Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Moving charges and — 
Difficulty: Moderate

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

1) $\frac{e\mu_{0}nIR}{2m}$
2) $\frac{e\mu_{0}nIR}{m}$
3) $\frac{2e\mu_{0}nIR}{m}$
4) $\frac{e\mu_{0}nIR}{4m}$

Correct option: 1

Solution:
$R_{\max} = \frac{R}{2} = \frac{mv_{\max}}{e\mu_{0}in}$
$V_{\max} = \frac{Re\mu_{0}in}{2m}$
Subject: Physics — Thermodynamics — Carnot engine
Difficulty: Moderate

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


Answer: 9

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

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


Answer: 9

Solution:
Hysteresis loss = 250 cycle
No. of cycle per hour = 50 ×60 = 3000
Hence total loss per hour = 250 × (volume) × (cycles)
$= 250 \times \frac{3.6}{7.5 \times 1000} \times 3000$
= 360 WH = 0.36 kWH
Subject: Physics — EM wave — Magnetic field form of the EM wave.
Difficulty: Moderate

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

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

Correct option: 3

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

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

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

Correct option: 3

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

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

4) None of these

Correct option: 1

Solution:
$\frac{dN}{dt} = t^{2} - \lambda N$
Subject: Physics — Centre of Mass — Position of COM
Difficulty: Moderate

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

4) none of these

Correct option: 2

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

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

1) $\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$
2) $\frac{\varepsilon_{0}a^{2}}{d}(1 + \frac{\alpha a}{2d})$
3) $\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{4d})$
4) $\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{3\alpha a}{2d})$

Correct option: 1

Solution:
$\Rightarrow c = \frac{\varepsilon_{0}a}{\alpha}\lbrack ln(d + \alpha x)\rbrack_{0}^{a}$
$= \frac{\varepsilon_{0}a}{\alpha}In(1 + \frac{\alpha a}{d}) \approx \frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$
Subject: Physics — Newton's Laws of Motion — Inclined plane with friction
Difficulty: Moderate

Question: Starting from rest, a body slides down a $45^{\circ}$ inclined
plane in twice the time it takes to slide down the same distance in the
absence of friction. The coefficient of friction between the body and
the inclined plane is

1) 0.33
2) 0.25
3) 0.75
4) 0.80

Correct option: 3

Solution:
$\mu = tan\theta\left( 1 - \frac{1}{n^{2}} \right)$
$\theta = 45^{\circ}$ and $n = 2$ (Given)
$\therefore\mu = tan45^{\circ}\left( 1 - \frac{1}{2^{2}} \right) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$
Subject: Physics — Thermodynamics — Thermodynamic process
Difficulty: Tough

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

1) $\Delta W = 0$
2) $\Delta Q = 1.08 \times 10^{5}J$
3) $\Delta U = 0$
4) $\Delta Q > \Delta W$

Correct option: 2

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

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

1) $10m/s$
2) $20m/s$
3) $15m/s$
4) $30m/s$

Correct option: 1

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

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

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

Correct option: 4

Solution:
Flux via ABCD ls ¶yDl
$\phi_{1} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
[IMAGE] Flux via BCEF ls ¶yDl
$= 16x,x = 3$
$\phi_{2} = 48\frac{N - m^{2}}{C};\phi_{2} - \phi_{1} = 48\frac{N - m^{2}}{C}$
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

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

3) 3V/4L counter clockwise
4) V/L counter clockwise

Correct option: 3

Solution:
Cons. of ang. momentum about P gives
$MV\frac{L}{2} = \frac{(2M)(2L)^{2}}{12}w$
$\Rightarrow \frac{V}{2} = \frac{2Lw}{3}$
$w = \frac{3V}{4L}$
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

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


Answer: 90

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

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

1) $- \left( \frac{Q}{2} \right)$
2) $\left( \frac{Q}{2} \right)$
3) Q
4) - Q

Correct option: 3

Solution:
At point D,[IMAGE] must be zero because potential
of left most plate is zero.
[IMAGE] So, vr x = 0
[IMAGE] (conductor pkyd)
x = y + Q
y = -Q
Subject: Physics — Rigid Body Dynamics — Torque
Difficulty: Moderate

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


Answer: 2

Solution:
${mgR}\left( 1 - \frac{2}{\pi} \right) = I\alpha$
$I = {mR}^{2} - m\left( \frac{2R}{\pi} \right)^{2} + m\left( R - \frac{2R}{\pi} \right)^{2}$
$I = 2{mR}^{2}\left( 1 - \frac{2}{\pi} \right)$
${mgR}\left( 1 - \frac{2}{\pi} \right) = 2\ m\ R^{2}\left( 1 - \frac{2}{\pi} \right)\alpha$
$\Rightarrow \alpha = \frac{g}{2R} = \frac{10}{2 \times 0.4} = 12.50rad/\sec^{2}$
Subject: Physics — Ray Optics — Optical instruments
Difficulty: Moderate

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


Answer: 5

Solution:
Subject: Physics — Rigid Body Dynamics — Torque
Difficulty: Moderate

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

1) The angular acceleration of the system is [IMAGE]
2) The angular acceleration of the system is [IMAGE]
3) The Tension in the string is [IMAGE]
4) The Tension in the string is [IMAGE]

Correct option: 3

Solution:
$\frac{mgl}{2} = \left( \frac{2ml^{2}}{12} + \frac{{ml}^{2}}{4} \right)\alpha$
$\alpha = \frac{6\text{g}}{5l}$
$\text{a}_{\text{cm}} = \frac{\alpha l}{6} = \frac{\text{g}}{5}$
Subject: Physics — Wave Optics — Diffraction
Difficulty: Moderate

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

1) $2t_{2}\lambda + t_{1}$
2) $\frac{2t_{2}\lambda}{t_{1}}$
3) $\frac{2t_{2}\lambda}{t_{1}} + t_{1}$
4) $\frac{2t_{2}\lambda}{t_{1}} - t_{1}$

Correct option: 3

Solution:
$t_{1}sin\theta = \lambda$
$\theta = \lambda/t_{1}$
Width of central Max. $y = 2t_{2}\theta + t_{1}$
$y = \frac{2t_{2}\lambda}{t_{1}} + t_{1}$
Subject: Physics — Capacitance — Energy stored in capacitor
Difficulty: Moderate

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

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

Correct option: 4

Solution:
[IMAGE] V_0 = 20 V
Heat loss gkfu = U_i - U_f
$= \frac{1}{2}CV_{0}^{2} - 2\left\lbrack \frac{1}{2}C\left( \frac{V_{0}}{2} \right)^{2} \right\rbrack$
$= \frac{CV_{0}^{2}}{4}$
$= \frac{\left( 60 \times 10^{- 12} \right)(20)^{2}}{4}J$
= 6 × 10^-9 J = 6 nJ
Subject: Physics — Capacitance — Distribution of charge on plates
Difficulty: Easy

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

1) $- \left( \frac{Q}{2} \right)$
2) $\left( \frac{Q}{2} \right)$
3) Q
4) - Q

Correct option: 2

Solution:
${\overrightarrow{E}}_{A} = 0$ (Conductor)
-x = x + Q + y - y
2x = - Q
$x = \frac{- Q}{2}$
$- x = \frac{Q}{2}$
Subject: Physics — Magnetic effect of current — Magnetic flux
Difficulty: Moderate

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

1) $\phi > \phi_{0}$
2) $\phi_{1} < \phi_{0}$
3) $\phi_{i} = - \phi_{0}$
4) $\phi_{1} = \phi_{0}$

Correct option: 3

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

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

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

Correct option: 4

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

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

1) Both statement (i) and (ii) are correct
2) Statement (i) is correct while statement (ii) is incorrect
3) Both statement (i) and (ii) are incorrect
4) Statement (i) is incorrect while statement (ii) is correct

Correct option: 3

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

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


Correct option: 1

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

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


Answer: 2

Solution:
Let [IMAGE] Right diode is reversed biased and left diode is forward biased
Subject: Physics — Newton's Laws of Motion — Friction
Difficulty: Moderate

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

1) 0.1
2) 0.2
3) 0.5
4) 1

Correct option: 2

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

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

Answer: 8

Solution:
and [IMAGE].....(2)
[IMAGE] Now TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate
Subject: Physics — Work Power & Energy — Work done by a variable force
Difficulty: Moderate

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

1) $- \mu mg\mathcal{l}$
2) $- 2\mu mg\mathcal{l}$
3) $- 3\mu mg\mathcal{l}$

Correct option: 3

Solution:
Work done by
friction$= - \int\mu mgcos\theta d\mathcal{l = - \int}\mu mgdx = - 3\mu mg\mathcal{l}$
Subject: Physics — Magnetic effect of current — Magnetic field due to current carrying conductor
Difficulty: Moderate

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

1) $\frac{\mu_{0}I}{8R} \otimes$
2) $\frac{3\mu_{0}I}{11R} \otimes$
3) $\frac{\mu_{0}I}{16R} \odot$
4) Zero

Correct option: 1

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

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


Correct option: 2

Solution:
Subject: Physics — Behaviour of perfect gases — Work done
Difficulty: Moderate

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


Answer: 1

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
Subject: Physics — Heat — Stefan's law of radiation
Difficulty: Tough

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

1) $+ \frac{\lambda_{0}}{4}$
2) $\frac{\lambda_{0}}{3}$
3) $–\frac{\lambda_{0}}{4}$
4) $- \frac{\lambda_{0}}{3}$

Correct option: 3

Solution:
$P = \sigma{AT}^{4}$
$\lambda_{m}T = b$
$T = \frac{b}{\lambda_{m}} \Rightarrow P = \frac{\sigma{Ab}^{4}}{\lambda_{m}^{4}}$
$\Rightarrow P \propto \frac{1}{\lambda_{m}^{4}} \Rightarrow \frac{P_{i}}{P_{2}} = {(\frac{\lambda_{m_{2}}}{\lambda_{m_{4}}})}^{4}$
$\frac{P_{0}}{\frac{256}{81}P_{0}} = {(\frac{\lambda_{m_{2}}}{\lambda_{0}})}^{4} \Rightarrow \lambda_{m_{2}} = \frac{3}{4}\lambda_{0}$
$\Delta\lambda = \frac{3}{4}\lambda_{0} - \lambda_{0} = - \frac{\lambda_{0}}{4}$
Subject: Physics — Relative motion — Minimum separation
Difficulty: Hard

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

1) $T_{0} = \frac{d}{5v}$
2) $R_{\min} = \frac{d}{\sqrt{5}}$
3) Graph of R vs time is straight line
4) Graph of R vs time is Circle

Correct option: 1

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