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

Question: The current $I_{1}$ (in A) flowing through $1\Omega$ resistor in
the following circuit is

1) 0.2
2) 0.4
3) 0.5
4) 0.25

Correct option: 1

Solution:
[IMAGE] Resistance of branch PQ = 2.5 [IMAGE] Apply Ohm's law:
I = = 0.4A \\
i = = 0.2A
Subject: Physics — Mechanical Properties of Fluids — Surface tension
Difficulty: Moderate

Question: A small spherical droplet of density d is floating exactly half
immerse in a liquid of density ρ and surface tension T. The radius of
the droplet is (take note that the surface tension applies an upward
force on the droplet)

1) $r = \sqrt{\frac{T}{(d + \rho)g}}$
2) $r = \sqrt{\frac{3T}{(2d + \rho)g}}$
3) $r = \sqrt{\frac{T}{(d + \rho)g}}$
4) $r = \sqrt{\frac{2T}{3(d + \rho)g}}$

Correct option: 2

Solution:
$\Rightarrow d \cdot \frac{4}{3}\pi r^{3}g = \rho \cdot \frac{2}{3}\pi r^{3}g + 2\pi rT$
$\Rightarrow \frac{2}{3}r^{2}g(2d - \rho) = 2T$
$\Rightarrow r = \sqrt{\frac{3T}{(2d - \rho)}}$
Subject: Physics — Mechanical Properties of Fluids — Hydrostatic pressure
Difficulty: Moderate

Question: A container of a large uniform cross-sectional area A resting on
a horizontal surface holds two immiscible, non-viscous and
incompressible liquids of densities and each of height
(1/2)H as shown. The smaller density liquid is open to atmosphere. A
homogeneous solid cylinder of length
[IMAGE] cross-sectional area (1/5) A is immersed such that it floats with its
axis vertical to the liquid-liquid interface with length (1/4) L in
denser liquid. If D is the density of the solid cylinder then D=5d/n.
Find n.,d yEck,d leku vuqizLFkdkV
gSa,d leku ?kuRo
jgk gS fd bldh dqy yEckbZ dk ?kuRo okys nzo A;fn D csyu


Answer: 4

Solution:
Subject: Physics — Rigid Body Dynamics — Reaction from the hinge
Difficulty: Easy

Question: One end of a uniform rod of mass m and length
[IMAGE] is clamped. The rod lies on a smooth
horizontalsurface and rotates on it about the clamped end at a uniform
angular velocity$\omega$. The force exerted bythe clamp on the rod has a
horizontal component

1) $m\omega^{2}\mathcal{l}$
2) zero
3) mg
4) $\frac{1}{2}m\omega^{2}\mathcal{l}$

Correct option: 4

Solution:
Subject: Physics — Mechanical Properties of Fluids — Hydrostatic pressure
Difficulty: Moderate

Question: The vessel shown in the figure has two sections of areas of
cross-section [IMAGE] and
[IMAGE]. A liquid of density ρ fills both the
sections, up to a height h in each. Neglect atmospheric pressure

1) the pressure at the base of the vessel is $2\ h\ \rho\ g$
2) the force exerted by the liquid on the base of the vessel is
$\ 2\ h\rho gA_{2}$
3) the walls of the vessel at the level X exert a downward force
$h\rho g\left( A_{2} - A_{1} \right)$ on the liquid.$\ $
4) All of the above

Correct option: 4

Solution:
$0 + \rho g(2h) =$Press at base
$F = A_{2}(\rho g2h)$
Subject: Physics — Rigid Body Dynamics — Angular impulse
Difficulty: Tough

Question: A uniform rod AB of mass m and length I at rest on a smooth
horizontal surface. An impulse P is applied to the end B. The time taken
by the rod to turn through a right angle is

1) $\frac{2\pi ml}{P}$
2) $\frac{\pi mI}{3P}$
3) $\frac{\pi ml}{12P}$
4) $\frac{2\pi ml}{3P}$

Correct option: 3

Solution:
Impulse = change in momentum
$\therefore\ \ \ \ \ \ \ P \cdot \frac{\mathcal{l}}{2} = \frac{mc^{2}}{12} \cdot \omega$
(about centre of AB)
$\Rightarrow \omega = \frac{6P}{m\mathcal{l}}$
$\frac{\pi}{2} = \omega t \Rightarrow t = \frac{\pi}{2\omega} = \frac{\pi m\mathcal{l}}{2 \times 6p}$
$\Rightarrow t = \frac{\pi m\mathcal{l}}{12p}$
Subject: Physics — Thermodynamics — Cyclic process
Difficulty: Moderate

Question: The figure shows the P-V plot of an ideal gas taken through a
cycle DCBAD then
dks pØ DCBAD

1) maximum temperature will occur during the process C → B → A.
2) negative work is done by the gas in the cycle DCBAD.
3) heat flows out of the gas during C → B → A.
4) All of the above.

Correct option: 4

Solution:
Since temperature at C and A are same so maximum temperature
occurs at a state between them during process CBA.
Since cycle in PV graph is anticlockwise so work done by gas in
negative.
During CBA work done by gas is negative and [IMAGE] so
[IMAGE]. So heat is released by gas.
D;ksafdC rFkk A ij rkieku leku gS rks eku bu izØe CBA
ds nkSjku gksxkA D;ksfd PV vkjs[k okekorZ
Subject: Physics —  — 
Difficulty: Moderate

Question: A boy sits in a chair connected to a rope that passes over a
frictionless pulley. He pulls the loose end of the string with such a
force that the spring scale reads 250 N. The boy's weight is 320 N, and
the chair weighs 160 N. The force that the boy exerts on the chair is
(Take g = 10 m/s2)

1) 250 N
2) 320 N
3) 160 N
4) [IMAGE] 
 250 N
 320 N 160 N 
 ( g = 10 m/s2)

Correct option: 4

Solution:
[IMAGE] TOPIC: NLM
SUB TOPIC: EQUILIBRIUM
Subject: Physics — Mechanical properties of matter — Breaking stress of rod
Difficulty: Moderate

Question: If $\rho$ is density of the material of a uniform rope of length
$\mathcal{l}$ and $\sigma$ is the breaking stress. It is rotated in
horizontal circular path then maximum angular velocity it can have for
circular path:
$\mathcal{l}$yEckbZ dh le:i jLlh ds inkFkZ dk ?kuRo $\rho$ gS rFkk
jLlh dh ruu lkeF;Z(breaking stress)$\sigma$gSA bl ry esa
o`Ùkh; iFk ij ? 
osx Kkr djks

1) $\frac{1}{\mathcal{l}}\sqrt{\frac{2\sigma}{\rho}}$
2) $\frac{1}{\mathcal{l}}\sqrt{\frac{\sigma}{2p}}$
3) $\frac{2}{\mathcal{l}}\sqrt{\frac{2\sigma}{\rho}}$
4) $\frac{2}{\mathcal{l}}\sqrt{\frac{\sigma}{\rho}}$

Correct option: 1

Solution:
$\omega^{2} = \frac{2\sigma}{\mathcal{l}^{2}\rho} \Rightarrow \omega = \frac{1}{\mathcal{l}}\sqrt{\frac{2\sigma}{\rho}}$
Subject: Physics — Simple Harmonic Motion — Position of particle
Difficulty: Moderate

Question: A particle is executing SHM between points
-[IMAGE] and [IMAGE], as shown in
figure-I. The velocity V(t) of the particle is partially graphed and
shown in figure-II. Two points A and B corresponding to time
are marked on the V(t) curve

1) At time [IMAGE], it is going towards
2) At time [IMAGE], its speed is increasing.
3) At time [IMAGE], its position lies in between
[IMAGE] and O
4) The phase difference ∆φ between points A and B must be expressed
as 90°
<∆φ< 180°.

Correct option: 3

Solution:
At time [IMAGE], velocity of the particle is
negative i.e. going towar. From the graph, at
time [IMAGE], its speed is decreasing. Therefore
particle lies in between [IMAGE] and
Subject: Physics — Work Power & Energy — Work energy theorem
Difficulty: Moderate

Question: A body is given a velocity $\sqrt{2gr}$ at the highest point of
a half circular smooth track which is joined by arough horizontal track
whose co-efficient of friction is $\mu = 0.5$. Then the distance
travelled by body before it stops on horizontal track is: (r=1m)

1) 1 m
2) 4 m
3) 6 m
4) none of these

Correct option: 4

Solution:
$W_{g} + W_{t} = 0 - \frac{1}{2}m(2gr)$
$mg2r - \mu mgx = mgr\ $ $x = \frac{r}{\mu} = \frac{1}{0.5} = 2m$.
Subject: Physics — Magnetism — Dimension of energy density in magnetic
Difficulty: Easy

Question: The dimensions of$\frac{B^{2}}{2\mu_{0}}$, where B is magnetic
field and $\mu_{0}$ is the magnetic permeability of vacuum, is

1) MLT^-2
2) ML^-1T^-2
3) ML^2T^-1
4) ML^2T^-2

Correct option: 2

Solution:
Energy density in magnetic field$= \frac{B^{2}}{2\mu_{0}}$
$= \frac{\ Force\ \times \ displacement\ }{(\ displacement\ )^{3}} = \frac{{MLT}^{- 2}L}{L^{3}} = {ML}^{- 1}T^{- 2}$
Subject: Physics — Electrostatics — Electric potential
Difficulty: Moderate

Question: Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of
radii$R_{1}$and $R_{2}$, respectively. The electric fields $E_{1}$(on
$S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that
$E_{1}/E_{2} = R_{1}/R_{2}$. Then the
ratio$V_{1}(onS_{1})/V_{2}(onS_{2})$ of the electrostatic potentials on
each sphere is

1) ${(\frac{R_{1}}{R_{2}})}^{3}$
2) $\frac{R_{1}}{R_{2}}$
3) $(\frac{R_{2}}{R_{1}})$
4) ${(\frac{R_{1}}{R_{2}})}^{2}$

Correct option: 4

Solution:
$\frac{E_{1}}{E_{2}} = \frac{r_{1}}{r_{2}}$
$\frac{V_{1}}{V_{2}} = \frac{E_{1}r_{1}}{E_{2}r_{2}} = \frac{r_{1}}{r_{2}} \times \frac{r_{1}}{r_{2}} = {(\frac{r_{1}}{r_{2}})}^{2}$
Subject: Physics — Rigid Body Dynamics — Work energy theorem
Difficulty: Moderate

Question: A bead of mass m is kept at a distance from the axis of
rotation of a smooth massless tube inside it. The tube is rotating on a
smooth horizontal surface about a vertical axis perpendicular to the
length L of the tube and passing through one of its end with
constant angular velocity[IMAGE].

1) The speed of the bead relative to tube when it is at a distance r
from axis of rotation is2[IMAGE]
2) The speed of the bead relative to tube when it is at a distance
'r' from axis of rotation is[IMAGE]
3) The work done by the centrifugal force in the frame of tube when
the bead reaches at a distance the axis of rotation starting
4) The work done by the centrifugal force in the frame of the tube
when the bead reaches at a from the axis of rotation

Correct option: 3

Solution:
$\frac{v^{2}}{2} = \omega^{2}\frac{\left( r^{2} - a^{2} \right)}{2}$
$V = \omega\sqrt{r^{2} - a^{2}}$
Applying WET equation
$W_{(Centrifugal\ force\ )} = \frac{1}{2}{mV}^{2}$
$W = \frac{1}{2}m\omega^{2}\left( r^{2} - a^{2} \right)$
Subject: Physics — Kinematics — Uniform accelerated motion
Difficulty: Moderate

Question: Consider a particle initially moving with a velocity 7 m/s
starts decelerating at a constant rate of$2m/s^{2}$.
The distance travelled in the fourth second is$\frac{x}{10}m$. Find
the value of x.


Answer: 5

Solution:
Here particle is decelerating so velocity of particle becomes
 after t=3.5 sec. the particle has a turning point at t=3.5 sec.
$0 = 7 - 2t$
$t = 3.5\sec$
$x_{1} = x_{3} - \frac{1}{2} \times 2 \times 3^{2} = 12m$
$y_{2} = 7 \times 3.5 - \frac{1}{2} \times 2 \times (3.5)^{2} = \frac{49}{4}m$
$\Rightarrow y_{2} - x_{1} = \frac{49}{4} - 12 = \frac{1}{4}m$.
Due to symmetry displacement of the particle at t=3 and t=4sec. is same.
so total distance traveled in the fourth sec
is$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}m$
Subject: Physics — Electronic devices — Diodes
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..........
ekusa rFkk vxzfnf'kd ¼QkjoMZ½ bu varjfufeZr foHkokUrj(built in potential), 0.7 VgSAfp=k
 (input) oksYVrk fcUnq A ij oksYVrk

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

Correct option: 2

Solution:
[IMAGE] Let [IMAGE] Right diode is reversed biased and left diode is forward biased
 + Ik'p
Subject: Physics — Magnetic effect of current — 
Difficulty: Moderate

Question: An electron gun is placed inside a long solenoid of radius R on
its axis. The solenoid has n turns/lengthand carries a current I. The
electron gun shoots an electron along the radius of the solenoid with
speedv. If the electron does not hit the surface of the solenoid,
maximum possible value of v is (all symbolshave 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 — Projectile Motion — Collision with wall
Difficulty: Tough

Question: An elevator of height ascends with constant acceleration
 When it crosses a platform bolt drops from the top of the
elevator. If the time for the bolt to hit the floor of the elevator
is$\sqrt{\frac{\lambda h}{g + a}}$ then find $\lambda$.


Answer: 2

Solution:
Choose the floor of the elevator as reference. The observer is
inside the elevator take the downwarddirection as positive.
Acceleration of bolt relative to elevators is
$a^{'} = g - ( - a) = g + a$
$h = \frac{1}{2}a^{'}t^{2} = \frac{1}{2}(g + a)t^{2}$
$t = \sqrt{\frac{2h}{g + a}}$
Subject: Physics — Dual Nature of Matter — Einstein's photoelectric equation
Difficulty: Moderate

Question: When photon of energy [IMAGE] strikes the surface
of a metal A, the ejected photoelectrons have maximum kinetic energy Ta
eV and de-Broglie wavelength[IMAGE]. The maximum kinetic
energy of photoelectrons liberated from another metal B by photon of
energy [IMAGE] is [IMAGE] If the de-Broglie
wavelength of these photoelectrons[IMAGE], then the work
function of metal B is

1) 2 eV
2) 3 eV
3) 1.5 eV
4) 4 eV

Correct option: 4

Solution:
Relation between De-Broglie wavelength and K. E. is
Subject: Physics — Electrostatics — 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 $\frac{a}{3}$ and 2a, respectively from the axis of
the wire is

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

Correct option: 4

Solution:
$B_{A} = \frac{\mu_{0}ir}{2\pi a^{2}} = \frac{\mu_{0}i\frac{a}{3}}{2\pi a^{2}} = \frac{\mu_{0}i}{\pi a^{2}}\frac{a}{6} = \frac{\mu_{0}i}{6\pi a}$
$B_{B} = \frac{\mu_{0}i}{2\pi(2a)}$
$\frac{B_{A}}{B_{B}} = \frac{4}{6} = \frac{2}{3}$
Subject: Physics — Kinetic Theory of Gases — P-T diagram
Difficulty: Moderate

Question: Pressure versus temperature graph of the ideal gas at constant
volume V of an ideal gas is shown by a straight line A. Now, pressure P
of the gas is doubled and the volume is halved: then the corresponding
pressure versus temperature graph will be shown by the line:-,d vkn"kZ xSLk ds fLFkj xSl dk ncko cuke rkieku xzkQ
rkieku xzkQ ykbu tk,xkA

1) A
2) B
3) C
4) none of these buesa

Correct option: 2

Solution:
$P = \frac{nRT}{V} = \frac{mRT}{MV}$
(Where[IMAGE] number of moles)
So, at constant volume pressure versus temperature graph is a straight
line passing through origin with slope mR/MV. As volume is halved,
slope becomes two times. Therefore, pressure versus temperature graph
will be shown by the line B.
 eku ds fo:) fLFkj <ky lkFk mRifÙk
 <yku nks xquk gks A ncko
cuke rkieku xzkQ tk,xkA
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Moderate

Question: A litre of hydrogen at [IMAGE] and
[IMAGE] pressure expands isothermally until its volume
becomes double and then expands adiabatically until it becomes four
times. [IMAGE]. Final temperature (in
27°C o 106 izlkj tc rd bl nqxquk gks rFkk fQj:)ks"e izØe }kjk
pkj xqus vk;ru rd izlkfjr A vfUre


Answer: 5

Solution:
Subject: Physics —  — 

Question: A pebble is thrown horizontally from the top of a 20 m high tower
with an initial velocity of 10 m/s. The air drag is negligible. The
speed of the pebble when it is at the same distance from top as well as
base of the tower (g = 10 m/s2)

3) 20

Correct option: 2

Solution:
Subject: Physics — Dual Nature of Matter — Photoelectric emission
Difficulty: Moderate

Question: A beam of electromagnetic radiation of intensity
[IMAGE] is comprised of wavelength,
[IMAGE] It falls normally on a metal (work
function[IMAGE] ) of surface area of
[IMAGE]. If one in [IMAGE] photons
ejects an electron, total number of electrons ejected
[IMAGE] then x is....

1) 12
2) 18
3) 11
4) 10

Correct option: 3

Solution:
Energy of photon. [IMAGE] (so photoelectric
effect will take place)
[IMAGE] No. of photons falling per second [IMAGE] No. of photoelectron emitted per second [IMAGE]
Subject: Physics — AC — RL circuit
Difficulty: Moderate

Question: In the figure shown switch S_1 remains connected and switch
S_2 remains open for a long time. Now S_2 is also closed. Assuming
$\varepsilon = 10V$ and L = 1H. Find the magnitude of rate of change of
current (in A/s) in inductor just after the switch S_2 is closed.


Answer: 10

Solution:
$i_{1} + i_{2} = \varepsilon/R$
$\frac{\varepsilon - X}{R} + \frac{- 3\varepsilon - X}{2R} = \frac{\varepsilon}{R}$
$\Rightarrow X = - \varepsilon$
Using ${kv}^{\mathcal{l}}$
$3\varepsilon + \frac{Ldi}{dt} - 2\varepsilon = 0$
$\frac{di}{dt} = - 10A/s$
Subject: Physics —  — 
Difficulty: Easy

Question: The potential energy for a force field is given by U (x, y) = sin
(x + y), the magnitude of force acting on the particle at (0, π/4) is

1) 1
4) 0
 U (x, y) = sin (x + y) 
(0, π/4)

Correct option: 1

Solution:
[IMAGE] = - cos (x + y) [IMAGE] -
cos (x + y) [IMAGE] TOPIC: WPE
SUB TOPIC: POTENTIAL ENERGY
Subject: Physics — Kinematics — Position-time graph
Difficulty: Moderate

Question: The position-time (x-t) graph of two particles moving along
x-axis are as shown
[IMAGE] The separation between them

1) First increases then decreases then increases then decreases then
increases
2) First increases then decreases then increases
3) First decreases then increases then decreases
4) First increases then decreases

Correct option: 1

Solution:
Separation first increases then decreases then increases then
decreases then increases.
Subject: Physics — Magnetic effect of current — Principle of superposition
Difficulty: Tough

Question: Two long conductors are arranged as shown above to form
overlapping cylinders, each of radius r,whose centers are separated by a
distance d. Current of density J flows into the plane of the page along
the shaded part of one conductor andan equal current flows out of the
plane of the page along the shaded portion of the other, as shown. What
are the magnitude and direction of themagnetic field at point A?

1) $\left( \frac{\mu_{0}}{2\pi} \right)\pi dJ,\ $in the +y- direction
2) $\left( \frac{\mu_{0}}{2\pi} \right)\frac{d^{2}}{r},$ in
the +y direction
3) $\left( \frac{\mu_{0}}{2\pi} \right)\frac{4d^{2}J}{r},$ in
the -y direction
4) $\left( \frac{\mu_{0}}{2\pi} \right)\frac{{Jr}^{2}}{d},$ in
the -y direction

Correct option: 1

Solution:
[IMAGE] Fill the over-lapping part with +J and -J current density
Using principle of super position
Magnetic field due to conductor 
$B_{M}(2\pi OA) = \mu_{0}J\pi(OA)^{2}$
$B_{M} = \frac{\mu_{0}J}{2\pi}\pi(OA)$ (in y-direction)
And magnetic field due to conductor 
$B_{N}\left( 2\pi O^{'}A \right) = \mu_{0}J\pi\left( O^{'}A \right)^{2}$
$B_{N} = \frac{\mu_{0}J}{2\pi}\pi\left( O^{'}A \right)$ in y-direction 
So Net is given by
$B_{Net} = B_{m} + B_{N} = \frac{\mu_{0}J}{2\pi}\pi\left\lbrack OA + O^{'}A \right\rbrack$
$= \frac{\mu_{0}J\pi d}{2\pi}$ in y -direction
Subject: Physics — Wave Optics — Polarization
Difficulty: Moderate

Question: A polarizer - analyzer set is adjusted such that the intensity
of light coming out of the analyzer is just 10 of the original
intensity. Assuming that the polarizer-analyzer set does not absorb any
light, the angle by which the analyzer needs to be rotated further to
reduce the output intensity to be zero is

1) 45º
2) 90º
3) 71.6º
4) 18.4º

Correct option: 4

Solution:
[IMAGE] Which is 0.707
So [IMAGE] and [IMAGE] so only one
option is correct
i.e. 18.4º
Angle rotated should be [IMAGE]
Subject: Physics — Gravitation — Acceleration due to gravity
Difficulty: Moderate

Question: Two balls A and B having masses m kg. and2 m kg. moving with
in opposite direction, collide head on.
After collision A moves with a speed of [IMAGE] in the
same direction. Then correct statementare-
 The velocity of B after collision is
[IMAGE] opposite to the direction before collision.
 The coefficient of restitution is 0.2.
 The loss of kinetic energy due to collision
 All of the above


Correct option: 4

Solution:
By COLM
Subject: Physics — Centre of Mass — Coordinates of center of mass
Difficulty: Moderate

Question: A projectile of mass 3 m is projected from ground with velocity
$20\ \sqrt{2}$m/s at an angle of 45º. At highestpoint it explodes into
two pieces. One of mass 2 m and the other of mass m. Mass 2 m falls at a
distance of 100 m from point of projection towards right of point of
projection. If the distance(inm) of second mass from point of projection
where it strikes the ground is n × 10, then what is the value of n.(g=10


Answer: 4

Solution:
Range of the projectile in the absence of explosion
The path of centre of mass of projectile will not change, i.e.,
x_COM is still 80 m, Now, from the definition of centre of mass
$x_{COM} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}$
or $80 = \frac{(m)\left( x_{1} \right) + (2m)(100)}{m + 2m}$
Solving this equation, we get
x_1 = 40m
Therefore, the mass m will fall at a distance x_1=40 cm from point of
projection.
Subject: Physics —  — 

Question: In figure a wire perpendicular to a long straight wire is
moving parallel to it with a speed
v = 10 m/s in the direction of the current flowing in the wire. The
current is 10A. What is the magnitude of the potential difference (in
µV) between the ends of the moving wire?


Answer: 0014

Solution:
Subject: Physics — Magnetic effect of current — Magnetic field due to current carrying conductor
Difficulty: Moderate

Question: Magnitude of Magnetic field at point due to given current
distribution is

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

Correct option: 2

Solution:
${\overset{\rightarrow}{B}}_{0} = \frac{\mu_{0}I}{4R}\overset{\hat{}}{j} + \frac{\mu_{0}I}{2\pi R}\overset{\hat{}}{k}$
$B = \sqrt{\left( \frac{\mu_{0}I}{4R} \right) + \left( \frac{\mu_{0}I}{4\pi R} \right)^{2}}$
$= \frac{\mu_{0}I}{4\pi R}\sqrt{\pi^{2} + 1}$
Subject: Physics — Rigid Body Dynamics — Torque-angular relation
Difficulty: Moderate

Question: A thin rod having mass m and length 4[IMAGE] is
free to rotate about horizontal axis passing through a point distant
[IMAGE] from one of its end, as shown in figure. It is
released, from the horizontal position as shown:
[IMAGE] What will be acceleration of centre of mass at this instant?

1) $\frac{3g}{7}$
2) $\frac{2g}{7}$
3) $\frac{3g}{5}$
4) $\frac{2g}{5}$

Correct option: 1

Solution:
Torque equation
$\tau_{Hinge\ } = I_{Hinge\ } \propto$
$mg\mathcal{l =}\left( \frac{m(4\mathcal{l})^{2}}{12} + m\mathcal{l}^{2} \right) \propto$
$\frac{3g}{7\mathcal{l}} = \propto$
Tangential acceleration$= \propto \mathcal{l =}\frac{3g}{7\mathcal{l}}$
Radial acceleration$= \omega^{2}\mathcal{l} = 0$
Subject: Physics — Electronic devices — Zener diode
Difficulty: Moderate

Question: The circuit shown below is working as a 8V dc regulated voltage
source. When 12V is used as input,the power dissipated (in mW) in each
diode is 8n. Find n. (considering both zener diodes are
identical)__________.

1) 5
2) 3
3) 10
4) 6

Correct option: 1

Solution:
$i = \frac{(12 - 8)}{(200 + 200)}A = \frac{4}{400} = 10^{- 2}A$
Power loss in each diode = (4)(10^-2 ) W = 40 mW
Subject: Physics — Rigid Body Dynamics — Pure rolling
Difficulty: Moderate

Question: An impulsive force F acts horizontally on a solid sphere of
radius R placed on a horizontal surface. The line of action of the
impulsive force is at a height h above the centre of the sphere. If the
rotational and translational kinetic energies of the sphere just after
the impulse are equal, then h is equal to$\sqrt{\frac{2}{n}}R$. Find n.

1) and
2) and
3) $h = \frac{2}{5}\frac{R^{2}\omega}{v}$...(iv)
4) h$= \sqrt{\frac{2}{5}}R$

Answer: 5

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

Question: A steel wire is rigidly fixed at both ends. Its length, mass and
cross sectional area are 1m, 0.01 kg and
[IMAGE] respectively. Then the temperature of the wire is
lowered by[IMAGE]. If the transverse waves are setup by
plucking the wire at [IMAGE] from one end and assuming
that wire vibrates with minimum number of loops possible for such a
case. The frequency of vibration is 44n Hz. Find n. [Coefficient of
linear expansion of steel [IMAGE] and Young's

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

Correct option: 3

Solution:
[IMAGE] The mechanical strain [IMAGE] The tension in wire[IMAGE] speed of wave in wire
[IMAGE] Since the wire is plucked at [IMAGE] from one end
The wire shall oscillate in [IMAGE] overtone (for minimum
number of loops)
TOPIC:Waves in a string
SUB TOPIC: Wire fixed at both ends
LEVEL:Moderate