Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

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

1) $60\Omega$
2) $120\Omega$
3) $100\Omega$
4) $80\Omega$

Correct option: 3

Solution:
[IMAGE] Potential gradient
$= \frac{5}{1000} = \frac{V_{p}}{1200}$
$V_{P} = 6V$
And $R_{P} = \frac{V_{P}}{I} = \frac{6}{60 \times 10^{- 3}} = 100\Omega$
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

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

3) $x_{1}:x_{2}:x_{3} = \sqrt{3}:\sqrt{6}:\sqrt{5}$
4) $x_{1}:x_{2}:x_{3} = \sqrt{2}:\sqrt{5}:\sqrt{3}$

Correct option: 1

Solution:
For case-1: fLFkfr-1 ds fy, [IMAGE] For case-2: fLFkfr-2 ds fy, [IMAGE] For case-3: fLFkfr-3
Subject: Physics — Mechanical Properties of Fluids — Velocity of efflux
Difficulty: Moderate

Question: The speed of water flowing out of orifice after the cylinder is
kept inside it

1) $\ \sqrt{3gH}$
2) $\ \sqrt{2gH}$
3) $\ \sqrt{\frac{3gH}{2}}$
4) $\ \sqrt{\frac{gH}{2}}$

Correct option: 3

Solution:
[IMAGE] Let h′ be height of water column just after putting cylinder,
$\ \rho h'\left( A - \frac{A}{3} \right) = \left( \frac{H}{2} \right)A\rho$
$\ \Rightarrow h^{'} = \frac{3}{4}H$
$\ V' = \sqrt{2{gh}^{'}} = \sqrt{\frac{3}{2}gH}$
Subject: Physics — Capacitance — KVL
Difficulty: Moderate

Question: In the figure shown $C_{1} = 11\ \mu F$ and $C_{2} = 5\ \mu F$
then at steady state

1) the potential difference across C_1 is 5V.
2) the potential difference across C_2 is 2V.
3) the potential difference$\ V_{a} - V_{b} = - 4V$.
4) the potential difference between terminals of 15V battery is 9V.

Correct option: 1

Solution:
[IMAGE] at steady state
$I(3) + 1(2) = 15$
$I = 3$
kvL
$C \rightarrow D \rightarrow E \rightarrow a \rightarrow b \rightarrow C$
$V_{c} - I(3) + \frac{q}{11} - 7 + \frac{q}{5} = V_{c}$
$\frac{q}{11} + \frac{q}{5} = 7 + 3 \times 3 = 16$
$q = 55\mu c$
Now kvL for $a \rightarrow b$
$v_{a} - 7 + \frac{q}{5} = v_{b}$
$\Rightarrow V_{a} - V_{b} = 7 - \frac{q}{5}$
$= 7 - \frac{55}{5}$
$= - \frac{20}{5} = - 4V$
Subject: Physics —  — 
Difficulty: Moderate

Question: A solid sphere of mass 10 kg is placed over two smooth inclined
planes as shown in figure. The normal reactions at 2 is 10x Newton. Find


Answer: 5

Solution:
[IMAGE] Solving above equation
[IMAGE] =10 x
x = 5
TOPIC:Newton's first Law
SUB TOPIC:Free body diagram
Subject: Physics — Behaviour of perfect gases — Ideal Gas
Difficulty: Moderate

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

1) The pressure of gas changes to [IMAGE]
2) The temperature of gas changes to [IMAGE].
3) The graph of above process on the P - T diagram is parabola.
4) The graph of above process on the P - T diagram is not rectangular
hyperbola.

Correct option: 2

Solution:
PT = constant
Subject: Physics — Rigid Body Dynamics — Angular velocity of rigid body
Difficulty: Easy

Question: When a person throws a meter stick it is found that the centre
of the stick is moving with speed 10 m/s and left end of stick with
speed 20m/s. Both points move vertically upwards at that moment. Then
angular speed of the stick is

4) None of these

Correct option: 1

Solution:
Angular
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Moderate

Question: A uniform disc of radius R lies in the x-y plane, with its centre
at origin. Its moment of inertia about z-axis is equal to its moment of
inertia about line y = x + c. The value of c will be

4) - R

Correct option: 2

Solution:
$I_{PQR} = \frac{1}{4}{MR}^{2} + M\left( \frac{C}{\sqrt{2}} \right)^{2}$.
But $I_{PQR} = \frac{1}{2}{MR}^{2}$
$C = \pm \frac{R}{\sqrt{2}}$.
Hence is (B) is correct.
Subject: Physics — EM wave — Lorentz force on a charged particle due to EM wave.
Difficulty: Tough

Question: The electric fields of two plane electromagnetic plane waves in
vacuum are given by:
${\overset{\rightarrow}{E}}_{1} = E_{0}\overset{\hat{}}{j}cos(\omega t - kx)\ and\ {\overset{\rightarrow}{E}}_{2} = E_{0}\overset{\hat{}}{k}cos(\omega t - ky)$
At t = 0, a particle of charge q is at origin with a velocity
j$\overset{\rightarrow}{v} = 0.8c\overset{\hat{}}{j}$(c is the speed
of the light in vacuum). The instantaneous force experienced by the
particle is

1) $E_{0}q(0.8\overset{\hat{}}{i} + \overset{\hat{}}{j} + 0.2\overset{\hat{}}{k})$
2) $E_{0}q( - 0.8\overset{\hat{}}{i} + \overset{\hat{}}{j} + \overset{\hat{}}{k})$
3) $E_{0}q(0.4\overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 0.8\overset{\hat{}}{k})$
4) $E_{0}q(0.8\overset{\hat{}}{i} - \overset{\hat{}}{j} + 0.4\overset{\hat{}}{k})\ $

Correct option: 1

Solution:
Magnetic field vectors associated with this electromagnetic
wave are given by
${\overset{\rightarrow}{B}}_{1} = \frac{E_{0}}{c}\overset{\hat{}}{k}cos(kx - \omega t)\&{\overset{\rightarrow}{B}}_{2} = \frac{E_{0}}{c}\overset{\hat{}}{i}cos(ky - \omega t)$
$\overset{\rightarrow}{F} = q\overset{\rightarrow}{E} + q\left( \overset{\rightarrow}{V} \times \overset{\rightarrow}{B} \right)$
$= q\left( {\overset{\rightarrow}{E}}_{1} + {\overset{\rightarrow}{E}}_{2} \right) + q\left( \overset{\rightarrow}{V} \times \left( {\overset{\rightarrow}{B}}_{1} + {\overset{\rightarrow}{B}}_{2} \right) \right)$
By putting the value of
${\overset{\rightarrow}{E}}_{1},{\overset{\rightarrow}{E}}_{2},{\overset{\rightarrow}{B}}_{1}\&{\overset{\rightarrow}{B}}_{2}$
The net Lorentz force on the charged particle is
$\overset{\rightarrow}{F} = {qE}_{0}\lbrack 0.8cos(kx - \omega t)\overset{\hat{}}{i} + cos(kx - \omega t)\overset{\hat{}}{j} + 0.2cos(ky - \omega t)\overset{\hat{}}{k}\rbrack$
At t = 0 and at x = y = 0
$\overset{\rightarrow}{F} = qE_{0}\lbrack 0.8\overset{\hat{}}{i} + \overset{\hat{}}{j} + 0.2\overset{\hat{}}{k}\rbrack$
Subject: Physics — Relative motion — Shortest distance
Difficulty: Hard

Question: Two particle A and B are located in x-y plane at points (0, 0)
and (0, 4 m). They simultaneously start moving with velocities.
${\overset{\rightarrow}{v}}_{A} = 3\overset{\hat{}}{j}m/s$and
${\overset{\rightarrow}{v}}_{B} = 4\overset{\hat{}}{i}m/s$. The shortest
distance between them will be

4) zero

Correct option: 1

Solution:
Assuming B to be at rest, A will move with velocity
${\overset{\rightarrow}{v}}_{AB}$ in the direction shown in figure. The
distance between them will first decrease from A to Cand then increase
beyond C.
[IMAGE] Minimum distance between them is BC which is equal to
$4\ sin53^{\circ} = 16/\mathbf{5}\mathbf{m}$
Subject: Physics — Atomic Physics — Bohr's model of hydrogen atom
Difficulty: Tough

Question: Suppose the potential energy between electron and proton at a
distance r is given by [IMAGE] where k is constant, e is
charge of proton and r is radius of orbit. Application of Bohr's theory
to hydrogen atom in this case shows that

1) energy of electron in [IMAGE] orbit is proportional
2) energy of electron in [IMAGE] orbit is proportional
to [IMAGE] mass of electron).
3) energy of electron in [IMAGE] orbit is proportional
4) energy of electron in [IMAGE] orbit is proportional

Correct option: 1

Solution:
[IMAGE] From (1) and (2)
Subject: Physics — Collision — Inelastic collision
Difficulty: Moderate

Question: The dependence of acceleration due to gravity g on the distance r
from the centre of the earthassumed to be a sphere of radius R of
uniformdensity is as shown in figure below:-
[IMAGE] The correct figure is

1) (i)
2) (ii)
3) (iii)
4) (iv)

Correct option: 4

Solution:
The acceleration due to gravity at a depth d below the surface
of earth is
Where R-d=r= distance of location from thecentre of the earth. When
[IMAGE] From [IMAGE] till R=r, for which
[IMAGE] For [IMAGE] or [IMAGE] Here, R+h=r Therefore, the variation of g with distance r from centre
of earth will be as shown in figure(4). Thus, option (4) is correct
Subject: Physics — Work Power & Energy — Conservation of mechanical energy
Difficulty: Easy

Question: A block of mass 1 kg initially at rest is dropped from a height
h on the a spring of force constant k=50N/m. The maximum compression in
the spring is x=1 m then height hof the block is

1) 3 m
2) $\frac{3}{2}$ m
3) $\frac{5}{2}$ m
4) 5 m

Correct option: 2

Solution:
Change in gravitational potential energy
= Elastic potential energy stored in-compressed spring
$\Rightarrow mg(h + x) = \frac{1}{2}kx^{2}$
$\Rightarrow 1 \times 10(h + 1) = \frac{1}{2} \times 50 \times 1^{2}$
$\Rightarrow h = \frac{3}{2}$
Subject: Physics — Newton's Laws of Motion — Static friction
Difficulty: Tough

Question: A block of mass 10 kg is to be pulled on a horizontal rough
surface (µ =0.75) the minimum magnitude of applied force is

1) 75 N
2) 80 N
3) 60 N
4) None of these

Correct option: 3

Solution:
$Fcos\theta - \mu N = 0$
$N = mg - {Fsin}\theta\ldots\ldots(2)$
From (1) and (2)
${Fcos}\theta = \mu\left( mg - {Fsin}\theta \right)$
$K = \frac{\mu mg}{cos\theta + \mu sin\theta}$
$F_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^{2}}}\left( \because y = Asin\theta + Bcos\theta;ymax = \sqrt{A^{2} + B^{2}} \right)$
$F_{\min} = \frac{0.75 \times 10 \times 10}{\sqrt{1 + (0.75)^{2}}}$
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Easy

Question: The internal energy of a system in adiabatic expansion

1) Decreases
2) Increases
3) Remains unchanged
4) Nothing can be said

Correct option: 2

Solution:
Adiabatic expansion is adiabatic cooling.
$\because\ $temp. Decreases $\Rightarrow$ internal energy decreases
Subject: Physics — Motion in 1D — Non-uniform acceleration
Difficulty: Moderate

Question: Position of a particle moving on x-axis as a function of time t
is given by
$x = \left( t^{3} - 6t^{2} + 3t + 4 \right)$meter.
What is the velocity of the particle when its acceleration is zero?

1) $- 12m/s$
2) $- 6m/s$
3) $- 3m/s$
4) $- 9m/s$

Correct option: 4

Solution:
$v = 3t^{2} - 12t + 3$
$a = 6t - 12 = 0$
$\Rightarrow t = 2\sec$
$v(t = 2sec) = - 9m/s$
Subject: Physics — Work Power & Energy — Work -energy theorem
Difficulty: Moderate

Question: A body of mass 6 kg is acted upon by a force which causes a
displacement in it given by $x = \frac{t^{2}}{4}$ metre where t is the
time in second. The work done by the force is 2 seconds is

1) 12 J
2) 9 J
3) 6 J
4) 3 J

Correct option: 4

Solution:
The velocity of the body a time t is given by
$= \frac{dx}{dt} = \frac{d}{dt}(\frac{t^{2}}{4}) = \frac{t}{2}$
At $t = 0, = u = 0\ and\ t = 2s, = 1{ms}^{- 1}$, Now, work done =
increase in KE
$= \frac{1}{2}m - \frac{1}{2}mu^{2} = \frac{1}{2}m - 0$
$= \frac{1}{2}m = \frac{1}{2} \times 6 \times (1)^{2} = 3J$
Subject: Physics — Thermodynamics — Carnot engine
Difficulty: Moderate

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


Answer: 6

Solution:
Subject: Physics —  — 

Question: A particle moving with initial velocity 3 m/s along x-axis
from origin. Its acceleration is varying with x as
[IMAGE] as shown in figure. At
[IMAGE] tangent to the graph makes an angle 60° with
positive x-axis as shown in diagram. Then at [IMAGE]


Correct option: 1

Solution:
TOPIC: Rectilinear motion
SUB TOPIC: Acceleration-displacement graph
LEVEL: Moderate
Subject: Physics — Center of mass — Centripetal force
Difficulty: Easy

Question: You are shown a photo of a car driven on the inside wall of a
huge vertical cylinder of radius 50 m. The coefficient of static
friction between the car tyres and the cylinder is $\mu_{s} = 0.8$. The
minimum speed in (m/s), at which the car can be driven like that is x.
Find x.

1) 23
2) 26
3) 25
4) 22

Correct option: 3

Solution:
F. B. D. Of car:-
[IMAGE] $\mu N = mg$....(1)
$N = \frac{mv^{2}}{R}$....(2)
From (1) and (2)
$V = 25m/s$
Subject: Physics — Wave on a string — Power delivered to the string
Difficulty: Moderate

Question: Sinusoidal waves 5.00 cm in amplitude are to be transmitted along
a string having a linear mass density [IMAGE]. If the
source can deliver an average power of 90 W and the string is under a
tension of 100 N, then the frequency at which the source can operate is
(take [IMAGE] )

1) 45.3 Hz
2) 50 Hz
3) 30 Hz
4) 62.3 Hz

Correct option: 3

Solution:
[IMAGE] using data f = 30 Hz
Subject: Physics — Work Power & Energy — Conservation of mechanical energy
Difficulty: Moderate

Question: A frictionless track ABCDE ends in a circular loop of radius R. A
body slides down the track from point A which is at height $h = 82cm$.
The maximum value of R for a body to complete the loop successfully is

1) 22.8 cm
2) 32.8 cm
3) 12.8 cm
4) 42.8 cm

Correct option: 2

Solution:
Using energy conservation:
$\frac{1}{2}mv_{B}^{2} = mgh$
$V_{B} = \sqrt{\frac{2mgh}{m}}$
$v_{B} = \sqrt{2hg}\ \ldots(1)$
Also to complete vertical circle $v_{B} = \sqrt{5gR}$...(2)
$\therefore R = \frac{2}{5}h = 32.8cm$
Subject: Physics — Kinetic Theory of Gases — Gases in equilibrium
Difficulty: Tough

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$\frac{\alpha}{48}$.


Answer: 4

Solution:
At equilibrium, pressure and temperature are same.
$\therefore P_{1} = P_{2}$ and $T_{1} = T_{2}$
Also, let cross-sectional area of tube be A and radius of ring be r.
$\therefore\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$
$\Rightarrow \frac{360 - \alpha}{n_{1}} = \frac{\alpha}{n_{2}}$
$\left( \therefore V_{1} = (360 - \alpha)rA \right.\ $ and $V_{2} = \alpha rA$
$\Rightarrow M_{1}(360 - \alpha) = M_{2}\alpha\lbrack\therefore n_{1} = \frac{m}{M_{1}}andn_{2} = \frac{m}{M_{2}}\rbrack$
$\alpha = \frac{360M_{1}}{M_{1} + M_{2}} = 192^{\circ}$
Subject: Physics — Work Power & Energy — Work done by variable force
Difficulty: Moderate

Question: A particle is moved under a force
$\overset{\rightarrow}{F} = x\overset{\hat{}}{j}N$(x in meter) from O to
A by two paths: path 1 is OBA andpath 2 is straight line OA. If w_1 and
w_2 are work done by the force $\overset{\rightarrow}{F}$ along the two
paths respectivelythe value of $\frac{w_{1}}{w_{2}}$is

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

Correct option: 1

Solution:
work done in path 2
$W_{2} = \int\overset{\rightarrow}{F} \cdot (dx\overset{\hat{}}{i} + dy\overset{\hat{}}{j}) = \int x\overset{\hat{}}{j} \cdot (dx\overset{\hat{}}{i} + dy\overset{\hat{}}{j})$as
the line: y = x
$= \int_{0}^{1}{xdx} = \frac{1}{2}J$
work done in path 1
$w_{OB} = 0$ (force is zero)
$W_{1} = \int\overset{\rightarrow}{F} \cdot dx\overset{\hat{}}{i} + \int\overset{\rightarrow}{F} \cdot dy\overset{\hat{}}{j}$
$= 0 + \int_{y = 0}^{1}{(1\overset{\hat{}}{j}) \cdot dy\overset{\hat{}}{j} = 1J}$
$\frac{w_{1}}{w_{2}} = 2$
Subject: Physics —  — 

Question: In an [IMAGE] the Kinetic energy of
[IMAGE] is 48 MeV and Q-value of the reaction is 50 MeV.
If the mass number of the mother nucleus is 100N. Find N: (Assume that
daughter nucleus is in ground state)


Answer: 1

Solution:
We have
[IMAGE] TOPIC:Modern physics
SUB TOPIC:Alpha decay
LEVEL: Moderate
Subject: Physics — Wave on a string — Speed wave
Difficulty: Moderate

Question: The particle displacement (in cm) in a stationary wave is given
by [IMAGE] Find the distance (in cm) between a node and
the next antinode.

1) 18.75 cm
2) 36.5 cm
3) 9.375 cm
4) 73 cm

Correct option: 1

Solution:
[IMAGE] compare with [IMAGE] λ = 75 cm
$\frac{\lambda}{4} = 18.75cm$
Subject: Physics — Current Electricity — Wheatstone bridge
Difficulty: Moderate

Question: Four resistances of [IMAGE] and
[IMAGE] respectively in cyclic order to form Wheat
stone's network. The resistance (in ohms) that is to be connected in
parallel with the resistance of [IMAGE] to balance the
network is2n. Find n.


Answer: 5

Solution:
on solving
Subject: Physics — Thermal properties of matter — Thermal expansion
Difficulty: Moderate

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


Answer: 6

Solution:
$V = 2\alpha_{2} + \alpha_{1}$
$= 10 \times 10^{- 6} + 5 \times 10^{- 5}$
$= 60 \times 10^{- 6}/\ ^{\circ}C$
Subject: Physics — Semiconductors — Diodes in circuit
Difficulty: Moderate

Question: Both the diodes used in the circuit shown are assumed to be
ideal and have negligible resistance whenthese are forward biased. built
in potential in each diode is 0.7 V. For the input voltages shown in
thefigure, the voltage (in volts) at point A is 6n. Find


Answer: 2

Solution:
Right diode is reversed biased and left diode is forward biased
$\therefore V_{E} = 12.7 - 0.7 = 12Volt$
Subject: Physics — Electronic devices — Gates
Difficulty: Moderate

Question: Which of the following gives a reversible operation ?


Correct option: 4

Solution:
A logic gate is reversible if we can recover input data from
the outputeg. NOT gate,d rdZ gksxk ladsr
izkIr fd;k tk ldsA
eg. NOT }kj
Subject: Physics — Mechanical Properties of Fluids — Velocity of efflux
Difficulty: Moderate

Question: The speed of water flowing out of orifice after the cylinder is
kept inside it
csyu dks VSd ckn fNnz


Correct option: 3

Solution:
[IMAGE] Let h' be height of water column just after putting cylinder,
ekuk csyu
Subject: Physics — Thermodynamics — Carnot engine
Difficulty: Moderate

Question: An engine has an efficiency of 0.25 when temperature of sink is
reduced by58ºC, if ts efficiency is doubled, then the temperature of the
source in °C is 29n. Find n.


Answer: 8

Solution:
Here,$\eta_{1} = 1 - \frac{T_{2}}{T_{1}}$
or $\ 0.25 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{1}{4} = 1 - \frac{T_{2}}{T_{1}}$
$\Rightarrow \frac{T_{2}}{T_{1}} = 1 - \frac{1}{4} = \frac{3}{4}$.....(i)
According to question,
$\eta_{2} = 2\eta_{1},$ and $T_{2} = T_{2} - 58^{\circ}C$
$\therefore 2 \times \frac{1}{4} = 1 - \frac{\left( T_{2} - 58^{\circ}C \right)}{T_{1}}$
$\Rightarrow 1 - \frac{1}{2} = \frac{T_{2} - 58^{\circ}C}{T_{1}}$
$\frac{1}{2} = \frac{T_{2}}{T_{1}} - \frac{58^{\circ}}{T_{1}} \Rightarrow \frac{3}{4} - \frac{1}{2} = \frac{58}{T_{1}}$
$\Rightarrow T_{1} = 232^{\circ}C$
Subject: Physics — Newton's Laws of Motion — Equilibrium of forces
Difficulty: Moderate

Question: A uniform sphere of weight w and radius 3 m is being held by a
string of length 2 m. attached to a frictionless wall as shown in the
figure. The tension in the string is $\frac{5w}{n}$ then find the value
of n.


Answer: 4

Solution:
[IMAGE] From geometry:
$\cos\theta = \frac{3}{5}$
$\sin\theta = \frac{4}{5}$
As sphere is at equilibrium
$T\sin\theta = w$
$T\left( \frac{4}{5} \right) = w$
$T = \frac{5w}{4}$
Subject: Physics — EMI — Lenz's law
Difficulty: Moderate

Question: A conducting circular loop is pulled with a constant velocity v
towards a region of uniform magnetic field of induction B as shown in
figure then current induced in the loop is (d > 2r).

1) clockwise while entering.
2) anti clockwise while entering.
3) zero when completely inside.
4) clockwise while leaving.

Correct option: 2

Solution:
By lenz's law induced current will oppose change in flux.
Subject: Physics — Newton's Laws of Motion — Impulse
Difficulty: Moderate

Question: The particle of mass 1 kg is acted upon by a
force$\overset{\rightarrow}{F} = F_{x}\overset{\hat{}}{i} + F_{y}\overset{\hat{}}{j}$
from t = 0 to t = 4 sec. Find the velocityof the particle at t=4 sec.,
if its initial velocity is
$\overset{\rightarrow}{u} = - \overset{\hat{}}{i} + \overset{\hat{}}{j}$and
the variation of applied force with time is as shown in the graphs

1) $\overset{\rightarrow}{v} = - \overset{\hat{}}{i} + \overset{\hat{}}{j}$
2) $\overset{\rightarrow}{v} = - \overset{\hat{}}{i} + 11\overset{\hat{}}{j}$
3) $\overset{\rightarrow}{v} = 10\overset{\hat{}}{i}$
4) $\overset{\rightarrow}{v} = 2\overset{\hat{}}{i} + 10\overset{\hat{}}{j}$

Correct option: 2

Solution:
$\int_{0}^{4}{F_{x}dt = \Delta P_{x} = m(v_{x} - u_{x})}$
$\Rightarrow 0 = m(v_{x} - u_{x})$.
$\Rightarrow V_{x} = u_{x} = - 1$
$\int_{0}^{4}{F_{y}dt = \Delta P_{y} = m(v_{y} - u_{y})}$
$\Rightarrow \frac{1}{2} \times 4 \times 5 = 1(v_{y} - 1) \Rightarrow v_{y} = 11$
$\overset{\rightarrow}{v} = - \overset{\hat{}}{i} + 11\overset{\hat{}}{j}$
Subject: Physics —  — 
Difficulty: Moderate

Question: A projectile is projected at an angle of 37° to the horizontal. When
it is projected at 53°, its time of flight on level ground would have
been 1 sec. more. At what angle should it be projected so that its range
is 10 m less than either of the situations ?

1) 26.5°
2) 30°
3) 18°
4) 15°
 37° 53° 
 1 
 10 m ?

Correct option: 1

Solution:
[IMAGE] T1 =[IMAGE] (sin 53° - sin 37°) = 1
R =[IMAGE] = 60 m
R' = 50 =[IMAGE] = sin 2θ
2θ = 53°
TOPIC:KINEMATIC
SUB TOPIC: 2 D
Subject: Physics — Wave Optics — Diffraction
Difficulty: Moderate

Question: Light of wavelength 600 mm is incident upon a single slit with
width[IMAGE]. The figure shows the pattern observed on a
screen positioned 2 m from the slits. Determine the distance s (in cm).


Answer: 6

Solution:
Subject: Physics — Capacitance — Equivalent capacitance
Difficulty: Easy

Question: Effective capacitance of parallel combination of two capacitors
C1 and C2 is 10$\mu F$. When these capacitorsare individually connected
to a voltage source of 1V, the energy stored in the capacitor C_2 is 4
times thatof C_1. If these capacitors are connected in series, their
effective capacitance will be

1) $4.2\mu F$
2) $8.4\mu F$
3) $3.2\mu F$
4) $1.6\mu F$

Correct option: 4

Solution:
Given C_1 + C_2$= 10\mu F$....(i)
$4\left( \frac{1}{2}C_{1}V^{2} \right) = \frac{1}{2}C_{2}V^{2}$
$\Rightarrow$ 4C_1=C_2....(ii)
from equation (i) and (ii)
$C_{1} = 2\mu F$
$C_{2} = 8\mu F$
If they are in series
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = 1.6\mu F$
Subject: Physics — Capacitance — Equivalent capacitance
Difficulty: Tough

Question: [IMAGE] A parallel plate capacitor has plates of area A separated by distance
 between them. It is filled with adielectric which has a
dielectric constant that varies as k(x) = K(1 + $\alpha$x) where 
is the distancemeasured from one of the plates. If ($\alpha$d) << 1,
the total capacitance of the system is best given by theexpression

1) $\frac{A \in_{0}K}{d}\left( 1 + \frac{\alpha^{2}d^{2}}{2} \right)b$
2) $\frac{AK \in_{0}}{d}\left( 1 + \frac{\alpha d}{2} \right)$
3) $\frac{A \in_{0}K}{d}\left( 1 + \left( \frac{\alpha d}{2} \right)^{2} \right)$
4) $\frac{AK\epsilon_{0}}{d}(1 + \alpha d)$

Correct option: 2

Solution:
Capacitance of element$= \frac{k\varepsilon_{0}A}{dx}$
[IMAGE] Capacitance of
element$C^{'} = \frac{K(1 + \alpha x)\varepsilon_{0}A}{dx}$
$\sum\frac{1}{C^{'}} = \int_{0}^{d}\mspace{2mu}\frac{dx}{K\varepsilon_{0}A(1 + \alpha x)}$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\mathcal{l}n(1 + \alpha d)$
 Given $- \ \alpha d < < 1$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\left( \alpha d - \frac{\alpha^{2}d^{2}}{2} \right)$
$\frac{1}{C} = \frac{d}{K\varepsilon_{0}A}\left( 1 - \frac{\alpha d}{2} \right)$
$C = \frac{K\varepsilon_{0}A}{d}\left( 1 + \frac{\alpha d}{2} \right)$