Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Vector — Equilibrium of forces
Difficulty: Moderate

Question: You have a ring balanced at the center of the table by three
forces ${\overset{\rightarrow}{F}}_{1},{\overset{\rightarrow}{F}}_{2}$
and ${\overset{\rightarrow}{F}}_{3}$. The forces
${\overset{\rightarrow}{F}}_{1}$ and ${\overset{\rightarrow}{F}}_{2}$
have
components.$F_{1x} = 10N,F_{1y} = - 50N,F_{2x} = - 40N,F_{2y} = 100N$
The force ${\overset{\rightarrow}{F}}_{3}$ is required to make the
ringstationary at the center of the table. Which one ofthe 4 vectors
in the diagram could represent${\overset{\rightarrow}{\ F}}_{3}$?

1) $\overrightarrow{A}$
2) $\overrightarrow{B}$
3) $\overrightarrow{C}$
4) $\overrightarrow{D}$

Correct option: 4

Solution:
${\overset{\rightarrow}{F}}_{1} + {\overset{\rightarrow}{F}}_{2} + {\overset{\rightarrow}{F}}_{3} = 0 \Rightarrow 10\overset{\hat{}}{i} - 50\overset{\hat{}}{j} - 40\overset{\hat{}}{j} + {\overset{\rightarrow}{F}}_{3} = 0$
$\Rightarrow {\overset{\rightarrow}{F}}_{3} = 30\overset{\hat{}}{i} - 50\overset{\hat{}}{j} \Rightarrow IVth\ quadrant.$
Subject: Physics — Unit and dimension — Dimensional analysis
Difficulty: Moderate

Question: If the velocity of light denoted by acceleration due to
gravity are taken as fundamental
units, then the dimensions of length will be

1) $c^{2}/g$
2) $c/p$
3) $g/c$
4) $c/g$

Correct option: 1

Solution:
Let $L\alpha c^{a}g^{b}p^{c}$
$L = kc^{a}g^{b}p^{c}$
Putting dimensions on both sides:
$\left\lbrack M^{0}LT^{0} \right\rbrack = \left\lbrack LT^{- 1} \right\rbrack^{a}\left\lbrack LT^{- 2} \right\rbrack^{b}\left\lbrack ML^{- 1}T^{- 2} \right\rbrack^{c}$
$\left\lbrack M^{0}LT^{0} \right\rbrack = \left\lbrack M^{c}L^{a + b - c}T^{- a - 2b - 2c} \right\rbrack$
Comparing dimensions of both sides
$c = 0,\ \ \ \ \ a + b - c = 1$
$- a - 2b - 2c = 0$
solving we get $a = 2,b = - 1$
hence$\lbrack L\rbrack = \left\lbrack c^{2}/g \right\rbrack$
Subject: Physics — Current Electricity — KVL
Difficulty: Moderate

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:
$i = \frac{20}{2} = 10A$
$1 = \frac{41}{5} - \frac{3i}{5}$
$= + \frac{i}{5} = 2A$
Subject: Physics — Dual Nature of Matter — Einstein's photoelectric equation
Difficulty: Moderate

Question: Figure is the plot of the stopping potential versus the
frequency of the light used in an experiment on photoelectric effect. If
the work function is $\frac{x}{5}($ in
[IMAGE] then value of x is.


Answer: 4

Solution:
Subject: Physics — EMI — Induced emf
Difficulty: Tough

Question: A uniform magnetic field B in positive z direction exists in a
circular region of radius R = 5 m. A loop of radius R = 5m lying in x -
y plane encloses the magnetic field at t = 0 and then pulled at uniform. The emf induced (in volts) is the
loop at t = 2 sec is 6V. Then magnitude of 4B is


Answer: 1

Solution:
$= VB2\sqrt{R^{2} - \left( \frac{vt}{2} \right)^{2}} = VB\sqrt{4R^{2} - V^{2}t^{2}}$
$6V = 4 \times B\sqrt{4 \times 25 - 16 \times 4}$
$B = 0.25T$
Subject: Physics — Dual Nature of Matter — De Broglie wavelength
Difficulty: Moderate

Question: An electron of mass m and magnitude of charge |e| initially at
rest gets accelerated by a constant electric field E. The rate of change
of de-Broglie wavelength of this electron at time t ignoring
relativistic effects is


Correct option: 3

Solution:
Subject: Physics — Projectile Motion — Projectile up the inclined plane
Difficulty: Tough

Question: A particle is projected up an inclined plane of inclination
$\beta$ at an elevation $\alpha$ to the horizontal. If the particle
strikes the inclined plane at right angles then $\tan\alpha =$

1) $(cot\beta - 2tan\beta)$
2) $(tan\beta + 2cot\beta)$
3) $(cot\beta + 2tan\beta)$
4) $(tan\beta - 2cot\beta)$

Correct option: 3

Solution:
$T = \frac{2usin(\alpha - \beta)}{gcos\beta}$......(1)
$O = ucos(\alpha - \beta) - (gsin\beta)T$
$\Rightarrow T = \frac{ucos(\alpha - \beta)}{gsin\beta}$......(2)
from (1) &(2)
$2tan(\alpha - \beta) = cot\beta$
$\Rightarrow tan\alpha = (cot\beta + 2tan\beta)$
Subject: Physics — Wave on a string — speed of wave
Difficulty: Tough

Question: A string of length L of variable density is clamped between two
rigid supports. Whose density variation is shown in figure, ρ = density
[IMAGE] Find the time after which the disturbance created at one end will reach
other end.
(A = area of cross section of string in [IMAGE], T =
tension in the string is constant in N)


Correct option: 1

Solution:
Subject: Physics —  — 

Question: For a particle moving in a straight line, the acceleration
versus displacement graph is as shown. The relation between which of the
following quantities cannot be a straight line?

1) Velocity vs displacement
2) Force vs velocity
3) Force vs displacement
4) Kinetic energy vs displacement


Solution:
D
α = kx
or Force = ma = kmx
∴ (C) is a straight line.
Again [IMAGE] ∴ (A) is a straight line.
∴ (B) is a straight line.
or [IMAGE] (D) is a parabola.
∴
Subject: Physics — Nuclear physics — Alpha decay
Difficulty: Moderate

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


Answer: 5

Solution:
Subject: Physics — Electrostatics — Electrical potential in an electric circuit
Difficulty: Tough

Question: In the show circuit diagram all capacitors are initially
uncharged and cells are ideal. If C_1 = 1 $\mu F$,C_2 = 2 $\mu F$ and
potential of earth is taken zero. Choose the correct options.
 izkjEHk rFkk
 A;fn C_1 = 1 $\mu F$,C_2 = 2 $\mu F$ rFkk i`Foh

1) Potential of point A is (4/3) volts
fcUnq A
2) Potential of point B is (-13/3) volts
fcUnq B
3) Potential of point C is (10/3) volts
fcUnq C
4) Potential of point A is (-4/3) volts
fcUnq Adk foHko (-4/3) oksYV gSA

Correct option: 3

Solution:
$V_{C} = 5 - \frac{5}{3} = \frac{10}{3}$V
$\frac{10}{3} - V_{B} = 8$
$V_{B} = \frac{- 14}{3}$V
$\frac{- 14}{3} - V_{A} = \frac{10}{3}$
$V_{A} = \frac{- 24}{3} = - 8$V
Subject: Physics — Ray Optics — Convex lens
Difficulty: Moderate

Question: If an object is placed at 20 cm in front of a half thin convex
lens of focal length 10 cm as shown in figure then taking object
position as shown the.

1) x- coordinate of image is 20 cm.
2) x-coordinate of image is 40 cm.
3) y- coordinate of image is 0.2 cm.
4) y-coordinate of image is 0.4 cm.

Correct option: 1

Solution:
$\frac{1}{v} + \frac{1}{- 20} = \frac{1}{10} \Rightarrow v = 20$
$m = - 1$
Subject: Physics — Unit and dimension — Dimensional analysis
Difficulty: Moderate

Question: A force F is given by: F = at + bt^2, where t represents time.
What are dimensions of a and b?


Correct option: 2

Solution:
$\lbrack F\rbrack = \lbrack at\rbrack;\ \lbrack a\rbrack = \left\lbrack \frac{F}{t} \right\rbrack = \left\lbrack \frac{{MLT}^{- 2}}{T} \right\rbrack = \left\lbrack {MLT}^{- 3} \right\rbrack$
$$\lbrack F\rbrack = \left\lbrack \lbrack b\rbrack = \left\lbrack \frac{F}{t^{2}} \right\rbrack = \left\lbrack \frac{{MLT}^{- 2}}{T^{2}} \right\rbrack = \left\lbrack {MLT}^{- 4} \right\rbrack$$
Subject: Physics — Magnetic effect of current — Magnetic field due to current carrying conductor
Difficulty: Easy / Moderate/ Tough

Question: Two identical wires A and B, each of length $\mathcal{l}$, carry
the same current I. Wire A is bent into acircle of radius R and wire B
is bent form a squareof side a. If B_Aand B_B are the values of
magnetic field at the centres of the circle and square respectively,
then the ratio B_A/ B_B is$\pi^{2}:n\sqrt{2}.\ $What is the value of
n.


Answer: 8

Solution:
B at centre of a circle $= \frac{\mu_{0}I}{2R}$
B at centre of a
square $= 4 \times \frac{\mu I}{4\pi \cdot \frac{\mathcal{l}}{2}}\left( sin45^{\circ} + sin45^{\circ} \right)$
$= 4\sqrt{2}\frac{\mu_{0}I}{2\pi\mathcal{l}}$
 Now, $R = \frac{L}{2\pi}\ $and $\mathcal{l =}\frac{L}{4}(\ as\ L = 2\pi R = 4\mathcal{l)}$
 Where, L= length of wire 
$\therefore{\ \ \ B}_{A} = \frac{\mu_{0}L}{2 \cdot \frac{L}{2\pi}} = \frac{\pi\mu_{0}I}{L} = \pi\left( \frac{\mu_{0}I}{L} \right)$
$B_{B} = 4\sqrt{2}\frac{\mu_{0}I}{2\pi\left( \frac{L}{4} \right)} = \frac{8\sqrt{2}\mu_{0}I}{\pi L} = \frac{8\sqrt{2}}{\pi}\left( \frac{\mu_{0}I}{L} \right)$
$\therefore{\ \ \ \ B}_{A}:B_{B} = \pi^{2}:8\sqrt{2}$
Subject: Physics — Magnetic effects of current — Magnetic field due to current carrying conductor
Difficulty: Tough

Question: A very long wire ABDMNDC is shown in figure carrying current I.
AB and BC parts are straight, longand at right angle. At D wire forms a
circular turn DMND of radius R. AB, BC parts are tangential tocircular
turn at N and D. Magnetic field at the centre of circle is

1) $\frac{\mu_{0}I}{2\pi R}\left( \pi + \frac{1}{\sqrt{2}} \right)$
2) $\frac{\mu_{0}I}{2R}$
3) $\frac{\mu_{0}I}{2\pi R}\left( \pi - \frac{1}{\sqrt{2}} \right)$
4) $\frac{\mu_{0}I}{2\pi R}(\pi + 1)$

Correct option: 1

Solution:
$\frac{\mu_{0}i}{4\pi R}\left\lbrack \sin 90^{\circ} - \sin 45^{\circ} \right\rbrack \otimes + \frac{\mu_{0}i}{2R} \odot + \frac{\mu_{0}i}{4\pi R}\left( \sin 45^{\circ} + \sin 90^{\circ} \right) \odot$
$= \frac{- \mu_{0}i}{4\pi R}\left\lbrack 1 - \frac{1}{\sqrt{2}} \right\rbrack + \frac{\mu_{0}j}{2R} + \frac{\mu_{0}i}{4\pi R}\left\lbrack \frac{1}{\sqrt{2}} + 1 \right\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\left\lbrack - 1 + \frac{1}{\sqrt{2}} + 2\pi + \frac{1}{\sqrt{2}} + 1 \right\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack\sqrt{2} + 2\pi\rbrack \odot$
$= \frac{\mu_{0}i}{2\pi R}\left\lbrack \frac{1}{\sqrt{2}} + \pi \right\rbrack \odot$
Subject: Physics — EM wave — Relation $\overset{\rightarrow}{E}$ =
Difficulty: Moderate

Question: If the magnetic field in a plane electromagnetic wave is given
by
$\overset{\rightarrow}{B} = 3 \times 10^{- 8}sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{j}\ T$,
then what will be expression for electric field?

1) $\overset{\rightarrow}{E} = \left( 3 \times 10^{- 8}sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{j}V/m \right)$
2) $\overset{\rightarrow}{E} = \left( 3 \times 10^{- 8}sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{i}V/m \right)$
3) $\overset{\rightarrow}{E} = \left( 60sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{k}V/m \right)$
4) $\overset{\rightarrow}{E} = \left( 9sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{k}V/m \right)$

Correct option: 4

Solution:
$\frac{E_{0}}{B_{0}} = C$(speed of light in vacuum)
$E_{0} = B_{0}C = 3 \times 10^{- 8} \times 3 \times 10^{8}$
= 9 N/C
So, $E = 9\sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)$
Subject: Physics — Work Power & Energy — Potential energy curve
Difficulty: Moderate

Question: The potential energy of a particle varies with x according to
the relation $U(x) = x^{2} - 4x$. The point x=2 is a point of

1) stable equilibrium
2) unstable equilibrium
3) neutral equilibrium
4) none of above

Correct option: 1

Solution:
$U(x) = x^{2} - 4x$
$F = 0$
$\frac{dU(x)}{dx} = 0$
$2x - 4 = 0\ gives\ \ \ x = 2$
$\frac{d^{2}U}{{dx}^{2}} = 2 > 0$ i.e. U is minimum hence x = 2 is a
point of stable equilibrium.
Subject: Physics — Electronic devices — Diodes
Difficulty: Moderate

Question: Figure shows two identical capacitors each of capacity C in a
circuit with two ideal diodes. A 100 voltbattery in connected to the
input terminals. The output voltage across MN is.

1) 50 V
2) 25 V
3) 100 V
4) 75 V

Correct option: 1

Solution:
When terminal x is positive the diode D2 gets reverse biased
and D1 forward biased therefore circuit is
Tc fljk x rks rFkk D_1vxzck;l gks Hence vr V_MN = 50 V
Subject: Physics — Capacitance — Dielectric strength
Difficulty: Moderate

Question: A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. Apotential difference of 2500 v
is applied between the plates. If dielectric strength of air is 3 ×
10^6 v/m then.

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 = 0.0002 E_1 + 0.0023 E_2
2E_1 + 23 E_2 = 25 x 10^6
Also; E_1 = 5E_2
 
E_1 = 3.788 x 10^6 V/m
As E_1> 3.6 x 10^6 Hence Air will breakdown
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Moderate

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


Answer: 1

Solution:
Subject: Physics — Rigid Body Dynamics — Angular impulse
Difficulty: Easy

Question: A constant torque acting on a uniform circular wheel changes its
angular momentum from 10 J-sec to60 J-sec in 4 sec. The torque in Nm is
2.5n. Find the value of n.


Answer: 5

Solution:
$\tau = \frac{dL}{dt} = \frac{60 - 10}{4} = 12.5$ Nm
Subject: Physics — Projectile Motion — Angle of projection
Difficulty: Moderate

Question: At a height 0.4 m from ground, the velocity of projectile in
vector form
is$(\overset{\rightarrow}{v} = 6\overset{\hat{}}{i} + 2\overset{\hat{}}{j})$.
The angle of projection is π/n. Find n.

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

Correct option: 1

Solution:
${(v_{y})}^{2} = u_{y}^{2} + 2a_{y}S$
$(2)^{2} = {(u_{y})}^{2} + 2 \times ( - 10) \times 0.4$
$4 + 8 = u_{y}^{2}$
$u_{y} = \sqrt{12}$
$tan\theta = \frac{u_{y}}{u_{x}}$
$= \frac{\sqrt{12}}{6}$
$= \frac{1}{\sqrt{3}}$
$\theta = 30^{\circ}$
Subject: Physics — Magnetism — Ampere's circuital law
Difficulty: Moderate

Question: A long, straight wire of radius a carries a current distributed
uniformly over its cross-section. The ratioof the magnetic fields due
to the wire at distance[IMAGE] and 2a, respectively from
the axis of the wire is n/3. Find n.


Answer: 2

Solution:
Using Ampere's circuital law:
$B_{B} = \frac{\mu_{0}i}{2\pi(2a)}$
$\frac{B_{A}}{B_{B}} = \frac{4}{6} = \frac{2}{3}$
Subject: Physics — Electric current — Converting galvanometer into voltmeter
Difficulty: Moderate

Question: A galvanometer having a coil resistance $100\Omega$ gives a full
scale deflection when a current of 1 mA is passed through it. What is
the value of the resistance which can convert this galvanometer into a
voltmeter giving full scale deflection for a potential difference of 10
V?

1) $8.9k\Omega$
2) $10k\Omega$
3) $9.9k\Omega$
4) $7.9k\Omega$

Correct option: 3

Solution:
$V_{g} = i_{g}R_{g} = 0.1V$
$V = 10V$
$R = R_{g}(\frac{V}{V_{g}} - 1)$ $= 100 \times 99 = 9.9K\Omega$
Subject: Physics — Mechanical Properties of Fluids — Archimedes principle
Difficulty: Moderate

Question: A vessel contains two immiscible liquids of densities
$\rho_{1} = 1000\ {kgm}^{- 3}$ and $\rho_{2} = 1500\ {kgm}^{- 3}$. A
solid block of volume $V = 10^{- 3}m^{3}$ and density
$d = 800{kgm}^{- 3}$ is tied to one end of a string and other is tied to
the bottom of the vessel as shown in figure. The block is immersed with
$\frac{2^{th}}{5}$ of its volume in the liquid of higher density and
$\frac{3^{th}}{5}$ in the liquid of lower density. The entire system is
kept in an elevator which is moving upwards with an acceleration of a =
g/2. Find the tension in the string in Newton. (Take $g = 10ms^{- 2}$)
(in N)


Answer: 6

Solution:
[IMAGE] We analysed this problem from the reference frame of Elevator. Total
buoyant force On the block,
$F_{B} = \left( \frac{2}{5}V\rho_{2} + \frac{3}{5}V\rho_{1} \right)(g + a)$
For equilibrium,
$F_{B} = T + Vd(g + a)$
or $T = F_{B} - Vd(g + a)$
$= (g + a)V\left\lbrack \frac{2}{5}\rho_{2} + \frac{3}{5}\rho_{1} - d \right\rbrack$
$T = \left( 10 + \frac{10}{2} \right) \times 10^{- 3}\left\lbrack \frac{2}{5} \times 1500 + \frac{3}{5} \times 1000 - 800\rbrack = 6\text{N} \right.\ $
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

Question: The balancing length for a cell is 560 cm in a potentiometer
experiment. When an external resistance of10[IMAGE] is
connected in parallel to the cell, the balancing length changes by 60
cm. If the internalresistance of the cell is [IMAGE],
where N is an integer then value of N/2 is ______.

1) After connecting the resistor
$\frac{\varepsilon \times 10}{10 + r} = 500x$......
2) from

Answer: 6

Solution:
Let the emf of cell is $\varepsilon$ internal resistance is
 and potential gradient is x
Only cell connected:
$\epsilon = 560\ x$......(1)
After connecting the resistor
$\frac{\varepsilon \times 10}{10 + r} = 500x$......(2)
from (1) and (2)
$\frac{560 \times 10}{10 + r} = 500s$
$56 = 50 + 5r$
$r = \frac{6}{5} = 1.2\Omega$
n = 12
Subject: Physics — Relative motion — River problem
Difficulty: Easy

Question: A man who can swim at the rate of $2km/hr$ (in still river)
crosses a river to a point exactly opposite on the other bank by
swimming at an angle of 60º perpendicular direction of river flow of the
water in the river. The velocity of the water current in $km/hr$ is

1) 1
2) 2
3) $1/2$
4) $\sqrt{3}$

Correct option: 4

Solution:
Let speed of current be u. For net velocity ofman to be normal
to river flow.
$2sin60^{\circ} = u$
or$u = \sqrt{3}km/hr$
Subject: Physics — Projectile Motion — Projectile projected obliquely from a height
Difficulty: Moderate

Question: A particle at a height from the ground is projected with
an angle 30º fromthe horizontal, it strikes the ground making angle 45º
with horizontal. It is again projected from the same point with the same
speed but with an angle of 60º with horizontal. The tangent of the angle
it makes with the horizontal is $\sqrt{n}$when it strikes the ground


Answer: 5

Solution:
Using $v = \sqrt{u^{2} + 2gh}$
$v = \sqrt{u^{2}\sin^{2}\theta + 2gh}$(Vertical comp. when striking)
$ucos\theta = \sqrt{u^{2}\sin^{2}\theta + 2gh}$
$u^{2}\cos^{2}\theta = u^{2}\sin^{2}\theta + 2gh$
$u^{2}\left( \frac{3}{4} - \frac{1}{4} \right) = 2gh$
$u^{2} = 4gh$
$u = 2\sqrt{gh}$
Subject: Physics — Wave Optics — Interference
Difficulty: Moderate

Question: In a double-slit experiment, at a certain point on the screen
the path difference between the two interfering waves is
[IMAGE] of a wavelength. The ratio of the intensity of
light at that point to that at the centre of a bright fringe is

1) 0.568
2) 0.760
3) 0.853
4) 0.672

Correct option: 3

Solution:
Subject: Physics —  — 

Question: Consider two black bodies A and B. The surface temperature
of black body A is twice that of B. The wavelength corresponding to
maximum spectral radiancy for bodies A and B are λ_A and λ_B
respectively. When radiations of wavelength λ_A is incident on a
metallic surface the kinetic energy of fastest photoelectron is K_1
and when radiations of wavelength λ_B is incident on the same metallic
surface the kinetic energy of fastest photoelectron is K_2. The work
function of the metal is



Solution:
D
Subject: Physics —  — 

Question: Find the time when the displacement of the particle becomes zero.
Assume velocity and position at t = 0 to be 0 and -10 m respectively.

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

Correct option: 3

Solution:
Subject: Physics — Mechanical properties of liquid — Hydrostatic pressure
Difficulty: Moderate

Question: The pressure gauge reading in meter of water column shown in the
given figure will be

1) 3.20 m
2) 2.72 m
3) 2.52 m
4) 1.52 m

Correct option: 4

Solution:
$P_{0} + \rho_{w}g1 + \rho_{wg}0.2 - \rho_{Hg}g0.2 = P_{0}$
$P - P_{0} = \rho_{Hg}g0.2 - \rho_{w}g(1.2) = 14896$
$14896 = \rho_{wgh}$
h = 1.52
Subject: Physics — Electrostatics — Electric field by uniformly charged sphere
Difficulty: Moderate

Question: Consider a sphere of radius R which carries a uniform charge
density $\rho$. If a sphere of radius$\frac{R}{2}$iscarved out of it, as
shown, the
ratio$\left| \frac{{\overset{\rightarrow}{E}}_{A}}{{\overset{\rightarrow}{E}}_{B}} \right|$
of magnitude of electric field of magnitude of electric field
${\overset{\rightarrow}{E}}_{A}$and ${\overset{\rightarrow}{E}}_{B}$,
respectively, atpoints A and B due to the remaining portion is

1) $\frac{9}{17}$
2) $\frac{21}{34}$
3) $\frac{17}{54}$
4) $\frac{18}{54}$

Correct option: 1

Solution:
For a solid sphere
$E = \frac{\rho r}{3\varepsilon_{0}}$
$E_{A} = \frac{- \rho R}{2\left( 3\varepsilon_{0} \right)}$
$\left| E_{A} \right| = \frac{\rho R}{6\varepsilon_{0}}$
Electric field at point B = E_B = E_1A + E_2A
E_1A = Electric Field Due to solid sphere of radius R at point
$B = \frac{\rho R}{3\varepsilon_{0}}$
E_2A = Electric Field Due to solid sphere of radius R/2 (which having
charge density -$\rho$)
Subject: Physics — Modern physics — De Broglie wavelength
Difficulty: Moderate

Question: An electron (mass m) with initial velocity
$\overset{\rightarrow}{v} = v_{0}\overset{\hat{}}{i} + v_{0}\overset{\hat{}}{j}$
is in an electric field
$\overset{\rightarrow}{E} = - E_{0}\overset{\hat{}}{k}$. If
[IMAGE] is initial de-Broglie wavelength of electron, its
de-Broglie wave length at time t is given by

1) $\frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$
2) $\frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{2m^{2}v_{0}^{2}}}}$
3) $\frac{\lambda_{0}}{\sqrt{2 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$
4) $\frac{\lambda_{0}\sqrt{2}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$

Correct option: 2

Solution:
Initially m$(\sqrt{2}v_{0}) = \frac{h}{\lambda_{0}}$
Velocity as a function of
time$= v_{0}\overset{\hat{}}{i} + v_{0}\overset{\hat{}}{j} + \frac{{eE}_{0}}{m}t\overset{\hat{}}{k}$
so wavelength
$\lambda = \frac{h}{m\sqrt{2v_{0}^{2} + \frac{e^{2}E_{0}^{2}}{m^{2}}t^{2}}}$
$\lambda = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}}{2m^{2}v_{0}^{2}}t^{2}}}$
Subject: Physics — Ray Optics — TIR
Difficulty: Moderate

Question: There is a small source of light at some depth below the surface
of water (refractive index = 4/3) in a tank of large cross sectional
surface area. Neglecting any reflection from the bottom and absorption
by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height h and
radius of curvature r is $2\pi rh$]

1) 50
2) 34
3) 17
4) 21

Correct option: 3

Solution:
Solid angle $d\Omega = 2\pi R^{2}(1 - cos\beta)$
Percentage of light
$= \frac{2\pi R^{2}(1 - cos\beta)}{4\pi R^{2}} \times 100$
$= \frac{1 - cos\beta}{2} \times 100 = (\frac{4 - \sqrt{7}}{8}) \times 100 \approx 17\%$
Subject: Physics — Thermodynamics — Work done in thermodynamic process
Difficulty: Moderate

Question: For two thermodynamic process temperature and volume diagram are
given. In first process, it is a straight line having initial and final
coordinates as (V_0, T_0) and (2V_0,2T_0), where as in second
process it is a rectangular hyperbola having initial and final
coordinates (V_0, T_0) and (2V_0, T_0/2). Then ratio of work done in
the two processes must be
nks izfØ fy, rkieku rFkk (V_0, T_0) rFkk (2V_0, 2T_0), gS] tgk¡ nwljs

1) 1: 2
2) 2: 1
3) 1: 1
4) None of thesebueas

Correct option: 2

Solution:
[IMAGE] First process is constant pressure izFke
Hence vr ] $W_{1} = nR\left( 2T_{0} - T_{0} \right) = nRT_{0}$
Equation of second process is$T = \frac{c}{V}$
Subject: Physics — Thermodynamics — 
Difficulty: Moderate

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

1) 500 J
2) 460 J
3) 300 J
4) 380 J

Correct option: 2

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