Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Moderate

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.

1) Moment of inertia of the system about x axis is
2) Moment of inertia of the system about y axis
3) Moment of inertia of the system about the diagonal axis
4) Moment of inertia of the system about z axis is

Correct option: 1

Solution:
Subject: Physics — Nuclear Physics — Radioactivity
Difficulty: Moderate

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

1) 62 min
2) 66 min
3) 72 min
4) 52 min

Correct option: 1

Solution:
$\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$
Subject: Physics — Kinetic theory of gasses — Heat Capacity
Difficulty: Moderate

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

1) $\frac{23}{15}$
2) $\frac{40}{27}$
3) $\frac{67}{45}$
4) $\frac{19}{13}$

Correct option: 4

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

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

1) 7.88 × 10^4 K
2) 3.2 × 10^4 K
3) 0.85 × 10^4 K
4) 13.6 ×
10^4 K

Correct option: 1

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
Subject: Physics — Electrostatics — Energy stored in capacitor
Difficulty: Tough

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

1) Energy supplied by the battery isCV^2.
cSVjh }kjk lIykbZ CV^2
2) Energy linked in the space between plates 1 and 2 is$\frac{1}{6}$
CV^2
3) Potential difference between plates 2 and 4 is V/2
IysV 2 rFkk 4
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

Correct option: 1

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
Subject: Physics — Work Power & Energy — Work energy theorem
Difficulty: Moderate

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

1) 1600 J
2) 160 J
3) 16 J
4) 1.6 J

Correct option: 3

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 = $\Delta K = 16\ J$
Subject: Physics —  — 
Difficulty: Tough

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

1) zero
4) [IMAGE] m a 
 A, B

Correct option: 3

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

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


Answer: 60

Solution:
TOPIC: CIRCULAR MOTION
SUB TOPIC: RADIUS OF CURVATURE
Subject: Physics — Rigid Body Dynamics — Pure rolling on a plank
Difficulty: Moderate

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

1) Zero

Correct option: 4

Solution:
FBD for sphere & block
$a_{2} = \frac{f_{r}}{m} = \frac{\mu mg}{m}$
${\overrightarrow{a}}_{1} = \mu g\overset{\hat{}}{i}$
${\overrightarrow{a}}_{2} = - \mu g\overset{\hat{}}{i}$
${\overset{\rightarrow}{a}}_{rel} = {\overset{\rightarrow}{a}}_{1} - {\overset{\rightarrow}{a}}_{2} = 2\mu g\overset{\hat{}}{i}$
$a_{rel} = 2\mu g$.
Subject: Physics — Mechanical Properties of Fluids — Terminal velocity
Difficulty: Moderate

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


Answer: 2

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
Subject: Physics — Center of mass — Collision
Difficulty: Tough

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.

1) $\frac{3}{4}V_{0}\sqrt{\frac{m}{k}}$
2) $\frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$
3) $\frac{3}{4}V_{0}\sqrt{\frac{2m}{5}k}$
4) $\frac{3}{5}V_{0}\sqrt{\frac{m}{k}}$

Correct option: 2

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.
$\frac{1}{2}m{(\frac{3}{4}V_{0})}^{2} = \frac{1}{2}m{(\frac{3}{5}V_{0})}^{2} + \frac{1}{2}m{(\frac{3}{10}V_{0})}^{2} + \frac{1}{2}kx_{max.}^{2} \Rightarrow x_{\max} = \frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$
Subject: Physics — Behaviour of perfect gases — Bulk modulus
Difficulty: Moderate

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

1) 32.30
2) 33.40
3) 33.20
4) 33.00

Correct option: 3

Solution:
Bulk modulus of elasticity at constant temperature ij
Subject: Physics — Modern physics — Radioactivity
Difficulty: Moderate

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

1) 62 min
2) 66 min
3) 72 min
4) 52 min

Correct option: 1

Solution:
[IMAGE] From (i)(ii)
$\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$
Subject: Physics — Atomic physics — Bohr's mode of hydrogen atom
Difficulty: Moderate

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


Correct option: 3

Solution:
Subject: Physics — Ray Optics — Refraction (Lens)
Difficulty: Moderate

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)


Answer: 3

Solution:
since [IMAGE] For first position of the lens
[IMAGE] From displacement method
Using
$f = \frac{D^{2} - \Delta d^{2}}{4D} = \frac{60^{2} - 20^{2}}{4 \times 60} = \frac{40}{3}\text{cm}$
Subject: Physics — Electrostatics — Electric potential due to non-conducting solid sphere
Difficulty: Moderate

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


Correct option: 3

Solution:
[IMAGE] Where, r → distance from circle
Subject: Physics — Mechanical Properties of Fluids — Surface tension
Difficulty: Moderate

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


Correct option: 2

Solution:
Subject: Physics — Simple Harmonic Motion — Position and velocity of a particle
Difficulty: Moderate

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


Answer: 2

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
Subject: Physics —  — Elongation in rod due to unbalanced force
Difficulty: Tough

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.

1) $\frac{2F}{YA}L$
2) $\frac{4F}{YA}L$
3) $\frac{F}{YA}L$
4) $\frac{3F}{2YA}L$

Correct option: 1

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 $= \int_{0}^{L}\mspace{2mu}\frac{T}{YA}dx = \frac{2F}{YA}L$
Subject: Physics — Newton's Laws of Motion — Newton's second law
Difficulty: Moderate

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

1) 2
2) 4
3) 5
4) 3

Correct option: 3

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)
$a_{2} = g\sin^{2}\theta$
$= 5m/s^{2}$
Subject: Physics — Electric current — Ohm's law
Difficulty: Easy

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

1) zero
2) 2 A
3) 0.4 A
4) 4A

Correct option: 2

Solution:
$i = \frac{20}{2} = 10A$
$I = \frac{4i}{5} - \frac{3i}{5} = + \frac{i}{5} = 2A$
Subject: Physics — Mechanical Properties of Fluids — Fluid in motion
Difficulty: Moderate

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


Answer: 5

Solution:
For free surface eqDr lrg
Subject: Physics — Thermodynamics — Thermodynamic processes
Difficulty: Moderate

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

1) Isothermal
2) Adiabatic
3) Isobaric
4) Isochoric

Correct option: 2

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


Correct option: 4

Solution:
Jx = Jx [IMAGE] I = JxA
evdBz = eEy
Ey = vdBz
neAvd = JxA
Ey = [IMAGE] Constant of proportionality = [IMAGE] =
Subject: Physics — Electrostatics — Electric field due to electric dipole
Difficulty: Tough

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

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

Correct option: 1

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]
Subject: Physics —  — 

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

1) meter
4) s-1

Correct option: 1

Solution:
Subject: Physics —  — 
Difficulty: Easy

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


Correct option: 2

Solution:
After reflection reflected ray passes through focus
So vector along reflected ray[IMAGE] TOPIC:Ray optics
SUB TOPIC:Vectors
Subject: Physics — Mechanical properties of matter — Elongation in rod due to unbalanced force
Difficulty: Tough

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

1) $\frac{2F}{YA}L$
2) $\frac{4F}{YA}L$
3) $\frac{F}{YA}L$
4) $\frac{3F}{2YA}L$

Correct option: 1

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+
Subject: Physics — Work Power & Energy — Potential energy
Difficulty: Moderate

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


Correct option: 3

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.
Subject: Physics —  — 
Difficulty: Tough

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


Answer: 08

Solution:
S = [IMAGE] + 2t
v = t + 2
[IMAGE] R = 8 m
TOPIC: CIRCULAR MOTION
SUB TOPIC: KINEMATIC
Subject: Physics — Mechanical Properties of Fluids — Application of Bernoulli's principle
Difficulty: Moderate

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

1) Level will rise above A Lrj A
2) Level will fall below A Lrj A
3) Level will remain at A Lrj A ij gh jgsxk
4) can not say dqN ugha dgk tk ldrk

Correct option: 1

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
Subject: Physics — Electromagnetic induction — LR circuit
Difficulty: Tough

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

1) $\frac{\varepsilon L}{R^{2}}(1 - \frac{1}{e})$
2) $\frac{\varepsilon R}{eL^{2}}$
3) $\frac{\varepsilon L}{R^{2}}$
4) $\frac{\varepsilon L}{eR^{2}}$

Correct option: 4

Solution:
$q = \int_{0}^{T_{C}}idt$
$= \frac{\varepsilon}{R}{\lbrack t - \frac{e^{- t/T_{C}}}{\frac{- 1}{T_{C}}}\rbrack}_{0}^{T_{C}}$;
$= \frac{\varepsilon}{R}\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\rbrack$
$= \frac{\varepsilon}{R} \times \frac{1}{e} \times \frac{L}{R}$
$= \frac{\varepsilon L}{R^{2}e}$
Subject: Physics — Thermodynamics — Thermodynamic processes
Difficulty: Moderate

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)


Answer: 7

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
$\frac{5(1 - x) + 2}{2(1 - x)} = 0\ or\ \frac{7 - 5x}{1 - x} = 0 \Rightarrow x = 1.4$
Subject: Physics —  — 
Difficulty: Easy

Question: [IMAGE] m =
 a = = 

1) meter
4) s-1

Correct option: 1

Solution:
α] =[IMAGE] TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS
Subject: Physics — Rigid Body Dynamics — Rolling
Difficulty: Tough

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.

1) Ratio of magnitude of acceleration of centre of mass of bigger and
smaller rollers is 1
2) Magnitude of friction between bigger roller and ground is more than
that between smaller roller and ground.
3) Magnitude of friction between plank and bigger roller is more than
that between plank and smaller roller.
4) All of the above

Correct option: 4

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
Subject: Physics —  — 
Difficulty: Moderate

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


Correct option: 3

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
Subject: 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 ?

1) -x
2) +x
3) -y
4) +y

Correct option: 3

Solution:
Subject: Physics —  — 

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


Answer: 02

Solution:
[IMAGE] Along agbfcd
V_a + 1 - 2 - 4 + 8 - 1 = V_d
⇒ V_a + 2 = V_d ⇒ V_d - V_a = 2 V
Subject: Physics — Communication system — Modulation
Difficulty: Moderate

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


Answer: 2

Solution:
A_max = A_m + A_c = 10
A_min = A_c - A_m = 2
A_c = 6 V
A_m = 4 V
$m = \frac{A_{m}}{A_{c}} = \frac{4}{6} = \frac{2}{3}$
Subject: Physics — System of particle — Collision
Difficulty: Moderate

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

1) The tension in the string just before and after the collision changes
by$\frac{m^{2}v^{2}}{(m + M)\mathcal{l}}$
2) The tension in the string just before and after the collision changes
by $m\lbrack g + \frac{mv^{2}}{(m + M)\mathcal{l}}\rbrack$
3) Tension in the string just after the collision
becomes$m\lbrack g + \frac{mv^{2}}{(m + M)\mathcal{l}}\rbrack$
4) Tension in the string will not change after the collision

Correct option: 2

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

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


Answer: 15

Solution:
[IMAGE] u cos 30° = V cos 60°
[IMAGE] TOPIC:KINEMATIC
SUB TOPIC: 2 D
Subject: Physics — Gravitation — Acceleration due to gravity at poles and equator
Difficulty: Moderate

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)

4) None of these

Correct option: 3

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]