Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Error analysis — Dimensional analysis
Difficulty: Moderate

Question: The velocity v (in cm/sec) of a particle is given in terms of
time t (in sec) by the equation:$v = at + \frac{b}{t\ + \ c}$
The dimensions of a, b and c are:
 ------------------------------------
 a b c


Correct option: 3

Solution:
$\lbrack v\rbrack = \lbrack at\rbrack;\ \lbrack a\rbrack = \left\lbrack \frac{v}{t} \right\rbrack = \left\lbrack \frac{{LT}^{- 1}}{T} \right\rbrack = \left\lbrack {LT}^{- 2} \right\rbrack$
$\lbrack t\rbrack = \lbrack c\rbrack = \lbrack T\rbrack$
$\left\lbrack \frac{b}{t + c} \right\rbrack = \lbrack v\rbrack\ \Rightarrow \ \frac{\lbrack b\rbrack}{\lbrack t\rbrack} = \lbrack v\rbrack \Rightarrow \lbrack b\rbrack = \lbrack vt\rbrack = \left\lbrack {LT}^{- 1}T \right\rbrack = \lbrack L\rbrack$
Subject: Physics — Electrostatics — Electric dipole in electric field
Difficulty: Moderate

Question: An electric dipole of moment
$\overset{\rightarrow}{p} = ( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k}) \times 10^{- 29}$
cm. is at the origin (0,0,0). The electric fielddue to this dipole at
$\overset{\rightarrow}{r} = ( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} + 5\overset{\hat{}}{k})$
(note that$\overset{\rightarrow}{r}.\overset{\rightarrow}{p} = 0$) is
parallel to

1) $( + \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$
2) $( - \overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$
3) $( + \overset{\hat{}}{i} - 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$
4) $( - \overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 2\overset{\hat{}}{k})$

Correct option: 1

Solution:
$\overset{\rightarrow}{p} \cdot \overset{\rightarrow}{r} = 0$
$\overset{\rightarrow}{E}$must be antiparallel
to$\overset{\rightarrow}{p}$
So, $\overset{\rightarrow}{E} = - \lambda(\overset{\rightarrow}{p})$
Where $\lambda$ is a arbitrary positive constant
Let
$\overset{\rightarrow}{A} = a\overset{\hat{}}{i} + b\overset{\hat{}}{j} + c\overset{\hat{}}{k}$
is parallel to $\overset{\rightarrow}{E}$
$\frac{a}{\lambda} = \frac{b}{3\lambda} = \frac{c}{- 2\lambda} = k$
So
$\overset{\rightarrow}{A} = \lambda k(\overset{\hat{}}{i} + 3\overset{\hat{}}{j} - 2\overset{\hat{}}{k})$
Subject: Physics — Electrostatics — Charge on capacitor
Difficulty: Moderate

Question: A $1\ µF$ capacitor is connected in the circuit shown below. The
e.m.f. of the cell is 3 volts and internalresistance is 0.5 ohms. The
resistors R_1and R_2have values 4 ohms and 1 ohm respectively.
Thecharge (in$µC)$ on the capacitor in steady state must be


Answer: 2

Solution:
In steady state no current flows through capacitor. The
potential difference across capacitor andresistor of resistance R_2is
same.
$\mathbf{\therefore}$ charge on capacitor
$= CV = C \times 3 = 1\mu F \times 3 = 3\mu C$.
LFkk;h voLFkk esa] la
Subject: Physics — Rigid Body Dynamics — Pure rolling down the incline
Difficulty: Moderate

Question: A solid cylinder and a solid sphere, both having the same mass
and radius, are released from an inclined plane having angle of
inclination $\theta$ one by one. They roll on the incline without
slipping. The statement that holds good in this motion is

1) the force of friction that acts on the two is the same
2) the force of friction is greater in case of a sphere than for a
cylinder.
3) the force of friction is greater in case of cylinder than for
sphere.
4) the force of friction will depend on the nature of the surface
of the body that is moving and that of the inclined surface, and is
independent of the shape and size of the moving body.

Correct option: 3

Solution:
Equation of motion
$mg\ sin\theta - f = ma$....(i)
$fR = I\alpha = I\frac{a}{R}$
$\Rightarrow f = I\frac{a}{R^{2}}$....(ii)
from (i) and (ii)
$mgsin\theta - f = m.\frac{fR^{2}}{I}$
$mgsin\theta = f(1 + m.\frac{R^{2}}{I}$)
$f = \frac{mgsin\theta}{1 + \frac{mR^{2}}{I}}$
Moment of inertia of solid cylinder is greater, so friction force f will
be greater.
Subject: Physics —  — 
Difficulty: Moderate

Question: A small sphere tied to a string of length 0.8 m is describing a
vertical circle so that the maximum and minimum tensions in the string
are in the ratio 3: 1. The fixed end of the string is at a height of
5.8 m above ground. Find the velocity of the sphere at the lowest
position. (Take g = 10 m/s^2)

1) Using conservation of energy
KE from bottom to the top =Gain in GPE
2) Using equations

Answer: 8

Solution:
let u and v be the speeds of the sphere at the bottom and the
top positions and m be themass
[IMAGE] Radius of circle = length of string =r=0.8 m
[IMAGE] =tension at the bottom or the maximum tension
[IMAGE] =tension at the top or the minimum tension
Using conservation of energy
KE from bottom to the top =Gain in GPE
Using equations (1) and (2) we get[IMAGE] TOPIC:Circular motion
SUB TOPIC:Vertical circle
Subject: Physics — Magnetism — Properties of ferromagnetic material
Difficulty: Moderate

Question: A uniform magnetic field parallel to the plane of paper, existed
in space initially directed from left to right. When a bar of soft iron
is placed in the field parallel to it, the lines of force passing
through it will be represented by figure


Correct option: 2

Solution:
When a bar of soft iron is placed in the uniform magnetic field
which is parallel to it because of large permeability of soft iron,
magnetic lines of force prefer to pass through it. Concentration of
lines in soft iron bar increases as shown in figure.
Subject: Physics — Heat — Specific heat
Difficulty: Moderate

Question: When 300 J of heat is added to 25 gm of sampleof a material its
temperature rises from 25ºC to45ºC. The thermal capacity of the sample
andspecific heat of the material are respectively given by

1) $15J/ºC,600J/kg - ºC$
2) $600J/ºC,15J/ºC - kg$
3) $150J/ºC,60J/kg - ºC$
4) None of these

Correct option: 1

Solution:
$\Delta\phi = ms\Delta T$
$\therefore$ Thermal
capacity $= ms = \frac{\Delta Q}{\Delta T} = \frac{300J}{45 - 25} = 15J/ºC$
 and specific
heat, $s = \frac{\Delta Q}{m\Delta T} = \frac{300}{25 \times 10^{- 3} \times (45 - 25)} = \frac{600J}{kg} - C$
Subject: Physics — Mechanical Properties of Fluids — Application of Bernoulli's principle
Difficulty: Moderate

Question: Figure shows a capillary tube C dipped in a liquid that wets it.
The liquid rises to a point A. If we blow air through the horizontal
tube H, what will happen to the liquid column in the capillary tube?

1) Level will rise above A
2) Level will fall below A
3) Level will remain at A
4) Cannot say

Correct option: 1

Solution:
Level rise because to compensate for the reduction in pressure
above A, due to high velocity (from Bernoulli's principle pressure
decreases if velocity increases.)
Subject: Physics — Mechanical properties of liquid — Surface tension
Difficulty: Moderate

Question: A long capillary tube of mass$'\pi^{'}\ gm$, radius 2 mm and
negligible thickness, is partially immersed in a liquid of surface
tension 0.1 N/m. Take angle of contact zero and neglect buoyant force of
liquid. The force required to hold the tube vertically, will be-
$\left( g = 10m/s^{2} \right)$
 eksVkbZ dh,d yEch
 i`"B ruko okys æo:i ls Mqcksrs
gSaA s.k ysrs gq, rFkk æo ds mRIykou cy dks ux.;
ekurs cy

1) $10.4\ \pi mN$
2) $10.8\ \pi mN$
3) $0.8\ \pi mN$
4) $4.8\ \pi mN$

Correct option: 2

Solution:
The free body diagram of the capillary tube is as shown in the
figure. Net force F required to hold tube is
F = force due to surface tension at cross-section
$\left( S_{1} + S_{2} \right) +$weight of tube.
dsf'kdk uyh dk cy js
fy, dqy cy F gS
F = vuqizLFk dkV ij i`"B ruko j.k cy
$\left( S_{1} + S_{2} \right) +$V~ Hkkj
$= 4\pi \times 2 \times 10^{- 3} \times 0.1 + \pi \times 10^{- 3} \times 10 = 10.8\pi mN$
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

Question: In the given circuit, resistance of potentiometer wire AB is
$10\Omega$ and its length is 1m. Rest of quantities is shown in figure.
The null point will be at a separation x m from point A, then find value
of 50 x.


Answer: 18.75

Solution:
$k = (\frac{E}{R + R_{p}})\frac{R_{p}}{\mathcal{l}} = \frac{2 \times 10}{(10 + 15) \times 1} = 0.8 \vee /m = \frac{8}{10}V/m$
$\frac{8}{10} \times AJ = 0.3$
$AJ = x = \frac{3}{8}m$
Subject: Physics —  — 
Difficulty: Tough

Question: Two uniform solid spheres of equal radii R, butmass M and 4M have
a centre to centre separation 6R, as shown in figure. A projectile of
mass m in projected from the surface of thesphere of mass M directly
towards the centre of the second sphere. The minimum speed ofthe
projectile so that it reaches the surface of the second sphere
is[IMAGE]. What is n?

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

Answer: 3

Solution:
[IMAGE] The projectile need to reach a point on the line joining the centers
of sphere where net electric force on it is zero.
Let the point N has no net electric field due to spheres.
Therefore, $\frac{GM}{x^{2}} = \frac{G.4M}{{(6R - x)}^{2}}$ x = 2R
Apply conservation of mechanical energy at projectile initial position
and particle at N
E_i = E_f
SUB TOPIC: Conservation of mechanical energy
Subject: Physics —  — 

Question: A person with a normal near point (25 cm) is using a compound
microscope in normal adjustment with objective focal length
[IMAGE] and an eye-piece of focal length
[IMAGE] can bring an object placed
[IMAGE] from the objective in sharp focus. The
magnifying power of the microscope is 11n. Find n


Answer: 8

Solution:
Angular magnification of the eye-piece for image at
[IMAGE] TOPIC: Optical instruments
SUB TOPIC: Microscope
LEVEL: Moderate
Subject: Physics — Rigid Body Dynamics — Conservation of mechanical energy
Difficulty: Moderate

Question: A uniform metre stick is held vertically with one end on the
floor and is allowed to fall. The speed of theother end when it hits the
floor assuming that the end at the floor does not slip

1) $\sqrt{4g}$
2) $\sqrt{3g}$
3) $\sqrt{5g}$
4) $\sqrt{g}$

Correct option: 2

Solution:
using energy conservation
$mg\frac{\mathcal{l}}{2} = \frac{1}{2}I\omega^{2}$
$mg\frac{\mathcal{l}}{2} = \frac{1}{2} \cdot \frac{m\mathcal{l}^{2}}{3}\omega^{2}$
$= 1m$ $\omega = \sqrt{\frac{3g}{}}$
$V_{A} = \omega = \sqrt{3g} = (\sqrt{3g})$
Subject: Physics — Kinetic Theory of Gases — Root mean square speed
Difficulty: Moderate

Question: Two moles of hydrogen are mixed with n moles of helium. The root
mean square speed of gas molecules in the mixture is $\sqrt{2}$ times
the speed of sound in the mixture. Then n is.

1) 3
2) 2
3) 1.5
4) 2.5

Correct option: 2

Solution:
$V_{rms} = \sqrt{\frac{3RT}{M_{mix}}}$
$V_{sound\ } = \sqrt{\frac{\gamma RT}{M_{mix}}}$
$V_{rms} = \sqrt{2}V_{sound\ }$
$\sqrt{\frac{3RT}{M_{mix}}} = \sqrt{2}\sqrt{\frac{\gamma RT}{M_{mix}}}$
$\gamma_{mix} = \frac{3}{2}$
$\gamma_{mix} = \frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$
$\frac{3}{2} = \frac{2 \times \frac{7R}{2} + n \times \frac{5R}{2}}{2 \times \frac{5R}{2} + n \times \frac{3R}{2}} \Rightarrow \frac{3}{2} = \frac{14 + 5n}{10 + 3n} \Rightarrow 30 + 9n = 28 + 10n \Rightarrow n = 2$
Subject: Physics — Mechanics — Escape velocity
Difficulty: Moderate

Question: Planet P has mass M_P and radius R_P. Planet Q has mass M_P/2.
If the escape velocities from the Planets P and Q are equal, then what
is the radius of planet Q?


Correct option: 4

Solution:
Subject: Physics — Error analysis — Percentage error
Difficulty: Easy

Question: Error in the measurement of radius of a sphere is1.What is the
percentage error in the calculated value of its volume?


Answer: 3

Solution:
$V = \frac{4}{3}\pi r^{3}$
Take natural log on both side
$ln(V) = ln\left( \frac{4}{3}\pi \right) + 3ln(r)$
Differentiate both side
$\frac{\Delta V}{V} = 3\frac{\Delta r}{r}$
$\%\frac{\Delta V}{V} = 3 \times 1 = 3\%$
Subject: Physics — Unit and dimension — Dimensional analysis
Difficulty: Moderate

Question: The van der Waals' equation of state for somegases can be
expressed as:
$$\left( P + \right)(V - b) = RT$$
Where P is the pressure, V the molar volume andT is the absolute
temperature of the given sampleof gas, a, b, and R constants. The
dimensions is


Correct option: 1

Solution:
$\lbrack P\rbrack = \left\lbrack \frac{a}{V^{2}} \right\rbrack\lbrack a\rbrack = \lbrack P\rbrack\left\lbrack V^{2} \right\rbrack = \left\lbrack M^{1}L^{- 1}T^{- 2} \right\rbrack\left\lbrack L^{6} \right\rbrack = \left\lbrack M^{1}L^{5}T^{- 2} \right\rbrack$
Subject: Physics — Centre of Mass — Variable mass system
Difficulty: Tough

Question: In the fig. shown a cart moves on a smooth horizontal surface
due to an external constant force ofmagnitude F. The initial mass of the
cart is M_0and velocity is zero. Sand falls on to the cart
withnegligible velocity at constant rate and
sticks to the cart. The velocity of the cart at time t is

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

Correct option: 1

Solution:
[IMAGE] Formula $F = m\frac{dv}{dt} + (V - u)\frac{dm}{dt}$
Here, u = velocity of sand = O
m = M_o[IMAGE] t = mass at time t
and [IMAGE] $\therefore$ $F = \left( M_{0} + \mu t \right)\frac{dv}{dt} + v\mu$
$(F - \mu v)dt = \left( M_{0} + \mu t \right)dv$
$\int_{0}^{t}\mspace{2mu}\frac{dt}{M_{0} + \mu t} = \int_{0}^{v}\mspace{2mu}\frac{dv}{F - \mu v}$
$\frac{1}{\mu}\left\lbrack \mathcal{l}n\left( M_{0} + \mu t \right) \right\rbrack_{0}^{t} = - \frac{1}{\mu}\mathcal{\lbrack l}n(F - \mu v)\rbrack_{0}^{v}$
$\mathcal{l}n\frac{\left( M_{0} + \mu t \right)}{M_{0}}\mathcal{= l}n\left\{ \frac{F}{F - \mu v} \right)$
$F - \mu v = \frac{M_{0}F}{M_{0} + \mu t} \Rightarrow v = \frac{Ft}{M_{0} + \mu t}$
Subject: Physics —  — 
Difficulty: Tough

Question: A bob of mass m is connected to a spring of spring constant k and a
string which is fixed to the ceiling as shown in figure. Original length
of the spring is L and the bob rotates in a horizontal circle centred at
the fixed rod. Assuming the spring remains horizontal, then elongation
in the spring is

4) [IMAGE] m k 
 L,

Correct option: 4

Solution:
[IMAGE] T cos θ = mg... (1)
T sin θ + kx = mw2 (L + x)... (2)
from (1) and (2)
x =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS
Subject: Physics — Simple Harmonic Motion — Time period of SHM
Difficulty: Moderate

Question: A disc of mass is attached to a spring of
stiffness During its motion, the disc rolls on the ground. When
released from some stretched position, the centre of the disc will
execute simple harmonic motion with a time period of


Correct option: 3

Solution:
Torque = Iα about the point of contact
Subject: Physics —  — 
Difficulty: Moderate

Question: The vertical height between two points A and B is 30 m. A particle
is projected downward from A and at the same time another particle is
projected upward from B with the same speed. Both particles reach the
ground simultaneously. The velocity of the projection of the particle is
(B is on ground)

4) A B 30m A 
 B: (B )

Correct option: 2

Solution:
S = 0 = ut -[IMAGE] gt2
[IMAGE] 30 = ut + [IMAGE] gt2 =[IMAGE] by adding
30 = 2ut =[IMAGE] u2 =[IMAGE] ⇒[IMAGE] = u
TOPIC:KINEMATIC
SUB TOPIC: MOTION UNDER GRAVITY
Subject: Physics — Mechanical properties of liquid — Surface tension
Difficulty: Moderate

Question: On heating water, bubbles beings formed at the bottom of the
vessel detach and rise. Take the bubbles to be spheres of radius R and
making a circular contact of radius r with the bottom of the vessel. If
r << R, and the surface tension of water is T, value of r just before
bubbles detach is: (density of water is
ikuh dks xeZ
gksdj dh vksj mBrs gSaA cqycqyksa dks f= xksyk eku ysa
ikuh dk i`"B ruko T gSa, rc cqycyksa ds cl foyXu gksus ls tjk igys r


Correct option: 1

Solution:
[IMAGE] The bubble will detach if -
Buoyant force ≥ Surface tension force
[IMAGE] Solving [IMAGE] cqycqyk vyx gksxk;fn -
mRIykou cy≥i`"B ruko
Subject: Physics — Mechanical properties of liquid — Hydrostatic pressure
Difficulty: Moderate

Question: The pressure gauge reading in metre of water column shown in the
given figure will be:
 fp= esa ty LrEHk

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

Correct option: 4

Solution:
$P_{0} + \rho_{w}g1 + \rho_{w}g0.2 - \rho_{Hg}g0.2 = P_{0}$
$P - P_{0} = \rho_{Hg}g0.2 - \rho_{w}g(1.2) = 14896$
$14896 = \rho_{w}gh$
h = 1.52
Subject: Physics — Rigid Body Dynamics — Pure rolling
Difficulty: Moderate

Question: A uniform disc of mass M and radius R is lifted using a string
as shown in the figure. Then,

1) Its linear acceleration is g upward
2) Its rate of change of angular momentum is MgR
3) Its angular acceleration is[IMAGE]
4) All of the above

Correct option: 1

Solution:
$\frac{3Mg}{2} + \frac{Mg}{2} - Mg = Ma$
$a = g \uparrow$
$\tau = I\alpha$
$\frac{3Mg}{2}R - \frac{Mg}{2}R = \frac{{MR}^{2}}{2} \cdot \alpha$
$\alpha = \frac{2g}{R}$
$\frac{dL}{dt} = \tau = MgR$
Subject: Physics —  — 
Difficulty: Moderate

Question: 40 m 
80 m

1) 45°
2) 90°
3) 37°
4) 53°

Correct option: 1

Solution:
With standard kinematics, we get
[IMAGE] y = [IMAGE] gt2, t = [IMAGE] = 2.86 s
x = vxt, vx =[IMAGE] = 28.0 m/s
vy2 - v02 = 2a(y - y0)
vy = -[IMAGE] = -[IMAGE] = 28.0 m/s
θ = A tan[IMAGE] = 315°
TOPIC:KINEMATIC
SUB TOPIC: 2 D
Subject: Physics — Rigid Body Dynamics — Pure rolling
Difficulty: Moderate

Question: A solid cylinder and a solid sphere, both having the same mass
and radius, are released from an inclined plane having angle of
inclination[IMAGE] one by one. They roll on the incline
without slipping. The statement that holds good in this motion is

1) The force of friction that acts on the two is the same
2) The force of friction is greater in case of a sphere than for a
cylinder.
3) The force of friction is greater in case of cylinder than for sphere.
4) The force of friction will depend on the nature of the surface of
the body that is moving and that of the inclined surface, and is
independent of the shape and size of the moving body.

Correct option: 3

Solution:
[IMAGE] Equation of motion
mg sin [IMAGE]....(i)
$fR = I\alpha = I\frac{a}{R}$
$\Rightarrow f = I\frac{a}{R^{2}}$....(ii)
From (i) and (ii)
$mgsin\theta - f = m\frac{{fR}^{2}}{I} \Rightarrow mgsin\theta = f\left\{ 1 + \frac{mR^{2}}{I} \right)$
$\Rightarrow$ $\ \ f = \frac{mgsin\theta}{1 + \frac{mR^{2}}{I}}$
Moment of inertia of solid cylinder is greater, so friction force f will
be greater.
Subject: Physics — Magnetic effect of current — 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

2) 2

Correct option: 4

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 — Atomic physics — Alpha particle scattering
Difficulty: Easy

Question: The graph which depicts the results of Rutherford gold foil
experiment with α -particles is:
α -particles is:
Y = Number of scattered alpha -particles detected
(Plots are schematic and not to scale)


Correct option: 1

Solution:
Subject: Physics — Kinetic Theory of Gases — R.M.S. speed
Difficulty: Moderate

Question: The following graph shows two isotherms for a fixed mass of an
ideal gas. The ratio of r.m.s speed of the molecules at temperatures
T_1 and T_2 is

3) 2
4) 4

Correct option: 3

Solution:
$PV = \frac{1}{3}M\left( v_{rms} \right)_{1}^{2} = RT$
On lower curve, we have V=1m^3, corresponding to P=1×10^5Pa = 1
atmosphere. At T_1, we have
de oØ tks P=1×10^5Pa = 1 okrkoj.k ds
vuq:i gSA T_1 ij] gekjs ikl 
$1 \times 1 = \frac{1}{3}M\left( v_{rms} \right)_{1}^{2} = {RT}_{1}$
or$\left( v_{rms} \right)_{1}^{2} = \frac{3}{M}$
Similarly, for upper curve we have
blh rjg] oØ ds fy, gekjs ikl gS
V=1m^3, P = 2 atmosphere
$\therefore 2 \times 2 = \frac{1}{3}M\left( v_{ans} \right)_{2}^{2} = {RT}_{2}$
$\therefore\left( v_{mas} \right)_{2}^{2} = \frac{4 \times 3}{M} = \frac{12}{M}$
$\therefore\ Required\ ratio\ = \sqrt{\frac{12/M}{3/M}} = 2$
Subject: Physics —  — 

Question: A hollow conducting sphere of inner radius R and outer radius
2R is given a charge Q as shown in the figure, then the

1) potential at A and B is different
2) potential at O and B is
different
3) potential at O and C is different
4) potential at A, B, C and O
is different

Correct option: 4

Solution:
[IMAGE] charge is on the outer surface hence [IMAGE] remains
constant.
Subject: Physics — Magnetism — 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 — Magnetism — Torque on current carrying loop in magnetic
Difficulty: Tough

Question: A 100 turns closely wound current carrying coilof radius 10 cm
is placed in a vertical plane. Thecoil carries a current of 3.2 A and
free to rotateabout a horizontal axis which coincides with itsdiameter.
A uniform horizontal magnetic field of2 Tesla exists in the region. Due
to magnetic torquecoil starts rotating. The angular velocity in
rad/secof the coilwhen it has rotated by $\frac{\pi}{2}$ is 10n. Find n.


Answer: 2

Solution:
$\tau = MBsin\theta$
$I\alpha = MBsin\theta$
$\alpha = \frac{MB}{I}\sin\theta$
$\frac{d\omega}{d\theta}*\frac{d\theta}{dt} = \frac{MB}{I}sin\theta$
$\int_{0}^{\omega}\mspace{2mu}\omega d\omega = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\frac{MB}{I}\sin{\theta d\theta}$
$\left( \frac{\omega^{2}}{2} \right)_{0}^{\omega} = \frac{MB}{I}\lbrack - cos\theta\rbrack_{0}^{\pi/2}$
$\omega = \sqrt{\frac{2MB}{I}} = \sqrt{\frac{2NIAB}{I_{rot}}}\ $
$= \sqrt{\frac{2 \times 100 \times 3.2 \times \pi \times 100 \times 10^{- 4} \times 2}{0.1}}$
= 20 rad/sec.
Subject: Physics —  — 

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]

Correct option: 2

Solution:
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.

1) 5
2) 6
3) 7
4) 8

Correct option: 1

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.
V=U+at
$0 = 7 - 2t$
$t = 3.5\sec$
$y_{2} = 7 \times 3.5 - \frac{1}{2} \times 2 \times (3.5)^{2} = \frac{49}{4}m$
$y_{2} - x_{1} = \frac{49}{4} - 12 = \frac{1}{4}m$.
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 — Centre of Mass — Collision
Difficulty: Moderate

Question: A ball of mass M collides elastically with another
identical ball at rest as shown in figure. Initially velocity of
ball After collision

1) Speed of ball B is u $\cos\theta$
2) Speed of ball A is u $\sin\theta$
3) Speed of ball B is u $\sin\theta$
4) None of these

Correct option: 3

Solution:
[IMAGE] Velocity along line of impact will get exchanged because both bodies are
identical. $v_{B} = ucos\theta$
$V_{A} = usin\theta$