.

Magnitude of friction between plank and bigger roller is more than
that between plank and smaller roller.

Solution:
$S = ut + \frac{1}{2}at^{2}$
a = 0.2 m/s^2
$\alpha_{1} = \frac{a_{1}}{2R},\ $ $\alpha_{2} = \frac{a_{2}}{2r}$
$\alpha_{1}:\alpha_{2} = 1:2$
$a_{1} = a_{2} = 0.1$
[IMAGE] Solving f_1 = 3N
f_3 = 1N
f_2 = 1.5 N
f_4 = 0.5 N

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
An object of melting point [IMAGE] is at
temperature of [IMAGE] initially. When
[IMAGE] amount of heat is supplied to it then find its
temperature. If it is known that heat is
required to completely melt it and the ratio of specific heat to latent
heat is 1: 240

Solution:
Let specific heat =S
Latent heat = 240 s
$$2\theta = m(240s) + ms(120) = 360ms$$
[IMAGE] Only '120ms' of heat is required to increase
temp of object to its melting point. Hence final temp is
TOPIC:Thermometry
SUB TOPIC:Heat transfer

Physics


Question:
An electromagnetic wave has an electric field given by the
expression (in Cartesian co-ordinates):
[IMAGE] What is the direction of the magnetic field at time t = 0 and position x
= 0 ?

Physics


Question: In the circuit shown V_d - V_a is ( in V )

Solution:
[IMAGE] Along agbfcd
V_a + 1 - 2 - 4 + 8 - 1 = V_d
⇒ V_a + 2 = V_d ⇒ V_d - V_a = 2 V

Physics


Question:
For an amplitude modulated wave maximum and minimum amplitude
are 10 V and 2 V respectively.Modulation index is n/3. Find n:
 10 V rFkk 2 V

Solution:
A_max = A_m + A_c = 10
A_min = A_c - A_m = 2
A_c = 6 V
A_m = 4 V
$m = \frac{A_{m}}{A_{c}} = \frac{4}{6} = \frac{2}{3}$

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A bullet of mass m hits the hanging block of mass M at its
centre completely inelastically with a velocity v and gets embedded in
it as shown, then choose the correct option

The tension in the string just before and after the collision changes
by$\frac{m^{2}v^{2}}{(m + M)\mathcal{l}}$

The tension in the string just before and after the collision changes
by $m\lbrack g + \frac{mv^{2}}{(m + M)\mathcal{l}}\rbrack$

Tension in the string just after the collision
becomes$m\lbrack g + \frac{mv^{2}}{(m + M)\mathcal{l}}\rbrack$

Tension in the string will not change after the collision

Solution:
As $T_{1} = Mg$
After hitting
$T_{2} = (M + m)g + \frac{m^{2}v^{2}}{(M + m)\mathcal{l}}$
$\therefore T_{2} - T_{1} = m\lbrack g + \frac{mv^{2}}{(m + M)\mathcal{l}}\rbrack$

Physics


Question:
A particle is projected from point P with velocity
[IMAGE] ms-1 perpendicular to the surface of a hollow
right angled cone as shown in figure. It collides at Q normally. The
velocity with which it collides at Q will be ( in m/sec )
[IMAGE] ms-1
P Q Q

Solution:
[IMAGE] u cos 30° = V cos 60°
[IMAGE] TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
A body stretches a spring by x_0 at the equator of the earth of
average radius R. The height at which it will stretch the spring by the
same amount at North Pole is (assume [IMAGE] to be
angular velocity of the earth and g is value of gravity on the surface
of the earth)

Solution:
$\frac{GM}{R^{2}} - m\omega^{2}R = kx_{0}$.............(i)
None at height h $\frac{GmM}{(R + h)^{2}} = kx_{0}$............(ii)
Equating 1 & 2[IMAGE]

Physics

Acceleration due to gravity at poles and equator

P-10 Physics paper 1 30 October Jee Main.docx


Question:
A non-isotropic solid metal cube has coefficients of linear
expansion as:$5 \times \ 10^{–5}\ /{^\circ}C$ along the x-axis and
$5 \times 10^{–6}\ /{^\circ}C$ along the y and the z-axis. If the
coefficient of volume expansion of the solid is
$C \times 10^{–5}\ /{^\circ}C$ then the value of C is.........
izdkj $5 \times \ 10^{–5}\ /{^\circ}C$ x- rFkk
izlkj xq.kkad $C \times 10^{–5}\ /{^\circ}C$ gks] rks C

Solution:
$V = 2\alpha_{2} + \alpha_{1}$
$= 10 \times 10^{- 6} + 5 \times 10^{- 5}$
$= 60 \times 10^{- 6}/\ ^{\circ}C$

Physics


Question:
A litre of dry air at STP expands adiabatically to a volume of 3
litres. If [IMAGE] the work done by air is:
 air to be an ideal gas]
1 yhVj gok tks fd ekud rki o nkcSTP ij gS]:)ks"e
çfØ çlkfjr gksdj 3 yhVjvk;ru

Physics


Question:
Two moles of hydrogen are mixed with n moles of helium. The root
mean square speed of gas molecules in the mixture is $\sqrt{2}$ times
the speed of sound in the mixture. Then n is.

Solution:
$V_{rms} = \sqrt{\frac{3RT}{M_{mix}}}$
$V_{sound\ } = \sqrt{\frac{\gamma RT}{M_{mix}}}$
$V_{rms} = \sqrt{2}V_{sound\ }$
$\sqrt{\frac{3RT}{M_{mix}}} = \sqrt{2}\sqrt{\frac{\gamma RT}{M_{mix}}}$
$\gamma_{mix} = \frac{3}{2}$
$\gamma_{mix} = \frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$
$\frac{3}{2} = \frac{2 \times \frac{7R}{2} + n \times \frac{5R}{2}}{2 \times \frac{5R}{2} + n \times \frac{3R}{2}}$
$\Rightarrow \frac{3}{2} = \frac{14 + 5n}{10 + 3n}$
$\Rightarrow 30 + 9n = 28 + 10n$
$\Rightarrow n = 2$

Physics


Question:
During an experiment an ideal gas is found to obey condition
gas). The gas is initially at a temperature [IMAGE],
pressure [IMAGE] and density [IMAGE].
The gas expands such that density changes to [IMAGE] then which of following is/are correct?,d iz;ksx

The temperature of gas changes to [IMAGE].

The graph of above process on the P - T diagram is parabola.

The graph of above process on the P - T diagram is not rectangular
hyperbola.

Solution: PT = constant

Physics


Question:
Two blocks of masses $M_{1}$ and $M_{2}$are connected to each
other through a light spring as shown in figure. If we push mass $M_{1}$
with force F and cause acceleration $a_{1}$ in mass $M_{1}$, what will
be the magnitude of acceleration in $M_{2}$?

Solution:
$kx = M_{2}a_{2}$ [Newton's II law for $M_{2}$]
By adding both equations.
$F = M_{1}a_{1} + M_{2}a_{2} \Rightarrow a_{2} = \frac{F - M_{1}a_{1}}{M_{2}}$

Physics


Question:
A uniform magnetic field B in positive z direction exists in a
circular region of radius R = 5 m. A loop of radius R = 5m lying in x -
y plane encloses the magnetic field at t = 0 and then pulled at uniform
velocity[IMAGE]. The emf induced (in volts) is the loop
at t = 2 sec is 6V. Then magnitude of 4B is

Solution:
[IMAGE] TOPIC:Electromagnetic induction
SUB TOPIC:Faraday's law
LEVEL:Moderate

Physics


Question:
70 calories of heat is required to raise the temperature of 2
moles of an ideal gas at constant pressure from 30°C to 35°C (R= 2 cal
/mol-°C) The gas may be:
ls] bl gks tkrk
gs rks xSl gks ldrh

Solution: [IMAGE] So gas is

Physics


Question:
The speed of water flowing out of the orifice before the
cylinder kept inside the tank
csyu dks VSd iwoZ fNnz

Physics


Question:
A satellite is orbiting around the earth in a circular orbit
very close to surface of earth then comparing with other satellites of
same mass moving in circular orbits around the earth.

The total energy of earth plus given satellite system is minimum

The time period of revolution of given satellite is minimum

The orbital speed of given satellite is maximum

Solution:
$v_{0} = \sqrt{\frac{GM}{(R + h)}}$
$T = \frac{2\pi(R + h)}{v_{0}} = \frac{2\pi(R + h)^{3/2}}{\sqrt{GM}}$
$E = \frac{- GMm}{2(R + h)}$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
Two seconds after projection, a projectile is travelling in a
direction inclined at 30º to the horizontal after one more second, it is
travelling horizontally. Then the speed of projection(u) and angle of
projection ($\theta$) are given by

Solution:
$T = 6sec = \frac{2usin\theta}{g} \Rightarrow usin\theta = 30$
$\tan 30^{\circ} = \frac{{usin}\theta - gt}{{ucos}\theta} = \frac{30 - 20}{{ucos}\theta}$which
gives $ucos\theta = 10\sqrt{3}$
$$tan\theta =

Physics


Question:
The magnifying power of a telescope with tube length 60 cm is5.
What is the focal length of its eye piece?

Physics


Question:
In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K=2. The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
the capacitor have area each. The energy of the capacitor is

Solution:
$U = \frac{1}{2}C_{eq}V^{2}$
$C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}$
and $C_{2} = \frac{\varepsilon_{0}\ A}{d/2}$
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \frac{(2\frac{\varepsilon_{0}}{d/2}) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}\ A}{d/2}}$
$= \frac{4}{3}\frac{\varepsilon_{0}\ A}{d}$ or
$C_{eq} = \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}$
$U = \frac{1}{2}(\frac{4}{3}\frac{\varepsilon_{0}A}{d}) \vee^{2} = \frac{2}{3}(\frac{\varepsilon_{0}A}{d})v^{2}$

Physics


Question:
The position of a projectile launched from the origin at t = 0 is
given by [IMAGE] at
t = 2s. If the projectile was launched at an angle θ from the
horizontal, then θ is (take g=10 ms-2)

[IMAGE] t = 0 t = 2s
  θ  θ  (g = 10 ms-2 )

Solution:
[IMAGE] 40 = Sx = V cos θ × t
50 = Sy = V sin θt - [IMAGE] gt2
t = 2 given
θ = tan-1 [IMAGE] TOPIC: KINEMATIC
SUB TOPIC: 2 D

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
A massless thread passes through a ring (smooth). The ring is in a
circular path with both part of the thread in tension. Which of the
following values of θ and α are possible?

Solution:
Tension will be same in both parts.
[IMAGE] Tcosθ = mg + Tcosα
⇒ α > θ

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
A particle is projected upwards with velocity v = 30 m/s. There are
two points A and B at heights hA = 25 m and hB = 40m. If the particle
crosses point B twice in a time interval ∆t. Then, it will also cross A
twice in a time interval equal to

2.5 ∆t
 v = 30 m/s. hA = 25 hB = 40
 A B B ∆t
 A

Solution:
hmax = [IMAGE] = 45m
Time to go from C to B,
5 = [IMAGE] gt2
t = 1s, So ∆t = 2s
Time to go from C to A,
20 =[IMAGE] gt2
t = 2s
∆tA = 2 × 2 = 4s = 2∆t
TOPIC: KINEMATIC
SUB TOPIC: MOTION UNDER GRAVITY

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
[IMAGE] As shown in the figure, a battery of emf $\varepsilon$ is connected to
an inductor L and resistance R in series. The switch is closed at t=0.
The total charge that flows from the battery, between t=0 and
$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is

Solution:
$q = \int_{0}^{T_{C}}idt$
$= \frac{\varepsilon}{R}{\lbrack t - \frac{e^{- t/T_{C}}}{\frac{- 1}{T_{C}}}\rbrack}_{0}^{T_{C}}$;
$= \frac{\varepsilon}{R}\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\rbrack$
$= \frac{\varepsilon}{R} \times \frac{1}{e} \times \frac{L}{R}$
$= \frac{\varepsilon L}{R^{2}e}$

Physics


Question:
In LC circuit the inductance L= 40 mH and
capacitance$C = 100\mu F$. If a voltage V(t)=10 sin (314 t) is applied
to the circuit, the current in the circuit is given as

Solution:
$R = 0$
$z = X_{C} - X_{L}$
$= \frac{1}{\omega C} - \omega L$
$= \frac{1}{314 \times 100 \times 10^{- 6}} - 314 \times 40 \times 10^{- 3}$
$= 31.84 - 12.56$
$= 19.28\Omega$
$X_{C} > X_{L}$

Physics


Question:
In the given diagram aproton is given a speed v at Pin the
horizontal direction then the time after which it comesout of the
magnetic field B. m is the mass of proton.

It will never come out of the magnetic field

Solution:
Radius of charge particle
$r = \frac{mV}{eB}$
here $r < \frac{d}{2}$, proton will complete a semicircle and comes out
of the magnetic field.

Physics


Question:
When an electron of hydrogen like atom jumps from a higher energy
level to lower energy level, then

angular momentum of the electron about nucleus remains constant

de-Broglie wavelength, associated with motion of electron
increases

Solution: Angular momentum [IMAGE] Kinetic energy [IMAGE] De-Broglie wavelength [IMAGE] So as v increases [IMAGE] decreases.

Physics


Question:
In the figure shown a block A of mass m is kept on block B of
mass m. A is given a velocity towards right as shown. The coefficient of
friction between A and B is $\frac{3\mu}{2}$ and that between B and
ground is$\ \mu$.

B will accelerate towards right with acceleration$\frac{\mu g}{2}$.

B will accelerate towards left with acceleration$\frac{\mu g}{2}$

B will move towards right with acceleration$\ \mu g$.

Solution:
as$f_{k^{'}\max} > \frac{3}{2}\mu g$
So it will not move

Physics


Question:
A Carnot engine operates between two reservoirs of temperatures
900 K and 300 K. The engineperforms 1200 J of work per cycle. The heat
energy (in J) delivered by the engine to the lowtemperature reservoir,
in a cycle, is 100n. Find n,d dkuksZ bUtu dks 900 K vkSj 300 K ds nks HkaMkjks a ds chp pyk;k
fuEu rki okys HkaMkj esa izfr pØ 100n (J esa) NksM+rk gSFind
n

Physics


Question:
If the mass of block is 1 kg and a force of [IMAGE] N
is applied horizontally on the block as shown in the figure. The
frictional force acting on the block is

Solution:
Net force along the plane
= mg sin 30 - FH cos 30
[IMAGE] So friction force is zero.
TOPIC: NLM
SUB TOPIC: FRICTION

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
A container of fixed volume has a mixture of one mole of hydrogen
and one mole of helium in equilibrium at temperature T. Assuming the
gases are ideal, the correct statement(s) is (are)

The average energy per mole of the gas mixture is 2RT.

The ratio of speed of sound in the gas mixture to that in helium
gas is $\sqrt{6/5}$

The ratio of the rms speed of helium atoms to that of hydrogen
molecules is $1/\sqrt{2}$

Solution:
$\ $For hydrogen,$\ n_{1} = 1,C_{v,} = \frac{5}{2}R$
 For mixture of gases,
$C_{v} = \frac{n_{1}C_{V,} + n_{2}C_{V_{2}}}{n_{1} + n_{2}}$
$= \frac{1 \times \frac{5}{2}R + 1 \times \frac{3}{2}R}{1 + 1} = 2R$
$C_{P} = C_{v} + R = 3R,\gamma_{mix} = \frac{C_{P}}{C_{v}} = \frac{3}{2}$
Also, $\gamma_{mx} = 1 + \frac{2}{f} \Rightarrow \frac{3}{2} = 1 + \frac{2}{f}\therefore f = 4$
$\therefore$ Average energy per mole $= \frac{1}{2}fRT = 2RT$
$M_{\max} = \frac{n_{1}M_{1} + n_{2}M_{2}}{n_{1} + n_{2}} = \frac{1 \times 2 + 1 \times 4}{1 + 1} = 3gm{ol}^{- 1}$
 Speed of sound in a gas, $v = \sqrt{\frac{\gamma RT}{M}}$
For a given value of T,v$\propto \sqrt{\frac{\gamma}{M}}$
 Rms speed of a gas molecule at temperature T is given by
$T,v \propto \sqrt{\frac{\gamma}{M}}$
 For a given value of T,$\ v_{rms} \propto \frac{1}{\sqrt{M}}$
$\frac{\left( v_{rms\ } \right)He}{\left( v_{rms\ } \right)H_{2}} = \sqrt{\frac{M_{H_{2}}}{M_{He}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$
 So, options (a), (b) and (c) are correct

Physics


Question:
In a Young's double slit experiment 15 fringes are observed on a
small portion of the screen when light of wavelength 500 nm is used. Ten
fringes are observed on the same section of the screen when another
light source of wavelength $\lambda$ is used Then the value of
$\\lambda\$ is (in nm)______.

Solution:
$15 \times 500 \times \frac{D}{d} = 10 \times \lambda_{2} \times \frac{D}{d}$
$\lambda_{2} = 15 \times 50nm$
$\lambda_{2} = 750nm$

Physics


Question:
In the given A.C. circuit, if battery voltage
is[IMAGE], power delivered by battery will have a power
factor of[IMAGE]. Find n.

Solution:
[IMAGE] Phasor diagram
[IMAGE] TOPIC: Alternating current
SUB TOPIC: LCR circuit
LEVEL: Moderate

Physics


Question:
A particle of mass m is made to move with uniform speed v along
the perimeter of a regular polygon of n sides inscribed in a circle of
radius a. The magnitude of impulse applied at each corner of the polygon
is

Solution:
[IMAGE] Magnitude of the impulse at each corner is simply the change in momentum
of the particle.
Magnitude of final momentum = mV
Magnitude of initial momentum = mV
Angle between two momentums =[IMAGE] Hence,
$\left| {\overset{\rightarrow}{P}}_{final\ } - {\overset{\rightarrow}{P}}_{initial\ } \right| = |mpulse = \left| \sqrt{P^{2} + P^{2} - 2P^{2}\cos\left( \frac{2\pi}{n} \right)} \right|$
$\Rightarrow$ Impulse $\left. \ = mV \mid \sqrt{2\left( 1 - cos\left( \frac{2\pi}{n} \right) \right.\ } \right)| = mv|\sqrt{4\sin^{2}\left( \frac{\pi}{n} \right)} \mid = 2mvsin\frac{\pi}{n}$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
A uniform rod of length L and mass M is placed on a smooth
horizontal surface. The rod is hinged at the end A and is free to rotate
in horizontal plane about a vertical axis passing through A. As shown in
figure there is a nail N at a perpendicular distance $\frac{L}{4}$ from
end A of rod. At t = 0 and impulse P_0 is applied at a distance
$\frac{3L}{4}$ from end A. The impulse is in the horizontal plane and is
perpendicular to the rod. At $t = \frac{8\pi ML}{27P_{0}}$, the rod
returns to its initial position. Find the impulse of the force exerted
by the nail on the rod.

Solution:
$P_{0}\frac{3L}{4} = \frac{M\mathcal{l}^{2}}{3}\omega_{0}$
$t_{1} = \frac{\pi}{2\omega_{0}}$
$t_{2} = \frac{8\pi ML}{27P_{0}} - t_{1}$
$\omega = \frac{\pi}{2t_{2}}$
$P\frac{L}{4} = \frac{M\mathcal{l}^{2}}{3}(\omega_{0} + \omega)$
$P = 12P_{0}$

Physics


Question:
The velocity time relation of an electron starting from rest is
given by v=K t, where K=2 m/s^2. The distance traversed in 3 sec is 3n
then the find the value of n.

Solution:
$a = \frac{dv}{dt} = k = 2m/s^{2}$
$S = 0 + \frac{1}{2}(2)(3)^{2} = 9m$

Physics


Question:
A horizontal spring-block system of mass 2kg executes S.H.M.
When the block is passing through its equilibrium position, an object of
mass 1kg is put gently on it and the two move together. The new
amplitude of vibration is (A being its initial amplitude)

Solution: Conserving momentum: 2V = 3V′ [IMAGE]

Physics


Question:
The maximum peak to peak voltage of an AM wave is
[IMAGE] and the
minimum peak to peak voltage is[IMAGE]. The modulation factor is 10x. Find x.

Physics


Question:
A uniform rod of length L and mass M is placed on a smooth
horizontal surface. The rod is hinged at the end A and is free to rotate
in horizontal plane about a vertical axis passing through A. As shown
infigure there is a nail N at a perpendicular distance
[IMAGE] from end A of rod. At t = 0 and impulse
[IMAGE] isapplied at a distance [IMAGE] from end A. The impulse is in the horizontal plane and is perpendicular
to therod. At [IMAGE] the rod returns to its initial
position. If the impulse of the force exerted by the nail on
the rod is $\frac{36\text{P}_{0}}{\text{x}},$then value of x is

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
A disc of radius R rolls on a horizontal surface with linear
velocity V and angular velocity[IMAGE]. There is a point
P on circumference of disc at angle [IMAGE] with upward
vertical diameter measured anticlockwise (see figure), which has a
vertical velocity. Here [IMAGE] is equal to.

$\pi \pm \sin^{- 1}\left( \frac{V}{\omega R} \right)$

$\pi \pm \cos^{- 1}\left( \frac{V}{\omega R} \right)$

$2\pi \pm \cos^{- 1}\left( \frac{V}{\omega R} \right)$

$\frac{3\pi}{2} \pm \cos^{- 1}\left( \frac{V}{\omega R} \right)$

Solution:
[IMAGE] For vertical velocity of point P
$\omega{Rcos}\theta = - V$
$\Rightarrow$
$\omega{Rcos}(\pi - \theta) = V\ or\ \omega{Rcos}(\pi + \theta) = V$
$\therefore$
$\theta = \pi + \cos^{- 1}\left( \frac{V}{\omega R} \right)$
or $\theta = \pi - \cos^{- 1}\left( \frac{V}{\omega R} \right)$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
For a system of two
 as shown in the
figure (both are at origin and perpendicular toeach other along x and y
axes respectively) (P_1 and P_2denotes magnitudes
 and r is quite large in
comparison to dimensions of dipoles,
[IMAGE] is resultant electric field and QPR is a quarter
of circle whose centre is at O)
fp=kuqlkj nks
fy, ¼nksuksa ewy fcUnq ij gS rFkk,d nwljs ds yEcor~ gS rFkk Øe'k x
o y chp dh nwjh dh rqyuk esa r
pkSFkkbZ Hkkx

Work done in taking electron from P to R on QPR = 0
P ls R rd QPR ij ys tkus

Solution:
$E_{1} = \frac{2k\frac{\left( P_{1} + P_{2} \right)}{\sqrt{2}}}{r3}$
$E_{2} = \frac{k\frac{\left( P_{1} - P_{2} \right)}{\sqrt{2}}}{r3}$
$tan\alpha = \frac{E_{2}}{E_{1}}$
$= \frac{P_{1} - P_{2}}{2\left( P_{1} + P_{2} \right)}$

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A uniform solid sphere of radius R and mass M purely rolls down
an inclined plane. The coefficient of friction between the sphere and
the inclined plane is[IMAGE]. If the maximum value of
angle of inclination for this to be possible is
[IMAGE] then n is: (Take[IMAGE] )

Solution:
[IMAGE] put in (1)
[IMAGE] Now [IMAGE] TOPIC:Rotaional dynamics
SUB TOPIC: Rolling down the incline
LEVEL: Moderate

Physics


Question:
In a double-slit experiment, at a certain point on the screen the
path difference between the two interfering waves is
[IMAGE] of a wavelength. The ratio of the intensity of
light at that point to that at the centre of a bright fringe is

Physics


Question:
A container of fixed volume has a mixture of one mole of
hydrogen and one mole of helium in equilibrium at temperature T.
Assuming the gases are ideal, the correct statement(s) is (are)

The average energy per mole of the gas mixture is 2RT.

The ratio of speed of sound in the gas mixture to that in helium
gas is $\sqrt{6/5}$

The ratio of the rms speed of helium atoms to that of hydrogen
molecules is $1/\sqrt{2}$

$\ $All of the above,d ik= esa 1 eksy gkbMªkstu rFkk 1

Solution:
$\ $For hydrogen,$\ n_{1} = 1,C_{v,} = \frac{5}{2}R$
gkbMªkstu ds fy, [IMAGE] For mixture of gases,
$C_{v} = \frac{n_{1}C_{V,} + n_{2}C_{V_{2}}}{n_{1} + n_{2}}$
$= \frac{1 \times \frac{5}{2}R + 1 \times \frac{3}{2}R}{1 + 1} = 2R$
[IMAGE] T ds fn, x, eku ds fy,[IMAGE] T ds fn, x, eku ds fy,[IMAGE] Rms speed of a gas molecule at temperature Tis given by
T rkieku ij xSl dhRms pky
[IMAGE] T

Physics


Question:
A rotor is rotating with a constant angular velocity ω about its
own axis. At the centre of the rotor a spring is fixed of natural length
(l0) whose other end is connected to a block of mass m. Find the
minimum value of coefficient of friction between block and rotor wall
for which spring remains horizontal. The radius of rotor is R (R >
l0).

Solution:
N = mω2R - K(R - l0)... (1)
For the block to remains in the vertical equilibrium,
[IMAGE] f = mg
⇒ N = mg
⇒ µ(mω2R - K (R - l0) = mg
∴ µ =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
M grams of steam at 100°C is mixed with 200g of ice at its
melting point in a thermally insulated container. If it produces liquid
water at 40°C and heat of
fusion of ice is 80 the value of M is______.
100°C rkieku dh M xzke ok"i dks 200g xzke cQZ esa,d crZu
 A ok"i igys cQZ eku vius xyukad ds
cjkcj FkkA izfØ vUr esa 40°C

Physics


Question:
A projectile is thrown, from the origin along X-Y plane to have
maximum possible horizontal range of 100 m. Then the position vector of
the projectile when its speed is minimum is (take Y on the vertical
axis)

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
A person projected a small ball with speed U along the floor from
point A. If x = 3R, determine the required speed U so that the ball
return to A after moving the circular surface in the vertical plane from
B to C and becoming projectile at C.

Solution:
V2 = V2 - 4gR
3R = V [IMAGE] V2 = U2 - 2g2R
U2 = 9gR / 4 + 4gR
U2 = 25 gR / 4
U =[IMAGE] TOPIC: WPE
SUB TOPIC: VERTICAL CIRCULAR MOTION

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
Statement-1: In a dynamo, it is necessary that coil should
be moving and magnet should be stationary and not vice versa.
Statement-2: Magnetic force is perpendicular to the direction of
motion of charges.
Which of the following options is correct?

Both the statements are correct and is the correct reason of

Both the statements are correct and is not the reason of

Physics


Question:
In a meter bridge, the wire of length 1m has a non-uniform
cross-section such that, the variation [IMAGE] of its
resistance R with length.
Two equal resistances are connected as shown in the figure. The
galvanometer has zero deflection when the jockey is at point P. What is
the eight times the lengthAP ?

Solution:
[IMAGE] Acording to Question
[IMAGE] TOPIC:Current and electricity
SUB TOPIC: Potentiometer
LEVEL:Moderate

Physics


Question:
A uniform solid hemisphere of mass m and radius R is attached to
the roof with a chord of torsional constant C and performing torsional
SHM. If time period (in seconds) of SHM is x, then find the value of 5x.

Physics


Question:
Four metal rods all of identical dimensions and made of same
material are welded together at a single point. The configuration is
such that any two rods are oriented at 120°. The far ends of three of
the rods are maintained at 90°C temperature and that of the fourth rod
is maintained at 30°C temperature. The temperature of the junction point
is ( in degree Celsius)

Solution:
For equilibrium condition, heat flow in and out of junction
should be same. Let k represent some constant representing heat content,
which may depend on dimensions and material of the rods.
k(T - 30) = 3k(90 - T)
4T = 300
T = 75°C

Physics


Question:
An object is gradually moving away from the focal point of a
concave mirror along the axis of the mirror. The graphical
representation of the magnitude of linear magnification (m) versus
distance of the object from the mirror (x) is correctly given by (Graphs
are drawn schematically and are not to scale)

Solution: At focus, magnification is[IMAGE].

Physics


Question:
One mole of an ideal gas at pressure P_0, volume V_0 and
temperature T_0 is expanded isothermally to twice its volume and then
compressed at constant pressure to (V_0/2) and the gas is brought back
to original state by a process in which P ∝ V (Pressure is directly
proportional to volume). The correct representation of process is
[IMAGE] rFkk rkieku nqxqus:i ls çlkfjr dh tkrh gSrFkk fQj lenkoh çØe rd laihfMr dh tkrh gS rFkk xSl dks okil çkjfEHkd
voLFkk esa,sls çØe lsykrs gS ftl ¼ lekuqikrh gksrk gS½ gS rks mijksä çØe

Solution:
Process AB is isothermal expansion BC is isobaric compression
and in process CA
izfØ;k AB lerkih izlkj gS rFkk BC lenkch laihMu

Physics


Question:
A simple pendulum A and a homogeneous rod B hinged at its end
are released from horizontal position as shown in figure

The times period of swing of simple pendulum is greater than that
of rod if L = [IMAGE]

The time period of swing of simple pendulum is equal to that of
rod if L = [IMAGE]

The time period of swing of simple pendulum is greater than that
of rod if [IMAGE]

The time period of swing of simple pendulum is equal to that of
rod if L[IMAGE]

Solution:
from conservation of energy
for simple pendulum
[IMAGE] for homogeneous rod
[IMAGE] When [IMAGE] at all time
so time periods of swing are equal.

Physics


Question:
India's Mangalyan was sent to the Mars by launching it into a
transfer orbit EOM around the sun. It leaves the earth at E and meets
Mars at M. If the radius of Earth's orbit is a_e = 1.5 × 10^11 m,
that of laws give the
estimate of time for Mangalyan to reach Mars from Earth to be close to
(The time period of revolution of earth around sun is 1 year)

Solution:
$\left( \frac{T_{1}}{T_{2}} \right)^{2} = \left( \frac{a_{1}}{a_{2}} \right)^{3}$
${\left( \frac{1}{T} \right)}^{2} = \left\lbrack \frac{1.5 \times 10^{11}}{2 \times 10^{11}} \right\rbrack^{3}$
(semi major axis for the path from EOM$= \frac{a_{E} + a_{M}}{2}$)
$T = \left( \frac{4}{3} \right)^{3/2}$ years

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
The velocity v (in cm/sec) of a particle is given in terms of
time t (in sec) by the equation:$v = at + \frac{b}{t\ + \ c}$
The dimensions of a, b and c are:
 ------------------------------------
 a b c

Solution:
$\lbrack v\rbrack = \lbrack at\rbrack;\ \lbrack a\rbrack = \left\lbrack \frac{v}{t} \right\rbrack = \left\lbrack \frac{{LT}^{- 1}}{T} \right\rbrack = \left\lbrack {LT}^{- 2} \right\rbrack$
$\lbrack t\rbrack = \lbrack c\rbrack = \lbrack T\rbrack$
$\left\lbrack \frac{b}{t + c} \right\rbrack = \lbrack v\rbrack\ \Rightarrow \ \frac{\lbrack b\rbrack}{\lbrack t\rbrack} = \lbrack v\rbrack \Rightarrow \lbrack b\rbrack = \lbrack vt\rbrack = \left\lbrack {LT}^{- 1}T \right\rbrack = \lbrack L\rbrack$

Physics


Question:
An electric dipole of moment
$\overset{\rightarrow}{p} = ( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k}) \times 10^{- 29}$
cm. is at the origin (0,0,0). The electric fielddue to this dipole at
$\overset{\rightarrow}{r} = ( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} + 5\overset{\hat{}}{k})$
(note that$\overset{\rightarrow}{r}.\overset{\rightarrow}{p} = 0$) is
parallel to

$( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

$( - \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

$( + \overset{\hat{}}{i} - 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

$( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k})$

Solution:
$\overset{\rightarrow}{p} \cdot \overset{\rightarrow}{r} = 0$
$\overset{\rightarrow}{E}$must be antiparallel
to$\overset{\rightarrow}{p}$
So, $\overset{\rightarrow}{E} = - \lambda(\overset{\rightarrow}{p})$
Where $\lambda$ is a arbitrary positive constant
Let
$\overset{\rightarrow}{A} = a\overset{\hat{}}{i} + b\overset{\hat{}}{j} + c\overset{\hat{}}{k}$
is parallel to $\overset{\rightarrow}{E}$
$\frac{a}{\lambda} = \frac{b}{3\lambda} = \frac{c}{- 2\lambda} = k$
So
$\overset{\rightarrow}{A} = \lambda k(\overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$

Physics


Question:
A $1\ µF$ capacitor is connected in the circuit shown below. The
e.m.f. of the cell is 3 volts and internalresistance is 0.5 ohms. The
resistors R_1and R_2have values 4 ohms and 1 ohm respectively.
Thecharge (in$µC)$ on the capacitor in steady state must be

Solution:
In steady state no current flows through capacitor. The
potential difference across capacitor andresistor of resistance R_2is
same.
$\mathbf{\therefore}$ charge on capacitor
$= CV = C \times 3 = 1\mu F \times 3 = 3\mu C$.
LFkk;h voLFkk esa] la

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A solid cylinder and a solid sphere, both having the same mass
and radius, are released from an inclined plane having angle of
inclination $\theta$ one by one. They roll on the incline without
slipping. The statement that holds good in this motion is

the force of friction that acts on the two is the same

the force of friction is greater in case of a sphere than for a
cylinder.

the force of friction is greater in case of cylinder than for
sphere.

the force of friction will depend on the nature of the surface
of the body that is moving and that of the inclined surface, and is
independent of the shape and size of the moving body.

Solution:
Equation of motion
$mg\ sin\theta - f = ma$....(i)
$fR = I\alpha = I\frac{a}{R}$
$\Rightarrow f = I\frac{a}{R^{2}}$....(ii)
from (i) and (ii)
$mgsin\theta - f = m.\frac{fR^{2}}{I}$
$mgsin\theta = f(1 + m.\frac{R^{2}}{I}$)
$f = \frac{mgsin\theta}{1 + \frac{mR^{2}}{I}}$
Moment of inertia of solid cylinder is greater, so friction force f will
be greater.

Physics


Question:
A small sphere tied to a string of length 0.8 m is describing a
vertical circle so that the maximum and minimum tensions in the string
are in the ratio 3: 1. The fixed end of the string is at a height of
5.8 m above ground. Find the velocity of the sphere at the lowest
position. (Take g = 10 m/s^2)

Solution:
let u and v be the speeds of the sphere at the bottom and the
top positions and m be themass
[IMAGE] Radius of circle = length of string =r=0.8 m
[IMAGE] =tension at the bottom or the maximum tension
[IMAGE] =tension at the top or the minimum tension
Using conservation of energy
KE from bottom to the top =Gain in GPE
Using equations (1) and (2) we get[IMAGE] TOPIC:Circular motion
SUB TOPIC:Vertical circle

Physics


Question:
If N Joules of heat is to be transferred to nitrogen in the
isobaric heating process for that gas to perform the work A =2.0 J then
the value of N is
[IMAGE] xSl dks nh x;h dk eku N twy gksrk gS] rks
N

Solution: W = 2J

Physics


Question:
A uniform magnetic field parallel to the plane of paper, existed
in space initially directed from left to right. When a bar of soft iron
is placed in the field parallel to it, the lines of force passing
through it will be represented by figure

Solution:
When a bar of soft iron is placed in the uniform magnetic field
which is parallel to it because of large permeability of soft iron,
magnetic lines of force prefer to pass through it. Concentration of
lines in soft iron bar increases as shown in figure.

Physics


Question:
When 300 J of heat is added to 25 gm of sampleof a material its
temperature rises from 25ºC to45ºC. The thermal capacity of the sample
andspecific heat of the material are respectively given by

Solution:
$\Delta\phi = ms\Delta T$
$\therefore$ Thermal
capacity $= ms = \frac{\Delta Q}{\Delta T} = \frac{300J}{45 - 25} = 15J/ºC$
 and specific
heat, $s = \frac{\Delta Q}{m\Delta T} = \frac{300}{25 \times 10^{- 3} \times (45 - 25)} = \frac{600J}{kg} - C$

Physics


Question:
Figure shows a capillary tube C dipped in a liquid that wets it.
The liquid rises to a point A. If we blow air through the horizontal
tube H, what will happen to the liquid column in the capillary tube?

Solution:
Level rise because to compensate for the reduction in pressure
above A, due to high velocity (from Bernoulli's principle pressure
decreases if velocity increases.)

Physics


Question:
A long capillary tube of mass$'\pi^{'}\ gm$, radius 2 mm and
negligible thickness, is partially immersed in a liquid of surface
tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of
liquid. The force required to hold the tube vertically, will be-
$\left( g = 10m/s^{2} \right)$
 eksVkbZ dh,d yEch
 i`"B ruko okys æo:i ls Mqcksrs
gSaA s.k ysrs gq, rFkk æo ds mRIykou cy dks ux.;
ekurs cy

Solution:
The free body diagram of the capillary tube is as shown in the
figure. Net force F required to hold tube is
F = force due to surface tension at cross-section
$\left( S_{1} + S_{2} \right) +$weight of tube.
dsf'kdk uyh dk cy js
fy, dqy cy F gS
F = vuqizLFk dkV ij i`"B ruko j.k cy
$\left( S_{1} + S_{2} \right) +$V~ Hkkj
$= 4\pi \times 2 \times 10^{- 3} \times 0.1 + \pi \times 10^{- 3} \times 10 = 10.8\pi mN$

Physics


Question:
In the given circuit, resistance of potentiometer wire AB is
$10\Omega$ and its length is 1m. Rest of quantities is shown in figure.
The null point will be at a separation x m from point A, then find value
of 50 x.

Solution:
$k = (\frac{E}{R + R_{p}})\frac{R_{p}}{\mathcal{l}} = \frac{2 \times 10}{(10 + 15) \times 1} = 0.8 \vee /m = \frac{8}{10}V/m$
$\frac{8}{10} \times AJ = 0.3$
$AJ = x = \frac{3}{8}m$

Physics


Question:
Two uniform solid spheres of equal radii R, butmass M and 4M have
a centre to centre separation 6R, as shown in figure. A projectile of
mass m in projected from the surface of thesphere of mass M directly
towards the centre of the second sphere. The minimum speed ofthe
projectile so that it reaches the surface of the second sphere
is[IMAGE]. What is n?

Solution:
[IMAGE] The projectile need to reach a point on the line joining the centers
of sphere where net electric force on it is zero.
Let the point N has no net electric field due to spheres.
Therefore, $\frac{GM}{x^{2}} = \frac{G.4M}{{(6R - x)}^{2}}$ x = 2R
Apply conservation of mechanical energy at projectile initial position
and particle at N
E_i = E_f
SUB TOPIC: Conservation of mechanical energy

Physics