Question:
A Comet is observed as it passes our solar system. Its closest
approach to the sun is r = 6.67 × 10^10 m,at that time it has velocity
v = 2 × 10^6 m/s. If G = 6.67 × 10^-11 N m^-2 kg^-2 and mass of the
sun is 2 × 10^30kg. Then which of the following is true in respect with
this comet?

The comet will never return to the solar system

Solution:
Total energy of the comet = U+K
$= \frac{- GM_{s}m}{r} + \frac{1}{2}m\left( 2 \times 10^{6} \right)^{2}$ where m mass
of the Comet
$= m\left\lbrack \left( - 2 \times 10^{9} \right) + \left( 2 \times 10^{12} \right) \right\rbrack$
= 1998 m 10^9 = +ve
[IMAGE] Comets will never returns to the solar system
because total energy is +ve.

Physics

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


Question:
A small block of mass m is released from position A, inside the
frictionless circular groove of radius 2 m on a fixed inclined plane as
shown in figure. The contact force between the block and inclined plane
at point B is[IMAGE]. Find x

Solution:
[IMAGE] TOPIC: Newton's law of motion
SUB TOPIC: Circular motion
LEVEL: Moderate

Physics


Question:
In the below figure, a bob a mass m is tied by a massless string
whose other end portion is wound on a fly wheel (disc) of radius r and
mass m. When released from rest the bob starts falling vertically. When
it has covered a distance of h, the angular speed of the wheel will be

Solution: [IMAGE] (no slipping)

Physics

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


Question:
A hinged construction consists of three rhombus with the ratio
of sides (5: 3: 2). Vertex A3 moves in the horizontal direction with
velocity V. Velocity of A2 will be

Solution:
$u = \frac{dy}{dt} = - 16\sin\theta\frac{d\theta}{dt}$
$\Rightarrow u = \frac{4}{5}v = 0.8v$
TOPIC: Newton's first law
SUB TOPIC: Constraint motion

Physics


Question:
Three harmonic waves having equal frequency v and same intensity
$I_{0}$, have phase angles 0, $\frac{\pi}{4}$ and$- \frac{\pi}{4}$
respectively. When they are superimposed the intensity of the resultant
wave is close to

Solution:
$I_{res} = (\sqrt{2} + 1)^{2}I_{0}$
$= (3 + 2\sqrt{2})I_{0} = 5.8I_{0}$

Physics


Question:
If acceleration of block B w.r.t. A is $2m/s^{2}$just after the
system is released from rest, then mass(in kg) of block C will be 5m.
Find m. [pulley and string both are massless, take

Solution:
[IMAGE] Given Condition is possible if$a_{B} - a_{A} = 2m/s^{2}$
And $a_{A} = \frac{30}{5} = 6m/s^{2} \Rightarrow a_{a} = 8m/s^{2}$B & C
will move with same acc.$a_{B}$
$\Rightarrow mg - 50 = (m + 5)a_{B}$
$10m - 50 = 8m + 40$
$\Rightarrow m = 45kg$

Physics


Question:
Consider a sphere of radius R which carries a uniform charge
density $\rho$. If a sphere of radius $\frac{R}{2}$ is carved out of it,
as shown, the
ratio$\left| \frac{{\overset{\rightarrow}{E}}_{A}}{{\overset{\rightarrow}{E}}_{B}} \right|$
of magnitude of electric field of magnitude of electric field
${\overset{\rightarrow}{E}}_{A}$and ${\overset{\rightarrow}{E}}_{B}$,
respectively, atpoints A and B due to the remaining portion is:,d R f= xksys esa leku /uRo $\rho$ forfjr gSA;fn bl
xksys $\frac{R}{2}$ f= xksykdkVDj fp=kuqlkj fudky rks
cps gq, Hkkx j.k fcUnqvksa A rFkk B ijfo|qr {ks=k ¼Ø ½

Solution:
For a solid sphere Bksl xksys ds fy, e
$E = \frac{\rho r}{3\varepsilon_{0}}$
$E_{A} = \frac{- \rho R}{2\left( 3\varepsilon_{0} \right)}$
$\left| E_{A} \right| = \frac{\rho R}{6\varepsilon_{0}}$
Electric field at point ¼fcUnq B ijkfo|qr {ks=k dh rhozrk½ B = E_B =
E_1A + E_2A
E_1A= Electric Field Due to solid sphere of radius R at
point $B = \frac{\rho R}{3\varepsilon_{0}}$
E_1A= R f= Bkslxksys j.k fcUnqBijfo|qr {ks=k dh
rhozrk$= \frac{\rho R}{3\varepsilon_{0}}$
E_2A = Electric Field Due to solid sphere of radius R/2 (which having
charge density -$\rho$)
E_2A = R/2 f= Bksl xksys ¼ftl ?kuRo- $\rho$ gS½

Physics

Electric field by uniformly charged sphere

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


Question:
Three simple harmonic motions in the same direction, having the
same amplitude A and same time period, are superimposed. If each differs
in phase from the next by 45^0, then

the phase of the resultant motion relative to the first is 90^o

the energy associated with the resulting motion is
[IMAGE] times the energy associated with any single
motion

the resulting motion is not simple harmonic.

Physics


Question:
The figure gives experimentally measured B vs. H variation in a
ferromagnetic material. The retentivity, coercivity and saturation,
respectively, of the material are

Solution:
[IMAGE] x = retentivity
y = coercivity
z = saturation magnetization

Physics


Question:
The figure gives experimentally measured B vs. H variation in a
ferromagnetic material. The retentivity, coercivity and saturation,
respectively, of the material are

Physics


Question:
The second overtone of an open organ pipe A and that of a closed
pipe B have the same frequency at a given temperature. It follows that
the ratio of the

fundamental frequencies of A and B is 5: 6

frequencies of first overtone of A and B is 10: 9

Physics


Question:
Block M slide down on frictionless inclined plane as shown. Find
frictional force acting on block m while it is at rest with respect to
the block M. (m = 2kg, M = 5kg, g = 10 m/sec2)

Solution:
a = g sin 37°
f = m × a × cos 37°
TOPIC: NLM
SUB TOPIC: NEWTON'S SECOND LAW

Physics

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


Question:
A thin uniform rod of mass is hinged at
one end. This rod is maintained in horizontal position by colliding very
tiny balls each of mass m/10 completely elastically 10 times per sec
striking at the opposite end as shown in figure. Find the speed of the

Solution: [IMAGE] Taking torque about hinge

Physics

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


Question:
A concave mirror of radius of curvature 40 cm is filled with
water (n=4/3) upto a height of 12 cm. A point object O is kept on the
principal axis of the mirror at height 13.5 cm from the water surface.
The final image formed after refraction at water surface, reflection at
mirror and again refraction at water surface in succession is situated
at.

Solution:
Refraction -1: Apparent depth from water surface
$d^{'} = 13.5 \times \frac{4}{3} = 18cm$
Reflection from concave mirror.
$\frac{1}{v} + \frac{1}{( - 30)} = \frac{1}{- 20}$
$v = - 60cm$
Image is formed 48 cm above water surface
Refraction-2: $d^{''} = \frac{48}{4/3} = 36cm$ above water surface

Physics


Question:
The magnitude of magnetic field at O (centre of the circular
part) of the current carrying coil as shownis

Solution:
Magnetic field due to circular
segment$= \frac{3}{4} \cdot \frac{\mu_{0}i}{2a}$
 Due to one straight wire
segment $= \frac{\mu_{0}i}{4\pi b}\left( sin45^{\circ} + sin0^{\circ} \right) = \frac{\mu_{0}i}{4\sqrt{2}\pi b}$
Net field
$= \frac{3\mu_{0}i}{8a} + 2 \times \frac{\mu_{0}i}{4\sqrt{2}\pi b} = \frac{\mu_{0}i}{4\pi}\left( \frac{3\pi}{2a} + \frac{\sqrt{2}}{b} \right)$.

Physics

Magnetic field due to current carrying conductor


Question:
When photon of energy
[IMAGE] strikes the surface of a metal A, the ejected photoelectrons have
maximum kinetic energy Ta eV and de-Broglie
wavelength[IMAGE]. The maximum kinetic energy of photoelectrons liberated
from another metal B by photon of energy
[IMAGE] is [IMAGE] If the de-Broglie wavelength of these
photoelectrons[IMAGE], then the work function of metal B is

Solution: Relation between De-Broglie wavelength and K. E. is

Physics


Question:
A non−uniform string of mass 45 kg and length 1.5 m has a
variable linear mass density given by μ = kx, where x is the distance
from one end of the string and k is a constant. Tension in the string is
15 N which is uniform. If the time (in second) required for a pulse
generated at one end of the string to travel to the other end is x, then
find the value of 2x.

Physics


Question:
t = 0
-10 m

Solution:
a = - 3t + 5
[IMAGE] ∆x = [IMAGE] + [IMAGE] ∆x = 0
⇒[IMAGE] t = 0s & t = 5s
TOPIC:KINEMATIC
SUB TOPIC: ACCERATION

Physics

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


Question:
At time t = 0 magnetic field of 1000 Gauss is passing
perpendicularly through the area defined by theclosed loop shown in the
figure. If the magnetic field reduces linearly to 500 Gauss, in the next
5 s, theninduced EMF in $\mu V$in the loop is 8n. Find n

Solution:
$\varepsilon = \left| - \frac{d\varphi}{dt} \right| = \left| - \frac{AdB}{dt} \right|$
$=.(16 \times 4 - 4 \times 2)\frac{(1000 - 500)}{5} \times 10^{- 4} \times 10^{- 4}$
$= 56 \times \frac{500}{5} \times 10^{- 8} = 56 \times 10^{- 6}V$

Physics


Question:
A train is moving with a constant speed 15 m/s along a circular
track. The length of the train makes an angle 90º at the centre. A siren
in its engine is emitting a sound of frequency 1.5kHz. Find the apparent
frequency of sound as heard by a passenger at the rear end of the train.

Physics


Question:
The current I_1 (in A) flowing through 1$\Omega$ resistor in
the following circuit is

Solution: $i = \frac{I}{2} = 0.2A$

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:
In the figure (i) a disc of mass M(kg) and radius R(m) is
rotating smoothly about a fixed vertical axis AB with angular speed 26
rad/s. A rod CD of length [IMAGE] and mass M(kg) is
hinged at one end at point on the disc. The rod remains in vertical
position and rotates along with the disc about axis AB. At some moment
the rod CD gets a very small impulse at point due to air due to
which the rod falls on the disc along one radius and sticks to the disc
as shown in figure (ii). Now find the angular velocity of the disc in

Solution:
MI of the system when rod is vertical
[IMAGE] MI of system when rod is horizontal
[IMAGE] From conservation of angular momentum of system about axis AB is

Physics

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


Question:
Two children are playing a game in which they try to hit a small
box using a spring-loaded marble gun, which is fixed rigidly to a table
is a height h above the top of the box. The spring has a spring constant
k and the edge of the box is some unknown horizontal distance
$\mathcal{l}$ away from the table. The first child compresses the spring
a distance x and finds that the marble falls short of its target by a
horizontal distance y. The second child compresses the spring by an
extra amount $\Delta x$ so that marble lands in the box. The value of
$\Delta x$is $y\sqrt{\frac{mg}{Nhk}}$ (Where N is integer) then find the
value of N.

Solution:
time taken or time of flight is same for both projectiles.
i.e. $t = \sqrt{\frac{2h}{g}}$
using energy conservation:
$\Rightarrow \frac{1}{2}{kx}^{2} = \frac{1}{2}{mv}^{2}$.......(1)
$\Rightarrow \frac{1}{2}k(x + \Delta x)^{2} = \frac{1}{2}mv^{'2}$.......(2)
$\mathcal{l -}y = vt$.......(3)
$\mathcal{l =}v't$.......(4)
On solving we get
$\Delta x = y\sqrt{\frac{mg}{2hk}}$
N = 2

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 qiven by (Graphs
are drawn schematically and are not to scale)

Solution: At focus, magnification is $\infty$

Physics

Magnification and position of object graph


Question:
Find momentum of inertia about an axis passing through COM of
system and it is perpendicular to theplane of 3 rods.

Solution:
$\because I_{2} = I_{3}$
$= \frac{(2m)(2L)^{2}}{12} + 2\left\lbrack \frac{{ML}^{2}}{12} + {Md}^{2} \right\rbrack$
$= {mL}^{2}\left( \frac{2 \times 2 \times 2}{12} \right) + 2\left( \frac{1}{12} + \frac{3}{4} \right){mL}^{2}$
${mL}^{2}\left( \frac{2}{3} \right) + 2\left( \frac{1 + 9}{12} \right){mL}^{2}$
${mL}^{2}\left( \frac{2}{3} \right) + 2\left( \frac{1 + 9}{12} \right){mL}^{2}$
$= {mL}^{2}\lbrack\left( \frac{2}{3} \right) + \frac{20}{12}\rbrack$
$I_{net\ } = {mL}^{2}\left( \frac{7}{3} \right)$

Physics


Question:
A man and car both starts moving from rest simultaneously along
the same straight line in such a way that acceleration of A and B are
given by
A will catch B in time T. Then T is

Solution:
Let study the motion of A with respect to B
[IMAGE] a = t
[IMAGE] TOPIC:Rectilinear motion
SUB TOPIC:Non-Uniform acceleration

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

Physics


Question:
A particle performs uniform circular motion with an angular
momentum L. If thefrequency of particle's motion is doubled and its
kinetic energy is halved, the angularmomentum becomes

Solution:
$L = I\omega$
$\omega^{'} = 2\omega$
$\frac{1}{2}\left( \frac{1}{2}I\omega^{2} \right) = \frac{1}{2}I^{'}\omega^{'2}$
$\frac{I\omega^{2}}{2} = I^{'}4\omega^{2}$
$I^{'} = \left( \frac{I}{8} \right) \Rightarrow L^{'} = I^{'}\omega^{'} = \frac{I}{8}2\omega = \frac{I\omega}{4} = (L/4)$

Physics


Question:
A river is flowing with velocity 5 km/hr relative to the ground
as shown in the figure. A boat starts from A and reaches the other bank
by covering shortest possible distance. Velocity of boat in still water
is 3 km/hr. The distance (in m) boat covers is s. The value

Solution:
For drift to be minimum
[IMAGE] time taken by boat to cross the river = t
 So, distance covered[IMAGE] TOPIC:Motion in a plane
SUB TOPIC:Relative motion

Physics


Question:
A monoatomic ideal gas follows the process:
The molar specific heat for this process is: ( R = gas constant)
bl çØe ds fy, eksyj

Solution: [IMAGE] For polytropic process, cgqijek.koh; izØe

Physics


Question:
The net power P of all the forces acting on a particle versus time
curve is shown. Work done upon the particle from A to B

Solution:
Area under the graph || [IMAGE] Area under graph increases, hence work done upon the particle from A to
B increases.
TOPIC: WPE
SUB TOPIC: POWER

Physics

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


Question: If two soap bubbles of different radii are connected by a tube:;fn fofHkUu f= nks lkcqu ds cqycqyksa

air flows from the bigger bubble to the smaller bubble till the
sizes become

air flows from bigger bubble to the smaller bubble till the sizes
are intercha

air flows from the smaller bubble to the bigger bubble

Solution:
The excess pressure inside the soap bubble in inversely
proportional to radius of soap bubble i.e. $P \propto 1/r$, r being the
radius of bubble. It follows that pressure inside a smaller bubble is
greater than that inside a bigger bubble. Thus, if these two bubbles are
connected by a tube, air will flow from smaller bubble to bigger bubble
and the bigger bubble grows at the expense of the smaller one.
cqycqys cqycqys dh f= Øekuqikrh gksrk gSA
$P \propto 1/r$] tgk¡ rcqycqys dh f= SA vr NksVs cqycqys esa cM+s
cqycqys dh rqyuk gskxk;fn nksuksa cqycqyksa dks
ufy tksM+k rks gok NksVs cqycqys

Physics


Question:
A thin walled hollow cylinder is closed from both the ends. The
mass of the curved part is m and the mass of each circular part is also
m. The radius of the cylinder is R and its length is 2R. What will be
the moment of inertia of this hollow cylinder about the axis,
perpendicular to its length and passing through its centre of mass?

Solution:
First find I of the hollow tube
[IMAGE] Where [IMAGE] So [IMAGE] Now moment of inertias of each circular plate about that axis.

Physics

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


Question:
Figure below shows two paths that may be taken by a gas to go
from a state A to a state C.
[IMAGE] In process AB, 400J of heat is added to the system and in process BC,
100 J of heat is added to the system. The heat absorbed by the system
in the process AC (in J) is 230n. Find n.;gk¡ nks iFk ftu
A ls voLFkk B rd ys tk;k tk ldrk gSA
 s AB] çØe esa 400 J rFkk çØe BC esa 100 J nh tkrh gSA
rks] çØe AC (in J) 230 n gSA rks n
Kkr dhft,A

Solution:
For a complete cycle
iw.kZ pfØ; çØe ds fy,
Q_cycle = W_cycle
$+ 400 + 100 + Q_{C \rightarrow A} = \frac{1}{2}\left( 2 \times 10^{- 3} \right)\left( 4 \times 10^{4} \right)$
⇒ Q~C→ A~ = - 460 J
⇒ Q_A→C = + 460 J

Physics


Question:
An observer moves towards a stationary source of sound with speed
one - fifth of speed of sound, then

Physics


Question:
The energy required to ionize a hydrogen like ion in its ground
state is 9Rydbergs. What is the wavelength of the radiation emitted when
the electron in this ion jumps from the second excited state to the
ground state

Physics


Question:
An electron (of mass m) and a photon have the same energy E in
the range of a few[IMAGE]. The ratio of the de-Broglie
wavelength associated with the electron and the wavelength of the photon
is (c = speed of light in vacuum)

Physics


Question:
A mass supported by a massless string wound around a
uniform hollow cylinder of mass m and radius R. If the string does not
slip on the cylinder, with what acceleration will the mass fall on
release?

Solution:
[IMAGE] mg - T = ma....(1)
T.R = mR^2$\frac{a}{R}$....(2)
$\frac{g}{2} = a$

Physics


Question:
Boolean relation at the output stage-Y for the following circuit
is

Solution:
First part of figure shown is OR gate and
Second part of figure shown is NOT gate

Physics


Question:
The following four wires are made of the same material. Which of
these will have the largest extension when the same tension is applied?
fuEu pkj rkj leku inkFkZ ds cus
vf/kdre gksxk tc leku ruko vkjksfir A

Solution:
$Y = \frac{FL}{\pi r^{2}\mathcal{l}}$
$\therefore\mathcal{l =}\frac{FL}{\pi r^{2}Y} \Rightarrow \mathcal{l \propto}\frac{L}{r^{2}}$
$\therefore\frac{L}{r^{2}}$is greatest for option A. fodYi A

Physics


Question:
The aperture diameter of a telescope is 5 m. The separation
between the moon and the earth is $4 \times 10^{5}$km. With light of
wavelength of 5893 A, the minimum separation between objects on the
surface ofmoon, so that they are just resolved, is close to

Solution:
$\theta = 1.22\frac{\lambda}{a}$
[IMAGE] Distance $= O_{1}O_{2} = d\theta = 1.22\frac{\lambda}{a}d$
Distance
$= O_{1}O_{2} = \frac{1.22 \times 5893 \times 10^{- 10} \times 4 \times 10^{8}}{5} \approx 57.5m$
answer from options = 58 m
(minimum distance)

Physics


Question:
Two moles of an ideal gas with [IMAGE] are mixed
with 3 moles of another ideal gas with [IMAGE] Thevalue
of [IMAGE] for the mixture is (10+n)/12. Find n.
eksy=

Solution: [IMAGE] On rearranging we getiqu ij

Physics


Question:
Under an adiabatic process, the volume of an ideal gas gets
doubled. Consequently the mean collisiontime between the gas molecules
changes from. If
[IMAGE] for this gas then a good estimate
for[IMAGE] is given by [IMAGE]:,d
bl j.k mlds v.kqvksa esa gksus okyh VDdjksa

Physics


Question:
The displacement x of a body of mass 1 kg on horizontal smooth
surface as function of time t is given by x =[IMAGE].
The work done in the first one second is

Solution:
dW = F.dx
x =[IMAGE] v =[IMAGE] = t3 ⇒ dx = t3dt
[IMAGE] = 3t2
[IMAGE] t3 dt
TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE

Physics

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


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

Solution:
Phase difference between [IMAGE] Since D>>d then apply binomial approximation: [IMAGE] Phase difference between[IMAGE] Now using phase diagram
SUB TOPIC:YDSE

Physics


Question: has four choices

Physics

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


Question:
The series combination of two batteries, both of the same emf
10V, but different internal resistance of20$\Omega$ and 5$\Omega$, is
connected to the parallel combination of two resistors 30$\Omega$ and
R$\ \Omega$. The voltagedifference across the battery of internal
resistance 20$\Omega$ is zero, the value of R (in$\ \Omega$) is ___
(1)
(2)
(3)
(4)

Solution:
V_1 = $\varepsilon_{1}$  - i. r_1
0 = 10 - i × 20
i = 0.5A
V_2 = $\varepsilon_{2}$ - ir_2
= 10 - 0.5 × 5
V_2 = 7.5V
$0.5 = 0.25 + \frac{7.5}{x}$
$0.5 = \frac{7.5}{30} + \frac{7.5}{x}$
$0.5 = 0.25 + \frac{7.5}{x}$
$\frac{7.5}{x} = 0.25;$
$x = \frac{7.5}{0.25} = 30$ohms

Physics


Question:
One end of a spring is fixed to the ceiling and other end is
attached to a block. The block is released when spring is relaxed. The
product of time period and amplitude is 8 S.I. units. If spring is cut
in two equal parts and the two springs are attached to the block as
shown in figure. The block is released when both springs are relaxed. If
the product of time period and amplitude in S.I. units is x, then find
the value of x.

Solution:
In first case,
[IMAGE] In second case,

Physics


Question:
An ideal fluid flows (laminar flow) through a pipe of
non-uniform diameter. The maximum and minimum diameters of the pipes are
6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the
maximum velocities of fluid in this pipe is n/16. Find n:,d vkn'kZ nzo cnyrs gq,,d ikbi izokg esa cg jgk gSA
ikbi o Øe'k 6.4 cm
cgus okys nzo xfr

Solution:
Using equation of continuity
lkrR;rk lehdj.k yxkus ij
$A_{1}V_{1} = A_{2}V_{2}$
$\frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = \left( \frac{4.8}{6.4} \right)^{2} = \frac{9}{16}$

Physics


Question:
Two
produce stationary wave. The minimum distance of antinode from the
origin along positive x-axis in terms of the wavelength λ is
(y_1 and y_2 are in meter)

Solution:
B
y = y_1 + y_2 = [IMAGE] Amplitude is maximum at antinode
[IMAGE] General solution
For minimum distance of antinode from origin put n = 1
SECTION - B [INTEGER ANSWER section contains

Physics


Question:
A ball was thrown at velocity of 10 m/s at 37° to the horizontal.
What is the rate of change of its speed when its radius of curvature is
51.2 m.

Solution:
[IMAGE] aN = g cos θ =[IMAGE] v cos θ = 10 cos 37°
v =[IMAGE] g cos3 θ =[IMAGE] cos3 θ =[IMAGE] cos θ =[IMAGE] θ = 60°
at =[IMAGE] = g sin θ = 2
TOPIC: CIRCULAR MOTION
SUB TOPIC: Radius of Curvature

Physics

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


Question:
P- V diagram of ideal monoatomic gas is as shown. Find net heat
given to the gas in the process ABC.$\ $

Solution:
$\Delta Q = \Delta U + \Delta W$
$W_{ABCA} = \frac{1}{2}\left( 2P_{0} \right)\left( 2V_{0} \right) = 2P_{0}V_{0}$
$W_{ABC} = 2P_{0}V_{0} - W = 2P_{0}V_{0} - \left( - 2P_{0}V_{0} \right) = 4P_{0}V_{0}$
$\Delta U = \frac{nf}{2}R\Delta T = \frac{f}{2}P\Delta V = \frac{3}{2}P_{0}\left( 2V_{0} \right) = 3P_{0}V_{0}$
$\therefore\Delta Q = 7P_{0}V_{0}$

Physics

P-V diagram and first law of thermodynamics


Question:
A thin horizontal movable plate is separated from two fixed
horizontal plates [IMAGE] and
[IMAGE] by two highly viscous liquids of coefficient of viscosity
[IMAGE] and [IMAGE] as shown, where $\eta_{2} = 4\eta_{1}$.
Area of contact of movable plate with each fluid is same. If the
distance between two fixed plates is h, then the distance
[IMAGE] of movable plate from upper fixed plate such that the movable plate can
be moved with a constant velocity by applying a minimum constant
horizontal force F on movable plate is h/n (assume velocity gradient to
be uniform in each liquid). Find n.

Solution:
Let v be the velocity of the movable plate and F is equal to
viscous force
$F = \left\lbrack n_{1}\frac{v}{h_{1}} + n_{2}\frac{v}{h - h_{1}} \right\rbrack A$
$\Rightarrow \frac{dF}{{dh}_{1}} = 0\therefore h_{1} = \frac{h}{3}$

Physics


Question:
The ratio of the longest and shortest wavelengths of the Lyman
series is.

Solution: [IMAGE] Equation dividing

Physics


Question:
A certain quantity of oxygen [IMAGE] is
compressed isothermally until its pressure is doubled
[IMAGE]. The gas is then allowed to expand adiabatically
until its original volume is restored. Then the final pressure (P) in
terms of initial pressure [IMAGE] is:
vkWDlhtu,d izkjfEHkd nkc

Solution: [IMAGE] (isothermal process) ¼lerkih

Physics


Question:
A ring shaped tube contains two ideal gases with equal masses
and atomic mass numbers M_1=32 and M_2=28. The gases are separated by
one fixed partition X and another movable conducting partition Y which
can move freely without friction inside the ring. The angle is $\alpha$,
in equilibrium as shown in thefigure (in degrees). Find the value
of$\frac{\alpha}{48}$.,d vkdkj dh Vîqc esa nks vkn'kZ xSl ftu vyx&vyx dh tkrh gS rFkk nqljk foektd,d
xfreku pkydfoHkktd Y tks fcuk ?k"kZ.k xfr dj ldrk gSA
 fp=kuqlkj dks.k α ¼fMxzh esa½ rks [IMAGE]

Solution:
At equilibrium, pressure and temperature are same.
 rFkk rki leku
 Also, let cross-sectional area of tube be A and radius of ring be r.
ekuk nkc

Physics


Question:
The length of a potentiometer wire is 120 cm and it carries a
current of 60 mA. For a cell of emf 5V and internal resistance of
20$\Omega$, the null point on it is found to be at 1000 cm. The
resistance of whole wire is

Solution:
[IMAGE] Potential gradient
$= \frac{5}{1000} = \frac{V_{P}}{1200}$
$V_{p} = 6V$
 and
 $R_{p} = \frac{V_{P}}{I} = \frac{6}{60 \times 10^{- 3}} = 100\Omega$

Physics


Question:
Two uniform solid spheres of equal radii R, butmass M and 4M have
a centre to centreseparation 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 centres
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


Question:
A force given by the relation F = 8t, acts on a body of mass 2 kg,
initially at rest. Find the work done by this force on the body during
first 2 seconds of its motion ( in joule )
m = 2 kg F = 8t
 2 sec. ( )

Solution:
F = 8t
m[IMAGE] = 8t
[IMAGE] mv = 4t2
or m [IMAGE] = 4t2 ⇒ dx =[IMAGE] W =[IMAGE] =[IMAGE] = 64 J
TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE

Physics

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


Question:
The " y ' co-ordinate of the centre of mass of the system of
three rods of length as shown in
figure is: (Assume all rods to be of uniform density)

Solution:
The y-coordinate of centre of mass is
$= \frac{2m(\frac{\sqrt{3}}{2}a) + 2m(0) + 2m(0) + m(\frac{\sqrt{3}}{4}a) + m(\frac{\sqrt{3}}{4}a)}{8m}$
$= \frac{9a}{16\sqrt{3}}$

Physics


Question: Dimensions of gravitational constant are

Solution: $G = \frac{Fr^{2}}{m_{1}m_{2}};\ \lbrack G\rbrack = \left\lbrack \frac{Fr^{2}}{m_{1}m_{2}} \right\rbrack = \left\lbrack \frac{MLT^{- 2}L^{2}}{M^{2}} \right\rbrack = \left\lbrack M^{- 1}L^{3}T^{- 2} \right\rbrack$

Physics


Question:
A train of mass M is moving on a circular track of radius ' R
' with constant speed V. The length of thetrain is half of the
perimeter of the track. The linear momentum of the train will be

Solution:
If we treat the train as a ring of mass then its COM will
be at a distance[IMAGE] from the centre of the circle.
Velocity of centre of mass is:
$V_{CM} = R_{CM} \cdot \omega$
= $\frac{2R}{\pi} \cdot \omega$ =
$\frac{2R}{\pi} \cdot \left( \frac{v}{R} \right)$
$\left( \because\omega = \frac{v}{R} \right)$
$\Rightarrow$ $V_{CM} = \frac{2V}{\pi}$ $\Rightarrow$
${MV}_{CM} = \frac{2MV}{\pi}$
As the linear momentum of any system = MV_CM
[IMAGE] The linear momentum of the train
$= \frac{2MV}{\pi}$

Physics


Question:
A particle moving with kinetic energy E has de Broglie
wavelength$\lambda$. If energy $\Delta E$ is added to its energy, the
wavelength become $\lambda/2$. Value of$\Delta E$, is

Solution:
$\lambda = \frac{h}{\sqrt{2(KE)m}} = \lambda \propto \frac{1}{\sqrt{KE}}$
$\frac{\lambda}{\lambda/2} = \sqrt{\frac{{KE}_{f}}{{KE}_{i}}}$
$4{KE}_{i} = {KE}_{f}$
$\Rightarrow \Delta E = 4KE_{i} - KE_{i} = 3KE = 3E$

Physics


Question:
A smooth ball of mass m moving with the velocity v_o makes an
elastic impact on
another smooth ball of mass M at rest. After impact the velocity
components v_1 and v_2 are

Solution:
B
Velocity components tangential to the direction
of impact remains unchanged before and after the collision.
∴ (B)

Physics


Question:
The radius of gyration of a uniform rod of length
[IMAGE], about an axis passing through a point
[IMAGE] away fromthe centre of the rod, and perpendicular
to it, is

Physics

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


Question:
A U tube filled with a liquid is accelerating horizontally with
an acceleration a. The acceleration (ms^-2) of the tube is

Solution:
For free surface $tan\theta = a/g$
or $a = gtan\theta = 10/2 = 5$

Physics


Question:
Final image of point object formed by the combination is
located

At a distance 20 cm from curved surface.

Solution:
1st refraction at curved surface
[IMAGE] Reflection of mirror,
[IMAGE] 2nd refraction at curved surface
[IMAGE] So, final image from on plane surface.

Physics


Question: Which of the following gives a reversible operation?

Solution:
A logic gate is reversible if we can recover input data from
the output eg. NOT gate

Physics


Question:
A thin circular ring of mass per unit length ρ and radius r is
rotating at an angular speed ω as shown in figure. The tension in the
ring is

Solution:
2T sin[IMAGE] = dmRω2
T = ρR2ω2
TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS

Physics

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


Question:
The centre of mass of a non uniform rod of length L whose mass
per unit length $\lambda$varies as
$\lambda = \frac{k \cdot x^{2}}{L}$where k is a constant &x is the
distance of any point on rod from its one end, is (from the same end)

Physics

Center of mass of rod of non-uniformlinear mass density


Question:
The acceleration of the blocks (A) and (B) respectively in
situation shown in the figure is: (pulleys & strings are massless)

Solution:
aA = 2aB...(1)
mAg - T = mAaA...(2)
2T - mBg = mBaB...(3)
Solving, we get aA =[IMAGE] & aB
TOPIC: NLM
SUB TOPIC: NEWTON'S SECOND LAW

Physics

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


Question:
In the arrangement shown in figure, wavelength of light used is
[IMAGE]. The distance between slits
[IMAGE] The distance between [IMAGE] and
[IMAGE] is [IMAGE] If the ratio of
maximum to minimum intensity received on screen P is K. Then K is

Solution:
If intensity of light incident on slits [IMAGE] and
The intensity through [IMAGE] will be
[IMAGE] The intensity through [IMAGE] will be
[IMAGE] TOPIC:Wave Optics
SUB TOPIC:YDSE
LEVEL:Moderate

Physics


Question:
There is a small source of light at some depth below the surface
of water (refractive index =4/3) in a tank of large cross sectional
surface area. Neglecting any reflection from the bottom and absorption
by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height h and
radius of curvature r

Solution: [IMAGE] Solid angle [IMAGE] Percentage of light [IMAGE]

Physics


Question:
A simple pendulum of length 50cm and bob's mass 100g is
oscillating in the air. A damping force f = 0.0693V acts due to air,
where V is speed of bob. At any time it has a amplitude A. How much time
it will take so that amplitude becomes half of initial value.

Solution:
$A = A_{0^{e}}\frac{- bt}{2m}$
$\frac{\text{A}_{0}}{2} = \text{A}_{0^{\text{e}}}\frac{- \text{bt}}{2\text{m}} \Rightarrow \frac{\text{bt}}{2\text{m}} = \ln 2$

Physics


Question:
Two men of masses m and m/2 starts climbing up on two massless
strings fixed at the ceiling with acceleration g and g/2 respectively.
The ratio of tensions in the two strings are 2x: y then find the value
of x + y.

Solution:
FBD of man of mass (m)
$T_{1} = 2mg$
FBD of man of mass (m/2)
$T_{2} = \frac{mg}{2}\left\lbrack \frac{3}{2} \right\rbrack = \frac{3mg}{4}$
$\Rightarrow T_{1}:T_{2}::8:3$

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 [IMAGE] and [IMAGE] TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate

Physics


Question:
Visible light of wavelength [IMAGE] falls normally
on single slit and produces a diffractionpattern. It is found that the
second diffraction minimum is at 60º C from the central maximum. If the
first minimum is produced at[IMAGE], then
[IMAGE] is close to

Solution:
For 2^ndminima
So for 1^st minima
[IMAGE] (From equation (i)
[IMAGE] (From sin table)

Physics


Question:
A thin uniform rod of mass is hinged at one
end. This rod is maintained in horizontal position by colliding very
tiny balls each of mass m/10 completely elastically 10 times per sec
striking at the opposite end as shown in figure. The speed of the balls
just before colliding rod is [IMAGE].Find n.

Solution:
[IMAGE] Taking torque about hinge:
[IMAGE] TOPIC:Rotaional dynamics
SUB TOPIC: Torque
LEVEL: Moderate

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
ls tqM+s gYds fiLVu ¼{ks=Qy
 rFkk bl 100K
xeZ

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
∴[IMAGE] Work done by gas
dk;Z[IMAGE] = 1.50 = 1.5 J

Physics


Question:
The rod is uniform & weigh 500 N. Find magnitude of tension so
that tension in both strings (S_1, S_2) aresame (strings ideal).

Solution:
[IMAGE] Taking torque about W.
- T_1 (0.4)+T_2 (0.3)+500(0.2)=0
T = 1000

Physics


Question:
Three polaroids are kept coaxially. Angle between the first and
third polaroid is 90°. Angle between the first and second polaroid is
60°. If unpolarized light energy incident on the first polaroid is I0.
Light energy that emerges from the system is

Solution:
According to Malus law, I = I0 cos2 θ
After 2nd Polaroid, I = [IMAGE] cos2 60° =
[IMAGE] After 3rd Polaroid, I = [IMAGE] cos2 30° =

Physics


Question:
A uniform disc of mass M and radius R, is resting on a table on
its rim. The coefficient of frictionbetween disc and table is
[IMAGE] Now the disc is pulled with aforce F as shown in
the figure. What is the maximum value of F for which the disc rolls
without slipping?

Solution:
Let a be the acceleration of the centre of massof disc. Then
If there is no slipping, angular acceleration ofthe disc,
[IMAGE] Now torque of the disc (Using 2)
Substituting this in Eq. (i), we get
 Since there is no slipping,

Physics

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


Question:
The speed of water flowing out of the orifice before the
cylinder kept inside the tank

Physics


Question:
A man throws a packet from the top of the tower directly aiming
at his friend who is standing on the ground at a certain distance from
the base of the tower and this distance is same as the height of the
tower. If packet is thrown with a speed of 4 m /s and it hits the ground
midway between the tower base and his friend then height of the tower
(in m) is 0.4k. Find k. $(g = 10m/s^{2})$

Solution:
$\frac{4}{\sqrt{2}} \cdot t = \frac{h}{2}$....(1)
&$h = (\frac{4}{\sqrt{2}})t + \frac{1}{2}{gt}^{2}$....(2)
$\frac{h}{2} = \frac{1}{2}gt^{2}$
from (1) & (2)

Physics


Question:
A transverse sinusoidal wave move along a string in the positive
x-direction at a speed 5cm/s. The wavelength of the wave 0.2m and its
amplitude is 5cm. At given instant velocity of P is
[IMAGE] when its displacement is 2.5 cm. Which of the
following is correct?

$\overset{\rightarrow}{\text{v}} \cdot \overset{\rightarrow}{\text{a}} = \frac{2\sqrt{3}}{32}\pi^{3}\text{cm}^{2}/\text{s}^{3}$

$\overset{\rightarrow}{\text{v}} \cdot \overset{\rightarrow}{\text{a}} = \frac{25\sqrt{3}}{3}\pi^{3}\text{cm}^{2}/\text{s}^{3}$

$\overset{\rightarrow}{\text{v}} \cdot \overset{\rightarrow}{\text{a}} = \frac{25\sqrt{5}}{32}\pi^{3}\text{cm}^{2}/\text{s}^{3}$

Physics


Question:
Two uniform circular coils x and y having equal number of turns
and carrying equal current in same sense, subtend same solid angle at
point O. if smaller coil x is midway between O and y, then, if we
represent magnetic field due to coil y as B_y and that due to coil x as
B_x then $(\frac{B_{y}}{B_{x}}) = K$ then 20 K is

Solution:
If radius of x is r, then radius of y is 2r
$B_{y} = \frac{\mu_{0}}{4\pi}\frac{2\pi I(2r)^{2}}{{\lbrack(2r)^{2} + (2d)^{2}\rbrack}^{3/2}}$
$B_{x} = \frac{\mu_{0}}{4\pi}\frac{2\pi{Ir}^{2}}{{\lbrack r^{2} + d^{2}\rbrack}^{3/2}}$

Physics

Magnetic field due to current carrying coil


Question:
The sphere at P is given a downward velocity
[IMAGE] and swings in a vertical plane at the end of a
rope of [IMAGE] attached to a support at O. The rope
breaks at angle 30° from horizontal, knowing that it can withstand a
maximum tension equal to three times the weight of the sphere. The value
of [IMAGE] is [IMAGE]. Find

Solution:
[IMAGE] TOPIC:Vertical circle
SUB TOPIC: Simple pendulum
LEVEL: Moderate

Physics


Question:
A rubber cord has a cross-section area 1 cm^2 and natural
length of [IMAGE] It is stretched by
[IMAGE] to fire a small object of
mass[IMAGE]. If the Young's modulus (Y) is
[IMAGE] Speed acquired by the object in (m/s) is 2n.
Find n. Assume Hook's law is valid and neglect gravity.,d jcj dh Mksjh dk vuqçLFk dkV kjk çkIr
osx (m/s) esa½2n. nKkr djsaA ekfu,sa fd gqd

Solution:
v = 10^3 × 10^-2
v = 10.00 m/s

Physics

Potential energy stored in elongated rubber cord


Question:
A block is partially immersed in a liquid and the vessel is
accelerating upwards with acceleration "a". The block is observed by two
observers $O_{1}$ and $O_{2}$one at rest and the other accelerating with
an acceleration "a" upward. The total buoyant force on the block is:,d CykWd nzo:i ls Mqck gqvk gS rFkk ik= Roj.k "a" ls,d izs{kd fojke ij gS rFkk nwljk Roj.k"a" ls dh vksj Rofjr
gSA CykWd ij dk;Zjr~ dqy mRIykou cy gksxkA

Solution:
Buoyant force mRIykou
cy$= F_{b} = V_{sub} \cdot \rho_{\mathcal{l}} \cdot g$
Where,$V_{sub},\rho_{\mathcal{l}}$and g all are same w.r.t.
$O_{1}$and$O_{2}$
Hence vr (1)

Physics


Question:
The angular frequency of oscillation of mass m connected
symmetrically to four spring each of force constant K, when it is
displaced very slightly along one of the springs is

Solution:
[IMAGE] TOPIC: Simple harmonic motion
SUB TOPIC: Angular frequency of oscillation
LEVEL:Moderate

Physics


Question:
110 J of heat are added to a gaseous system where change in
internal energy is 40 J, then the amount of external work done is

Solution:
$\Delta W = \Delta u + \Delta W$
$\Rightarrow 110 = 40 + \Delta W$
$\Rightarrow \Delta W = 110 - 40 = 70J$

Physics


Question:
A train of mass M is moving on a circular track of radius ' R
' with constant speed V. The length of thetrain is half of the
perimeter of the track. The linear momentum of the train will be

Solution:
If we treat the train as a ring of mass then its COM will
be at a distance[IMAGE] from the centre of the circle. Velocity
of centre of mass is:
$V_{CM} = R_{CM} \cdot \omega$
= $\frac{2R}{\pi} \cdot \omega$ =
$\frac{2R}{\pi} \cdot \left( \frac{v}{R} \right)$
$\left( \because\omega = \frac{v}{R} \right)$
$\Rightarrow$ $V_{CM} = \frac{2V}{\pi}$ $\Rightarrow$
${MV}_{CM} = \frac{2MV}{\pi}$
As the linear momentum of any system = MV_CM
[IMAGE] The linear momentum of the train $= \frac{2MV}{\pi}$

Physics


Question:
A solid cylinder of mass m is placed over the trolley as shown
in the figure. The trolley starts to move forward with constant
horizontal acceleration a. The frictional force acting on the solid
cylinder will be (Assume no slipping)

Solution:
From the reference frame of trolley, the forces acting on the
cylinder is as shown.
and [IMAGE] (ii) From (i) and (ii),

Physics


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
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 $2x\pi\eta\text{r}\text{v}_{0}$.
Find x

Solution:
Relative to liquid, the velocity of sphere is
${\text{2}\text{v}}_{0}$ upwar Viscous force on sphere
$= 6\pi\eta\text{r}\text{v}_{0}$ downward
$= 12\pi\eta\text{r}\text{v}_{0}$ downward

Physics


Question:
Boolean relation at the output stage-Y for the following circuit
is

Solution:
First part of figure shown is OR gate and
 izFke Hkkx OR
Second part of figure shown is NOT gate
rFkk Hkkx NOT
So vr Y_p = OR + NOT = NOR gate

Physics

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


Question:
A penguin of mass m stands at the right edge of a sled of mass 3m
and legnth [IMAGE]. The sled lies onfrictionless ice. The
penguin starts moving towards left, reaches the left end and jumps with
a velocity uat an angle[IMAGE] with horizontal relative to
ground. (Neglect the height of the sled)

Till the penguin reaches the left end, the sled is displaced

After jumping, it will fall on the ground at a
distance[IMAGE] from the left end of the sled.

After jumping, it will fall on the ground at a distance
[IMAGE] from the left end of the sled.

Solution:
$S_{CM} = \frac{m_{1}S_{1} + m_{2}S_{2}}{m_{1} + m_{2}}$
$0 = \frac{(3m)( - x) + (m)(\mathcal{l -}x)}{3m + m}\ $
$x = \frac{\mathcal{l}}{4}$
Displacement of sled in this time
$= \left( \frac{{ucos}\theta}{3} \right)\left( \frac{2{usin}\theta}{g} \right) = \frac{1}{3}\left( \frac{u^{2}\sin{2\theta}}{g} \right)$
Total distance between penguin and left end of
sled$= \frac{4}{3}\left( \frac{u^{2}sin2\theta}{g} \right)$

Physics

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


Question:
Two forks A and B when sounded together produce 4 beats/s. The
fork A is in unison with 30cm length of a sonometer wire and B is in
unison with 25cm length of the same wire at the same tension. What can
be frequency of forks.

Physics


Question:
A block of mass 2 kg is given a push horizontally and then the
block starts sliding over a horizontal plane. The graph shows the
velocity time graph of the motion. The co-efficient of friction between
the plane and the block is

Solution:
a = slope of v-t graph = - 2 m/s2
∴ Ft = μmg = m × 2
⇒μ × 2 × 10 = 2 × 2 ⇒ μ = 0.2
TOPIC: NLM
SUB TOPIC: FRICTION

Physics

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


Question:
As soon as the valve is opened, the mercury level at B in the
tube

Solution:
In narrow passage velocity is greater.
Hence static pressure will be less at B. So the level moves up.

Physics


Question:
In a common-base configuration of transistor, α = 0.98, I_B =
0.02mA, R_L = 5 kΩ. Output voltage across load is

Solution:
[IMAGE] I_E 0.98 = I_C
I_E = I_B + I_C
I_E = 1 mA
I_C = 0.98 mA
V_0 = R_L × I_C

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 = 5 m lying x-y
plane encloses the magnetic field at t = 0 and then pulled at uniform
velocity$\ \overset{\rightarrow}{v} = 4\overset{\hat{}}{i}m/s$. The emf
induced (in volts) is the loop at t = 2 sec is 6 V. Then magnitude of 45
B is

Solution:
$\varepsilon = |(\overset{\rightarrow}{v} \times \overset{\rightarrow}{B}) \cdot \overset{\rightarrow}{\mathcal{l}}|$
$= VB2\sqrt{R^{2} - {(\frac{vt}{2})}^{2}} = VB\sqrt{4R^{2} - V^{2}t^{2}}$
$6V = 4 \times B\sqrt{4 \times 25 - 16 \times 4}$

Physics


Question:
A container of a large uniform cross-sectional area A resting on
a horizontal surface holds two immiscible, non-viscous and
incompressible liquids of densities and each of height
(1/2)H as shown. The smaller density liquid is open to atmosphere. A
homogeneous solid cylinder of length $\left( L < \frac{1}{2}h \right)$
cross-sectional area (1/5) A is immersed such that it floats with its
axis vertical to the liquid-liquid interface with length (1/4) L in
denser liquid. If D is the density of the solid cylinder then

Solution:
$D \times L \times Ag = 2d \times \frac{L}{4} \times Ag + d \times \frac{3L}{4} \times Ag$
$D = \frac{5d}{4}$

Physics


Question:
20 m ehukj,d dadM+ 10 izkjfEHkd osx Qsa o gksxh (g = 10 m/s2)
20 m 10 m/s, (g=10 m/s^2)

Solution:
vy2 - uy2 = 2 × g × s
[IMAGE] vy2 = 0 + 2 × 10 × 10 = 10
TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

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


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 ?

Solution:
For rectangular hyperbola
XY = constant
$U\rho =$ constant
$nC_{V}T\rho =$ constant
$n\frac{5}{2}RT\frac{M}{V} =$constant
$\frac{5}{2}P\frac{VM}{V} =$constant $\Rightarrow$ P = constant
(isobaric process)
$\Delta U_{B \rightarrow A} = 37 - 2 = 35J = \frac{5}{2}nR\Delta T$
$14J = nR\Delta T$
$\therefore$ $W = 14J$

Physics


Question:
Two steel wires having same length are suspended from a ceiling
under the same load. If the ratio oftheir energy stored per unit volume
is 1: 4, the ratio of their diameters is:
leky yEckbZ ds nks LVhy ds rkjksa ij leku Hkkj cka/kdj bUg a NRk ls
vuqikr 1: 4

Physics


Question:
The potential energy of a particle moving along x axis, starting
from x = 0 at rest is given by
U = (ax - b) x
 b are positive constant]

Solution:
Kmax ⇒ Umin
⇒[IMAGE] = 0
⇒ 2ax - b = 0
x =[IMAGE] Ki + Ui = Kf + Uf
0 + 0 = Kf +[IMAGE] ⇒ Kf = [IMAGE] TOPIC: WPE
SUB TOPIC: ENERGY CONSERVATION

Physics

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


Question:
A car is initially at rest, 330 m away from a stationary
observer. It begins to move towards the observer with an acceleration of
[IMAGE], sounding its horn continuously. 20 second
later, the driver stops sound the horn. The velocity of sound in air is
330 m/s. The observer will hear the sound of the horn for a duration of
[IMAGE] sec. Find n.

Solution:
[IMAGE] In 20 seconds, distance travelled by the Source is:
[IMAGE] Let, t = 0 be the starting time.
The sound wave started at t = 0 from reaches the observer at
'B' after [IMAGE] ie. observer started hearing the
sound at t = 1 sec. At t = 20 sec. the source reaches at 220 m
The last sound wave starting from reaches Hence, the observer do not hear the sound for whole 20 sec.
but for: [IMAGE] TOPIC: Sound wave
SUB TOPIC: Doppler effect
LEVEL: Moderate

Physics


Question:
Block A is given an acceleration $12{ms}^{- 2}$ towards left as
shown in Figure. Assume block B always remains horizontal, find the
acceleration $\left( \text{ in }{ms}^{- 2} \right)$ of B.

Solution: $\mathcal{l}_{1} + \mathcal{l}_{2} = C \Rightarrow \mathcal{l}_{1}^{''} + \mathcal{l}_{2}^{''} = 0$

Physics


Question:
A perfectly elastic uniform string is suspended vertically with
its upper end fixed to the ceiling and the lower end loaded with the
weight. If a transverse wave is imparted to the lower end of the string,
the pulse will

not travel along the length of the string

travelled upwards with variable acceleration

Physics


Question:
A uniform triangular plate ABC of moment of inertia I (about an
axis passing through A and perpendicular to plane of the plate) can
rotate freely in the vertical plane about point as shown in
figure. The plate is released from the position shown in the figure.
Line AB is horizontal. The acceleration of centre of mass just after the
release of plate is

Solution:
Torque about A:
$\Rightarrow$ $\alpha = \frac{mga}{2I}$
$\Rightarrow$ Acceleration
$= \frac{a}{\sqrt{3}}\alpha = \frac{m\ g{\ a}^{2}}{2\sqrt{3}\ I}$

Physics

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


Question:
Planet A has mass M and radius R. Planet B has half the mass and
half the radius of Planet A. If the escape velocities from the Planets A
and B are [IMAGE] and [IMAGE],
respectively, then [IMAGE] The value of n is

Physics

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


Question:
In a corner on a smooth horizontal surfaces we kept two masses
[IMAGE] and [IMAGE] attached to a spring
in natural length as shown. The spring has a spring constant 150 N/m.
Now [IMAGE] is suddenly given a velocity towards the
left at t = 0. After this spring attains natural length first at time
[IMAGE] sec and then next at time
[IMAGE]. If mass of [IMAGE] (in kg) is K
then K is

Solution:
Solving (i) and (ii), [IMAGE] TOPIC: Simple harmonic motion
SUB TOPIC: Block and spring system
LEVEL: Moderate

Physics


Question:
A mass of material exists in its solid form at its melting
temperature 10°C. The following processes then occur to the material:
Process-1: An amount of thermal energy Q is added to the material
and [IMAGE] of the material melts.
Process-2: An identical additional amount of thermal energy Q is
added to the material and the material is now a liquid at 50°C.
What is the ratio of the latent heat of fusion to the specific heat of
the liquid for this material?

Solution:
[IMAGE] mL = Q
Q = [IMAGE] + mS × 40
[IMAGE] = 4mS
[IMAGE] = 80

Physics


Question:
The variation of velocity with time of a toy car moving along a
straight line is as shown below. Which of the following graphs correctly
represents the variation of acceleration with time for the toy car?

Solution:
for 0 to 1 seconds velocity increases linearly
$a = \frac{4 - 2}{1 - 0} = 2m/s^{2}$
$\Rightarrow$a vs t graph is st. line parallel to x-axis (+ve
acceleration)
for 1 to 1.5 sec.$v = constant \Rightarrow a = 0$
for $t > 1.5\ sec.\ $a is negative

Physics

Velocity-time and acceleration-time graph


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