Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Mechanical Properties of Fluids — Surface tension
Difficulty: Moderate

Question: The surface tension of a liquid is 5 Newton per meter. If a film
of this liquid is held on a ring of area $0.02m^{2}$, if its surface
energy is about $\left( n \times 10^{- 1} \right)$ Joule then n is

1) $
$= 5frac times 2 times 0.02m_ = 0.2J$

Answer: 2

Solution:
We know that surface energy
$U_{s} = T \times \ Area\ $
Here, as 2 films are formed because of ring.
So $U_{s} = T \times 2 \times (A)$
$= 5\frac{N}{m} \times 2 \times 0.02m_{2} = 0.2J$
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 —  — 

Question: If the dipole moment vector[IMAGE] of a dipole
and the electric field[IMAGE] (uniform) are oriented as
shown then which of the following statement is incorrect?

1) potential energy in I = potential energy in II
2) potential energy in I = negative of potential energy in II
3) in I rotates anticlockwise
4) in II rotates clockwise

Correct option: 2

Solution:
[IMAGE] where α is always the shorter angle
between[IMAGE] and [IMAGE] ∴ in both cases I and II, α = π − θ ⇒ (cos α)_I = (cos α)_II
∴ U_I = U_II ⇒ (B) is wrong.
(C) and (D) are correct as direction of [IMAGE]
Subject: Physics — Simple Harmonic Motion — Time period of oscillation
Difficulty: Moderate

Question: The minimum time in which the potential energy of a particle
executing SHM changes from maximum to minimum is 5 s. Then the time
period of SHM is

1) 5 s
2) 10 s
3) 15 s
4) 20 s

Correct option: 4

Solution:
P.E. is maximum at extreme position and minimum at mean
position.
Time to go from extreme position to mean position is,
 where T is time period of SHM.
Subject: Physics — Projectile Motion — Angle of projection
Difficulty: Moderate

Question: The speed of a projectile at the highest point is $v_{1}$ and at
the point half the maximum height is $v_{2}$. If
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}}$, then the angle of projection
is π/n. Find n.


Answer: 3

Solution:
$0 = {(v_{2}sin\beta)}^{2} - 2g\frac{H}{2}$
$v_{1}^{2} = v_{1}^{2} - gh$
Also $\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}}$
From above
$v_{1} = \sqrt{\frac{2gH}{3}}$
$v_{2} = \sqrt{\frac{5gH}{3}}$
$H = \frac{u^{2}\sin^{2}\theta}{2g} \Rightarrow sin2\theta = \frac{2gH\cos^{2}\theta}{v_{1}^{2}}$
$\tan^{2}\theta = \frac{2gH}{v_{1}^{2}}{or\ tan}^{2}\theta = \frac{2gH \times 3}{2gH}$
$tan\theta = \sqrt{3}$
$\theta = 60^{\circ}$
Subject: Physics — Work Power & Energy — Work -energy theorem
Difficulty: Moderate

Question: A block of mass 60 kg is released from rest when compression in
the spring is 2 m (natural length of spring is 8 m). Surface AB is
smooth while surface BC is rough. Block travels x distance before coming
to complete rest. Value of x is: $\lbrack g = 10m/s^{2}\rbrack$

1) 2 m
2) 3 m
3) 1 m
4) 4 m

Correct option: 2

Solution:
By Work Energy
Theorem$W_{net} = \Delta K.E = 0 \Rightarrow \frac{1}{2}k(x_{0}^{2} - x^{2}) - \mu mgx = 0$
$\Rightarrow \frac{1}{2} \times 200(2^{2} - x^{2}) = \frac{1}{2} \times 60 \times 10x$
$\Rightarrow x = 1m$
Also at this moment $f_{\max} > kx$
So, block will not move so total distance travelled =2+1=3 m.
Subject: Physics —  — 

Question: The diagram below shows the lowest four energy levels for an
electron in a hypothetical atom. The electron is excited to the -1eV
level of the atom and transitions to the lowest energy state by emitting
only two photons. Which of the following energies could not belong to
either of the photons ?

1) 2 eV
2) 4 eV
3) 5 eV
4) 6 eV

Correct option: 2

Solution:
3eV to -7eV is not possible.
Subject: Physics — Dual Nature of Matter — De Broglie wavelength
Difficulty: Moderate

Question: An electron (mass m) with initial velocity [IMAGE] is in an electric field [IMAGE] If [IMAGE] is initial de-Broglie wavelength of electron, its de-Broglie wave length
at time t is given by


Correct option: 2

Solution:
Initially [IMAGE] Velocity as a function of time [IMAGE] so wavelength [IMAGE]
Subject: Physics — Rotational motion — Angularimpulse
Difficulty: Tough

Question: An impulsive force F acts horizontally on a solid sphere of
radius R placed on a horizontal surface. Theline of action of the
impulsive force is at a height h above the centre of the sphere. If the
rotational andtranslational kinetic energies of the sphere just after
the impulse are equal, then h is equal to

1) $\frac{R}{2}$
2) $\frac{2}{5}R$
3) $\frac{R}{\sqrt{2}}$
4) $\sqrt{\frac{2}{5}}R$

Correct option: 4

Solution:
Transitional k.e.= rotational k.e.
$\frac{1}{2}{mv}^{2} = \frac{1}{2}I\omega^{2}$
$V^{2} = \frac{2}{5}\omega^{2}R^{2}$...(1)
Impulse eq. Fdt = mV...(2)
$\ Fhdt\ = I\omega = \frac{2}{5}mR^{2}\omega$...(3)
Dividing (2) and (3)$h = \frac{2}{5}\frac{R^{2}\omega}{v}$...(4)
Solving (1) and (4)$h = \sqrt{\frac{2}{5}}R$
Subject: Physics —  — 

Question: Two liquids of densities 1 are mixed in equal
volumes. A cube of external dimension 20 cm has an inner cubical portion
of side 10 cm whose density is twice that of the outer portion. If this
cube is just floating in the liquid mixture, find the density of the
inner portion.


Correct option: 2

Solution:
ρ = [IMAGE] = 2 mg = 2ρ1 × 103 + ρ1 × (203 - 103)
= 9ρ1 × 103 g = ρ × 8 × 103 g
Subject: Physics — Kinetic Theory of Gases — Heat capacity
Difficulty: Moderate

Question: Consider a mixture of n moles of helium gas and 2 n moles of
oxygen gas (molecules taken to be rigid) as an ideal gas. Its
[IMAGE] value will be:
ds 2n eksYl dh feJ.k xSl eku s rks bl feJ.k


Correct option: 4

Solution:
Subject: Physics — Newton's Laws of Motion — Constraint motion
Difficulty: Tough

Question: Find out velocity of block B in the given figure: (all the
pulleys are ideal and string are inextensible massless)

1) $80\ m/s\ \widehat{j}$
2) $- 40\ m/s$
3) $0\ m/s\ \widehat{j}$
4) $- 80\ m/s\ \widehat{j}$

Correct option: 3

Solution:
$\mathcal{l}_{1}^{'} + \mathcal{l}_{2}^{'} = 0$
$x = 4$`<!- ->`0
$\mathcal{l}_{3}^{'} + \mathcal{l}_{4}^{'} = 0$
$40 - y + 40 - 80 = 0$
$y = 0$
Subject: Physics —  — 

Question: In the figure shown switch [IMAGE] remains
connected and switch S2 remains open for a long time. Now
[IMAGE] is also closed. Assuming
[IMAGE] and [IMAGE]. The magnitude of
rate of change of current is 2n A/s in inductor just after the switch S2
is closed. Find n


Answer: 5

Solution:
[IMAGE] Using [IMAGE] TOPIC: Electromagnetic induction
SUB TOPIC: Self induction
LEVEL: Moderate
Subject: Physics — Elasticity — 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 — Variable mass
Difficulty: Tough

Question: In the fig. shown a cart moves on a smooth horizontal surface
due to an external constant force of magnitude F. The initial mass of
the cart is $M_{0}$and velocity is zero. Sand falls on to the cart with
negligible velocity at constant rate$\mu\ kg/s$ and sticks to the cart.
The velocity of the cart at time t is

1) $\frac{Ft}{M_{0} + \mu t}$
2) $\frac{F}{\mu}\mathcal{l}n\frac{m_{0} + \mu t}{m_{0}}$
3) $\frac{Ft}{M_{0}}$
4) $\frac{Ft}{M_{0} + \mu t}e^{\mu t}$

Correct option: 1

Solution:
[IMAGE] Formula $F = m\frac{dv}{dt} + (V - u)\frac{dm}{dt}$
Here u=velocity of sand =0
$m = M_{o} - \mu t = mass\ at\ time\ t$
and$\frac{dm}{dt} = \mu$
$\therefore F = (M_{0} + \mu t)\frac{dm}{dt} + v\mu$
$(F - \mu v)dt = (M_{o} + \mu t)dv$
$\int_{0}^{t}\frac{dt}{M_{0} + \mu t} = \int_{0}^{v}\frac{dv}{F - \mu v}$
${\lbrack log(M_{0} + \mu t)\rbrack}_{0}^{t} = \frac{1}{\mu}\lbrack log(F - \mu v)\rbrack_{0}^{v}$
$log\frac{(M_{0} + \mu t)}{M_{0}} = log(\frac{F}{F - \mu v})$
$F - \mu v = \frac{M_{o}F}{M_{0} + \mu t} \Rightarrow v = \frac{Ft}{M_{0} + \mu t}$
Subject: Physics —  — 
Difficulty: Moderate

Question: Power delivered to a particle varies with time as P = (3t2 - 2t)
watt, where is in seconds, then change in its kinetic energy
between time t = 1 to t = 3s.
 P = (3t2 - 2t),, t = 1 t = 3s


Answer: 18

Solution:
[IMAGE] W =[IMAGE] = ∆ K.E.
TOPIC: WPE
SUB TOPIC: POWER
Subject: Physics — Mechanical properties of liquid — Hydrostatic pressure
Difficulty: Moderate

Question: A simple open U-tube contains mercury. Now water is poured
slowly upto height 27.2 cm in the left arm. How high (in cm) does the
mercury rise in the right arm from its initial level in equilibrium
state?
(Take density of mercury $= 13600\ kg/m^{3}$ and that of water
$= 1000\ kg/m^{3}$and $g\ = 10\ m/s^{2}$)
 27.2cm rd Hkjk fLFkfr 
Hkqtk es a blds çkjfEHkd Lrj ls ikjk fdruh (cm esa) rd tk;sxkA
ikjs dk ?kuRo $= 13600\frac{kg}{m^{3}}$ rFkk ikuh


Answer: 1

Solution:
[IMAGE] Pressure at A that is at interface of water and mercury and B is same as
both are at same horizontal level in same liquid. Thus
Aij vFkkZr~ ikuh dh vkUrfjd lrg rFkk ikjs ij rFkkBij nkc leku gksxk
D;ksafd nksauks leku nzo esa leku kjk çkIr 
$= \frac{x}{2} = 1\ cm$
Subject: Physics — Thermodynamics — Efficiency of a cyclic process
Difficulty: Moderate

Question: The efficiency $\eta$ of the cyclic process
$\left( \gamma = \frac{5}{3} \right)$ is given by $\frac{50n}{23}$ then
find out the value of n.
pØh; izØe $\eta\frac{50n}{23}$gS rksn

1) 12.5
2) 13.5
3) 18.5
4) 11.5

Correct option: 1

Solution:
$\eta = \frac{W}{+ Q}$
Temperature at A is T_0 then$P_{0}V_{0} = {nRT}_{0}$
 T_0gS rks$P_{0}V_{0} = {nRT}_{0}$
at$T_{A} = T_{0},T_{B} = 2T_{0},T_{C} = 6T_{0},T_{D} = 3T_{0}$ 
$W = 2P_{0}V_{0}$
$+ Q = Q_{AB} + Q_{BC}$
$= n \cdot C_{v}dT + nC_{p}dT$
$= \frac{3}{2}{nRT}_{0} + \frac{5}{2}nR\left( 4T_{0} \right)$
$= \frac{23}{2}{nRT}_{0} = \frac{23}{2}P_{0}V_{0}$
$\eta = \frac{2P_{0}V_{0}}{\frac{23}{2}P_{0}V_{0}} = \frac{4}{23}$
Subject: Physics — Heat — Bulk modulus
Difficulty: Moderate

Question: 32 g of O_2 is contained in a cubical container of side 1 m and
maintained at a temperature of 127ºC. The isothermal bulk modulus of
elasticity of the gas in terms of universal gas constant R is

1) 127 R
2) 400 R
3) 200 R
4) 560 R

Correct option: 2

Solution:
Bulk modulus of elasticity at constant temperature
$B = \frac{dP}{dV/V} - = - V\left( \frac{dP}{dV} \right)_{T} = + P = \frac{nRT}{v}$
$= \frac{1 \times R \times 400}{1^{3}} = 400R$
Subject: Physics — Capacitance — Dielectric
Difficulty: Moderate

Question: A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. A potential difference of 2500
v is applied between the plates. If dielectric strength of air is
$3 \times 10^{6}v/m$ than

1) Air will not breakdown
2) Air will breakdown
3) Potential gradient strength in fiber sheet exceeds dielectric
strength of air
4) All the above are correct

Correct option: 2

Solution:
As $V = V_{1} + V_{2}$
$2500 = 0.0002E_{1} + 0.0023E_{2}$
$2E_{1} + 23E_{2} = 25 \times 10^{6}$
Also $E_{1} = 5E_{2}$
Solving $E_{2} = 0757 \times 10^{8}V/m$
As $E_{1} > 3.6 \times 10^{6}$ Hence Air will breakdown
Subject: Physics — Simple Harmonic Motion — Time period of pendulum
Difficulty: Moderate

Question: Two particle of mass m each are fixed to a massless rod of
length [IMAGE]. The rod is smoothly hinged at one end to
a ceiling. It performs oscillation of small angle in vertical plane. The
length of the equivalent simple pendulum is


Correct option: 3

Solution:
Subject: Physics —  — 
Difficulty: Tough

Question: A shot is fired at an angle θ to the horizontal such that it strikes
the hill while moving horizontally Find initial angle of projection θ.

1) tan θ =[IMAGE]
2) tan θ =[IMAGE]
3) tan θ =[IMAGE]
4) none of these
 (shot) θ 
 

Correct option: 3

Solution:
[IMAGE] sin θ =[IMAGE] 4 sin θ = 8 sin θ - 6 cos θ
4 sin θ = 6 cos θ ⇒ tan θ = 
TOPIC:KINEMATIC
SUB TOPIC: 2 D
Subject: Physics — Electrostatics — Gauss"s law
Difficulty: Moderate

Question: An electric field
$\overset{\rightarrow}{E} = 4 \times \overset{\hat{}}{i} - (y^{2} + 1)jN/C$
passes through the box shown in figure. The flux of the electric field
through surface ABCD and BCGF and marked as $\phi_{I}$ and $\phi_{\Pi}$
respectively. The difference between
$(\phi_{I} - \phi_{\Pi})is\ (in{Nm}^{2/C})$

1) 47
2) 46
3) 42
4) 48

Correct option: 4

Solution:
Flux via ABCD
$\phi_{1} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
Flux via BCEF
$\phi_{2} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
$\phi_{2} = \overset{\rightarrow}{E} \cdot \overset{\rightarrow}{A} = (4 \times \overset{\hat{}}{i} - (y^{2} + 1)\overset{\hat{}}{j}) \cdot 4\overset{\hat{}}{i}$
$= 16x,x = 3$
$\phi_{2} = 48\frac{N - m^{2}}{C};\phi_{1} - \phi_{2} = - 48\frac{N - m^{2}}{C}$
Subject: Physics —  — 
Difficulty: Moderate

Question: A point moves according to the law, x = at and y = at (1- αt),
where a and α are positive constants and t is time. Find the moment at
which angle between velocity vector and acceleration vector

1) α sec
4) Not possible
 x = at y = at (1- αt) a α
 t 
 [IMAGE]

Correct option: 2

Solution:
x = at ⇒ vx = a ⇒ ax = 0
y = at - aαt2 ⇒ vy =[IMAGE] = a - 2αat
ay =[IMAGE] = - 2αa
[IMAGE] = av cos θ
= (2αa) [IMAGE] Solve to get, [IMAGE] TOPIC:KINEMATIC
SUB TOPIC: ACCERATION
Subject: Physics — Work Power & Energy — Work done by a variable force
Difficulty: Moderate

Question: The force exerted by a compression device is given by
$F(x) = kx(x - \lambda)$ for $0 \leq x \leq \lambda$, where $\lambda$
is the maximum possible compression, x is the compression and k is a
constant. The work required to compress the device by a distance d will
be maximum when

1) $d = \lambda/4$
2) $d = \lambda/\sqrt{2}$
3) $d = \lambda/2$
4) $d = \lambda$

Correct option: 4

Solution:
For W to be maximum 
i.e. $F(x) = 0 \Rightarrow x = \lambda,x = 0$
Clearly for $d = \lambda$,the work done is maximum.
Subject: Physics — Wave on string — Phase difference
Difficulty: Moderate

Question: A sinusoidal travelling wave in a string has a velocity of
propagation 300 m/sec. The time period of oscillations of the particles
of the string is 0.04 sec. The phase difference between the oscillations
of two points at distances 10 m and 16 m respectively from source of
oscillation is nπ. Find n.


Answer: 1

Solution:
Subject: Physics — Elasticity — Extension in compound rod
Difficulty: Moderate

Question: Three bars having lengths [IMAGE] and
[IMAGE] and area of cross section A, A/2, 2A are joined
rigidly end toend. The compound rod is subjected to a stretching force
F. It young's modules of the material of eachrod is Y then the increase
in length of the rod is.
 yEckbZ dh o vuqizLFk dkV
{ks=kQyA,[IMAGE], 2Adh rhu NM+s n`<+:i ls fljs ls fljs
NM+ xq.kkadY

1) $\frac{13F\mathcal{l}}{2AY}$
2) $\frac{F\mathcal{l}}{AY}$
3) $\frac{9F\mathcal{l}}{5AY}$
4) $\frac{19F\mathcal{l}}{2AY}$

Correct option: 1

Solution:
$\ \Delta l = \Delta l_{1} + \Delta l_{2} + \Delta l_{3}$
$\frac{F}{\frac{YA}{\mathcal{l}}} + \frac{F}{\frac{YA}{4\mathcal{l}}} + \frac{F}{\frac{2YA}{3\mathcal{l}}} = \frac{F\mathcal{l}}{YA}\left( 1 + 4 + \frac{3}{2} \right) = \frac{13F\mathcal{l}}{2YA}$
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

Question: A non-viscous ideal fluid is flowing vertically downwards in a
pipe. The area of cross-section at section is double that at
section Vertical distance between section A and section B is h
and the height of water column in tube (2) is more than that in tube

1) by $\frac{h}{2}$ distance. If the velocity(m/s) of the fluid at section
A is x, then find the value of x. (if h = 2.7 m)
2) is more than that in tube

Answer: 3

Solution:
Applying Bernoulli's equation between section A and B
$P_{A} + \frac{1}{2}\rho v_{A}^{2} + \rho{gh}_{A} = P_{B} + \frac{1}{2}\rho v_{B}^{2} + \rho{gh}_{B}$
$P_{A} + \frac{1}{2}\rho v^{2} + \rho gh = P_{B} + \frac{1}{2}\rho(2v)^{2} + 0$
$P_{B} - P_{A} = \rho gh - \frac{3}{2}\rho v^{2}\ $
and $P_{B} - P_{A} = \rho g\frac{h}{2}$
$\rho g\frac{h}{2} = \rho gh - \frac{3}{2}\rho v^{2}$
$\Rightarrow v = \sqrt{\frac{gh}{3}} = 3$
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

Question: The emf of cell E_2 is

1) 0.5 V
2) 1 V
3) 1.5 V
4) 2 V

Correct option: 1

Solution:
$i = \frac{2}{15 + 10}$
$V_{AB} = i10 = \frac{20}{25}$
$E_{2} = \frac{V_{AB}}{50} \times 31.25$
$= \frac{20}{25} \times \frac{31.25}{50}$
=0.5V
Subject: Physics — Kinetic Theory of Gases — RMS speed
Difficulty: Moderate

Question: The ratio of r.m.s. speed to the r.ms. angular speed of a
diatomic gas at certain temperature is:(assume m= mass of one molecule
mass I = moment of inertia of the molecules)
ewy ekfu, m =,d v.kq 
vk.kfod nzO;eku I = v.kqvksa

1) $\sqrt{\frac{3}{2}}$
2) $\sqrt{\frac{3I}{2M}}$
3) $\sqrt{\frac{3I}{2m}}$
4) 1

Correct option: 3

Solution:
$\frac{1}{2}{mV}^{2} = \frac{3}{2}kT$
$\frac{1}{2}I\omega^{2} = \frac{2}{2}kT$
$\frac{V}{\omega} = \sqrt{\frac{3I}{2m}}$
Subject: Physics — Electromagnetic induction — Self inductance
Difficulty: Moderate

Question: In a fluorescent lamp choke (a small transformer) 100 V of
reverse voltage is produced when the choke current changes uniformly
from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the
choke(in mH) is estimated to be 2.5n. Find n.


Answer: 4

Solution:
$100 = \frac{L(0.25)}{0.025} \times 10^{3}$
$\therefore L = 100 \times 10^{- 4}H$
$= 10mH$
Subject: Physics — electronic device — Diodes
Difficulty: Easy

Question: In the circuit, if the forward voltage drop for the diode is
[IMAGE] the current in the circuit will be (approximately)

1) 3.4 mA
2) 2 mA
3) 2.5 mA
4) 3 mA

Correct option: 1

Solution:
Subject: Physics —  — 

Question: The frequency of incident light falling on photo sensitive
substance is doubled. If E_1 is the kinetic energy of emitted electron
in first case and E_2 is the kinetic energy in second case, then

1) E_1 = E_2
2) E_2 = 2E_1
3) E_2 < 2E_1
4) E_2 > E_1

Correct option: 4

Solution:
. [IMAGE] Dividing both equations,
[IMAGE] Hence, E_2 > 2E_1
Subject: Physics — Centre of Mass — Momentum
Difficulty: Moderate

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

1) Zero
2) $\frac{2MV}{\pi}$
3) MVR
4) MV

Correct option: 2

Solution:
If we treat the train as a ring of mass then its COM will
be at a distance$\frac{2R}{\pi}$from the centre of thecircle. 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
$= \frac{2MV}{\pi}$.
Subject: Physics — Electronic devices — Diodes
Difficulty: Moderate

Question: In the circuit, if the forward voltage drop for the diode is
0.5V, the current in the circuit will be(approximately) -
ifjiFk fy, vxzfoHko iru 0.5V gS] rc ifjiFk

1) 3.4 mA
2) 2 mA
3) 2.5 mA
4) 3 mA

Correct option: 1

Solution:
Subject: Physics — Centre of Mass — Position of COM
Difficulty: Moderate

Question: A rod of length L has non-uniform linear mass density given by
[IMAGE] where a and b are constants
and[IMAGE]. The value of x for the centre of mass of the
rod is at


Correct option: 2

Solution:
$x_{cm} = \frac{1}{M}\int_{0}^{l}x \cdot dM$
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

Question: A non-viscous ideal fluid is flowing vertically downwards in a
pipe. The area of cross-section at section is double that at
section Vertical distance between section A and section B is h
and the height of water column in tube (2) is more than that in tube

1) by [IMAGE] distance. If the velocity(m/s) of the
fluid at section A is x, then find the value of x. (if h = 2.7 m),d nzo uyh esa uhps dh rjQ çokfgr gSAHkkx A ij
vuqçLFk dkV
chp nwjh h
2) is more than that in tube

Answer: 3

Solution:
Applying Bernoulli's equation between section A and B
ArFkk B
Subject: Physics — Modern physics — Photoelectric effect
Difficulty: Moderate

Question: Radiation, with wavelength falls on a metal surface to
produce photoelectrons. The electronsare made to enter a uniform
magnetic field of $3 \times 10^{- 4}$T. If the radius of the largest
circular pathfollowed by the electrons is 10 mm, the work function of
the metal is close to

1) 1.8 eV
2) 0.8 eV
3) 1.6 eV
4) 1.1 eV

Correct option: 2

Solution:
${KE}_{\max} = E - \phi$
$= \frac{12400}{\lambda(inÅ)} - \phi$ (in eV)
$r = \frac{\sqrt{2mKE}}{eB}$
$KE_{\max} = \frac{r^{2}e^{2}B^{2}}{2m}$ (in J)
$= \frac{r^{2}eB^{2}}{2m}$ (in eV)
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)
$\frac{1}{2} = \frac{- V_{1} + V_{2}}{V_{0}}$
$- V_{1} + V_{2} = \frac{V_{0}}{2}$.....(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 — Mechanical Properties of Fluids — Fluid in motion
Difficulty: Moderate

Question: A container is filled with a liquid of density ρ. Container is
accelerating on a horizontal surface with acceleration $\frac{3g}{4}$
towards right hand side and liquid is at rest with respect to container
as shown in figure. If there is a point A in liquid then, which of the
following is correct: (assume atmospheric pressure is zero and AB is
vertical and AC is horizontal line)

1) Pressure at A is $\rho g\mathcal{l}_{1}$
2) Pressure at A is $\rho\frac{3g}{4}\mathcal{l}_{2}$
3) $\theta = \tan^{- 1}\left( \frac{3}{4} \right)$
4) All of the above

Correct option: 4

Solution:
for point A from top surface
$P_{A} = \rho g \cdot \mathcal{l}_{1}$and from right wall
$P_{A} = \frac{\rho^{3}g}{4} \cdot \mathcal{l}_{2}$
$\mathcal{l}_{1} = \frac{3}{4}\mathcal{l}_{2}$ and
Subject: Physics — Thermodynamics — Internal energy
Difficulty: Moderate

Question: [IMAGE] gas follows the process,
[IMAGE] constant. If then the ratio of work done by gas
to change in itsinternal energy is [IMAGE] then n is:
vuqikr [IMAGE] gks rks n


Answer: 4

Solution: