Grass-G/jee-main-questions huggingface dataset:
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. [IMAGE]


Answer: 4

Solution:
Subject: Physics —  — 

Question: A uniform cylinder floats with its axis vertical with
[IMAGE] parts of its length in the upper liquid and
[IMAGE] part of its length in lower liquid. If the
densities of upper and lower liquids are [IMAGE] then
the density of the material of the cylinder is


Correct option: 1

Solution:
Let [IMAGE] be the be the length of the cylinder
then
 is the density of the cylinder.
TOPIC: Hydrostatics
SUB TOPIC:Bouyancy
LEVEL: Moderate
Subject: Physics — Thermodynamics — P-V diagram and first law of thermodynamics
Difficulty: Moderate

Question: P-V diagram of ideal monoatomic gas is as shown. Find net heat
given to the gas in the process ABC.


Correct option: 4

Solution:
Subject: Physics — Current Electricity — Meter bridge
Difficulty: Moderate

Question: n a meter bridge, the wire of length 1m has a non-uniform
cross-section such that, the variation$\frac{dR}{d^{\mathcal{l}}}$ of
its resistance R with
length$\mathcal{l}$ is $\frac{dR}{d\mathcal{l}} \propto \frac{1}{\sqrt{\mathcal{l}}}$.
Two equal resistances are connected as shown in thefigure. The
galvanometer has zero deflection when the jockey is at point P. What is
the length AP ?

1) 0.2 m
2) 0.3 m
3) 0.25 m
4) 0.35 m

Correct option: 3

Solution:
According to Question
$\int_{0}^{\mathcal{l}}\mspace{2mu} c\frac{d\mathcal{l}}{\sqrt{\mathcal{l}}} = \int_{\mathcal{l}}^{1}\mspace{2mu} c\frac{d\mathcal{l}}{\sqrt{\mathcal{l}}}$
$(2\sqrt{\mathcal{l}})_{0}^{\mathcal{l}} = (2\sqrt{\mathcal{l}})_{t}^{1}$
$2\sqrt{\mathcal{l}} = 2 - 2\sqrt{\mathcal{l}}$
$2\sqrt{\mathcal{l}} = 2 - 2\sqrt{\mathcal{l}}$
$4\sqrt{\mathcal{l}} = 2$
$\mathcal{l =}\frac{1}{4} = 0.25m$.
Subject: Physics —  — 
Difficulty: Tough

Question: A block of weight 10 N produces an extension of 9 cm when it is
hung by an elastic spring of length 60 cm and is in equilibrium. The
spring is cut into two parts, one of length 40 cm and the other of
length 20 cm. Another weight of 20 N hangs in equilibrium supported by
both parts as shown in the figure. Find the total extension of the two
parts of the spring (in cm) now.
10N 60 cm 9cm, 40
cm 20 cm 20N 
 (cm )


Answer: 04

Solution:
10 = k ×[IMAGE] k =[IMAGE] k1 × 4 = k × 60 =[IMAGE] × 60
k1 =[IMAGE] k2 × 20 = k × 60 ⇒ k2 =[IMAGE] kq = k1 + k2 = 500
10 = 500 x ⇒ x =[IMAGE] m = 2 cm
TOPIC: NLM
SUB TOPIC: SPRING
Subject: Physics — Communication system — Modulation
Difficulty: Moderate

Question: The maximum peak to peak voltage of an AM wave is 24 mV and the
minimum peak to peak voltage is8 mV. The modulation factor is.

1) 10
2) 20
3) 25
4) 50

Correct option: 4

Solution:
Here ¼;gk¡½, $V_{\max} = \frac{24}{2}$=12mV and¼rFkk½
$V_{\min} = \frac{8}{2} = 4mV$
Subject: Physics — Rigid Body Dynamics — Kinematics in rotation
Difficulty: Easy

Question: A particle moving on a circular path travels first one third
part of circumference in 2 sec & nextone third part in 1 sec. Average
angular velocity of the particle is $\frac{n\pi}{9}$ (in rad/sec). Then
valueof n.


Answer: 4

Solution:
$\omega_{avg} = < \omega > = \frac{\ total\ \theta}{\ total\ \ time\ }$
Subject: Physics — Electrostatics — Electric field due to infinite sheet of charge
Difficulty: Tough

Question: Two infinite planes each with uniform surface charge density
$+ \sigma$ are kept in such a way that the anglebetween them is 30°. The
electric field in the region shown between them is given by
vuur yEckbZ vkSj pkSM+kbZ okys nks leryksa

1) $\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$
2) $\frac{\sigma}{2\epsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$
3) $\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$
4) $\frac{\sigma}{\epsilon_{0}}\left\lbrack \left( 1 + \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

Correct option: 2

Solution:
$\overset{\rightarrow}{E} = \frac{\sigma}{2\varepsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{1}{2}\overset{\hat{}}{x} \right\rbrack$.
Subject: Physics — Thermodynamics — Heat engine
Difficulty: Moderate

Question: A coal based thermal power plant producing electricity operates
between the temperatures 27ºC and227ºC. The plant works at 80 of its
maximum theoretical efficiency. Complete burring of 1 kg of coalyields
3600 KJ of heat. A house needs 10 units of electricity each day. Coal
used for supplying theamount of energy for the house in one year is,d 1 kg iw.kZ tyus ij 3600 KJ mRiUu gksrh,d o"kZ rd nsus

1) 1141 kg
2) 580 kg
3) 605 kg
4) 765 kg

Correct option: 1

Solution:
Maximum efficiency of engine
$= \left( 1 - \frac{T_{2}}{T_{1}} \right) \times 100 = 40\%$
$= \left( 1 - \frac{T_{2}}{T_{1}} \right) \times 100 = 40\%$
$\therefore$overall efficiency = 0.4 x 0.8 = 0.32
$\therefore$dqy n{krk = 0.4 x 0.8 = 0.32
Requirement of house in one year,d lky esa ?kj 
= 10 × 365 units = 3650 kwh = 3650 × 3600 kJ
$\therefore$ in put =$\frac{3650 \times 3600}{0.32}$kJ
$\therefore$ coalrequired =$\frac{\ In\ put\ }{36000}$=1141kg
Subject: Physics —  — 

Question: In a Young's double slit experiment sources of equal
intensities are used. Distance between slits is d and wavelength of
light used is λ(λ << d). Angular separation of the nearest points on
either side of central maximum where intensities becomes half of the
maximum value is


Correct option: 2

Solution:
[IMAGE] I + I + 2 [IMAGE] cos φ
2I = 2I + 2I cos φ 
cos φ = 0
φ = [IMAGE] ∆x = [IMAGE] x = [IMAGE] θ =[IMAGE]
Subject: Physics — Mechanical properties of solid — Thermal expansion
Difficulty: Moderate

Question: A thin metal rod of length L_0 is shaped into a ring with a
small gap x as shown. On heating the system

1) x decreases, r and d increase
2) x and r increase, d decreases
3) x, r and d all increase
4) Data insufficient to arrive at a conclusion
L_0 yackbZ dh,d iryh NM+ dks vYi varjky x

Correct option: 3

Solution:
On uniform heating of an object which is free to expand,
distance between any two points increases.
oLrq tks çlkfjr gksus ds fy, LorU= gS]
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

Question: A uniform disc of mass m and radius R is resting on a smooth
horizontal surface and it is free to rotateabout its own axis which is
vertical. A small particle of mass m collides its periphery with a
velocity of Vin tangential direction. If the collision is perfectly in
elastic and the particle sticks to the periphery, theangular velocity of
the system just after the collision will be

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

Correct option: 2

Solution:
Applying angular momentum conservation about the rotating axis
L_i = L_f
O + mv R= $\left( \frac{mR^{2}}{2} + mR^{2} \right)\omega_{f}$
$\omega = \frac{2V}{3R}$
Subject: Physics — Kinematics — Relative motion
Difficulty: Moderate

Question: Two particle A and B are located in x-y plane at points (0,0)
and (0,4 m). They simultaneously start moving with velocities.
${\overset{\rightarrow}{v}}_{A} = 2\overset{\hat{}}{j}m/s$and${\overset{\rightarrow}{v}}_{B} = 2\overset{\hat{}}{i}m/s$.
Select the correct alternative(s).

1) Time after which they are at minimum distance is 1 s
2) The distance between them first decreases and then increases
3) The shortest distance between them is $2\sqrt{2}m$
4) All of the above

Correct option: 4

Solution:
${\overset{\rightarrow}{v}}_{AB} = {\overset{\rightarrow}{v}}_{A} - {\overset{\rightarrow}{v}}_{B} = 2(j - \overset{\hat{}}{i})m/s\ \ \ or\ |{\overset{\rightarrow}{v}}_{AB}| = 2\sqrt{2}m/s$
Assuming B to be at rest, A will move with velocity
${\overset{\rightarrow}{v}}_{AB}$ in the direction shown in figure.
The distance between them will first decrease from A to Cand then
increase beyond C.
[IMAGE] Minimum distance between them is BC which is equal to
$\frac{4}{\sqrt{2}}$ or $2\sqrt{2}$ and the time after which they are
at closest distance is:
$t = \frac{AC}{|{\overset{\rightarrow}{v}}_{AB}|} = \frac{2\sqrt{2}}{2\sqrt{2}} = 1s$
Subject: Physics — Thermodynamic — Thermodynamic processes
Difficulty: Moderate

Question: A certain quantity of oxygen [IMAGE] is
compressed isothermally until its pressure is doubled
[IMAGE]. The gas is then allowed to expand adiabatically
until its original volume is restored. Then the final pressure(P) in
terms of initial pressure [IMAGE] is:
vkWDlhtu,d izkjfEHkd nkc

4) None of these

Correct option: 2

Solution:
[IMAGE] (isothermal process) ¼lerkih
Subject: Physics —  — 
Difficulty: Moderate

Question: Radiation, with wavelength [IMAGE] falls on a
metal surface to produce photoelectrons. The electronsare made to enter
a uniform magnetic field of[IMAGE]. If the radius of the
largest circular pathfollowed by the electrons is 10 mm, the work
function of the metal is closest to what integral value in eV


Answer: 1

Solution:
[IMAGE] (in eV)
[IMAGE] (in eV)
[IMAGE] TOPIC:Dual nature of particle
SUB TOPIC:Photoelectric effect
Subject: Physics —  — 

Question: A sound wave of frequency n travels horizontally to right. It is
reflected from a large vertical wall moving to the left with a speed V.
The speed of sound in medium is C. Then

1) the frequency of reflected wave for stationary listener is
2) the frequency of reflected wave for stationary listener is
3) the frequency of reflected wave for stationary listener is
4) the number of beats heard by stationary listener to the left of
reflecting wall is [IMAGE]

Correct option: 2

Solution:
[IMAGE] (number of waves striking the wall per sec)
[IMAGE] (frequency observed by stationary listener)
TOPIC:Sound waves
SUB TOPIC:Doppler effect
LEVEL:Moderate
2. In YDSE arrangement, white light is used to illuminate the slits. At
point on the screen directly in front of slits, which of these
wavelength is missing?
([IMAGE] = wavelength of light used)
(d = distance between slits)
(D = distance between slits and screen)
(given d << D)
Subject: Physics — Elasticity — Potential energy stored in elongated rubber cord
Difficulty: Moderate

Question: A rubber cord has a cross-section area 1 cm^2 and natural
length of [IMAGE] It is stretched by
[IMAGE] to fire a small object of
mass[IMAGE]. If the Young's modulus (Y) is
[IMAGE] Speed acquired by the object in (m/s) is 2n. Find
n. Assume Hook's law is valid and neglect gravity.,d jcj dh Mksjh dk vuqçLFk dkV kjk çkIr
osx (m/s) esa½2n.nKkr djsaA ekfu,sa fd gqd


Answer: 5

Solution:
v = 10^3 × 10^-2
v = 10.00 m/s
Subject: Physics —  — 

Question: A small ball of mass 1 kg strikes a wedge of mass 4kg
horizontally with a velocity of 10 m/s. Friction is absent everywhere
and collision is elastic. Select the correct answers

1) The linear momentum of the ball along common normal and common
tangent will not change during the collision.
2) The linear momentum of the ball along common normal and common
tangent will change during the collision.
3) The linear momentum of the ball along common normal will not
change but remain same along common tangent during the collision.
4) The linear momentum of the ball along common normal will change
but remain same along common tangent during the collision.

Correct option: 4

Solution:
Since impulse acts along common normal the linear momentum of the
ball along common normal will change. There is no impulse along common
tangent therefore linear momentum of the ballwill not change during the
collision.
TOPIC: Collision
SUB TOPIC:Elastic collision
LEVEL:Moderate
Subject: Physics — Motion in 1D — Motion under gravity
Difficulty: Moderate

Question: A particle is projected vertically upwards. It will be at
$\left( \frac{3}{4} \right)$^th of its greatest height at time
$t = t_{1}$and $t = t_{2}$. $\left( t_{1} < t_{2} \right)$. Then the
ratio $\left( t_{1}:t_{2} \right)$ is

1) $\frac{1}{3}$
2) $\frac{1}{2}$
3) $\frac{1}{4}$
4) $\frac{1}{9}$

Correct option: 1

Solution:
$H = \frac{u^{2}}{2g}$;$\frac{3}{4}\left( \frac{u^{2}}{2g} \right) = ut - \frac{1}{2}gt^{2} \Rightarrow t = \frac{u}{g} \pm \frac{u}{2g}$
$\Rightarrow t_{1} = \frac{u}{2g}$and$t_{2} = \frac{3u}{2g} \Rightarrow t_{1}:t_{2} = 1:3$
Subject: Physics —  — 

Question: For D1 forward bias resistance is 20 Ohm and reverse bias
resistance is 600 Ohm. For D2 forward bias resistance is 30 Ohm and
reverse bias resistance is 200 Ohm. Find the equivalent resistance
between A and B if (i) A is at a higher potential (ii) B is at a higher
potential

3) 50 Ω and 800 Ω
4) 220 Ω and 630 Ω

Correct option: 2

Solution:
Subject: Physics — Magnetics — Cross field
Difficulty: Tough

Question: A charged particle of mass moving under
the influence of uniform electric field $E\overset{\rightarrow}{i}$and a
uniform magnetic field $B\overset{\rightarrow}{k}$follows a trajectory
from point P to Q as shown in figure. The velocitiesat P and Q are
respectively,
$v\overset{\rightarrow}{i}$and$- 2v\overset{\rightarrow}{j}$. Then which
of the following statements (A, B, C, D) are thecorrect ? (Trajectory
shown is schematic and not to scale)

1) $E = \frac{3}{4}(\frac{mv^{2}}{qa})$
2) Rate of work done by the electric field at P
is$\frac{3}{4}\left( \frac{{mv}^{3}}{a} \right)$and at Q is zero.
3) The difference between the magnitude of angular momentum of the
particle at P and Q is 2mav
4) . All of the above

Correct option: 4

Solution:
(A) by work energy theorem
$0 + {qE}_{0}2a = \frac{3}{2}{mv}^{2}$
$$E = 
(B) Rate of work done at A= power of electric force
$= {qE}_{0}V$
$= \frac{3}{4}\frac{{mv}^{3}}{a}$
(C) at$Q,\frac{dW}{dt} = 0$for both forces
(D)
$\Delta\overset{\rightarrow}{L} = ( - m2v2a\overset{\hat{}}{k}) - ( - mva\overset{\hat{}}{k})$
$|\Delta\overset{\rightarrow}{L}| = 3mva$
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

Question: A uniform rod of mass 200 gram and length
[IMAGE] is initially at rest in vertical position. The rod
is hinged at centre such that it can rotate freely without friction
about a fixed horizontal axis passing through its centre. Two particles
of mass m = 100 gram each having horizontal velocity of equal magnitude
u = 17 m/s strikes the rod at top and bottom simultaneously as shown and
stick to the rod.
The angular speed (in rad/s) of rod when it becomes horizontal is 8.5n.
Find n


Answer: 9

Solution:
From conservation of angular momentum.
$mu\frac{L}{2} + mu\frac{L}{2} = \left\lbrack 2m\frac{L^{2}}{12} + m\left\{ \left( \frac{L}{2} \right)^{2} + \left( \frac{L}{2} \right)^{2} \right\} \right\rbrack\omega$
$muL = \left\lbrack \frac{{mL}^{2}}{6} + \frac{{mL}^{2}}{4} + \frac{{mL}^{2}}{4} \right\rbrack\omega$
$= \frac{2{mL}^{2}}{3}\omega$
or $\omega = \frac{3u}{2L}$
$= 76.50$
Subject: Physics — Magnetics — Cross field
Difficulty: Tough

Question: A charged particle of mass moving under
the influence of uniform electric field $E\overset{\rightarrow}{i}$and a
uniform magnetic field $B\overset{\rightarrow}{k}$follows a trajectory
from point P to Q as shown in figure. The velocitiesat P and Q are
respectively,
$v\overset{\rightarrow}{i}$and$- 2v\overset{\rightarrow}{j}$. Then which
of the following statements (A, B, C, D) are the correct ? (Trajectory
shown is schematic and not to scale)

1) $E = \frac{3}{4}(\frac{mv^{2}}{qa})$
2) Rate of work done by the electric field at P
is$\frac{3}{4}\left( \frac{{mv}^{3}}{a} \right)$and at Q is zero.
3) The difference between the magnitude of angular momentum of the
particle at P and Q is 2mav
4) . All of the above

Correct option: 4

Solution:
(A) by work energy theorem
$0 + {qE}_{0}2a = \frac{3}{2}{mv}^{2}$
$$E = 
(B) Rate of work done at A= power of electric force
$= {qE}_{0}V$
$= \frac{3}{4}\frac{{mv}^{3}}{a}$
(C) at$Q,\frac{dW}{dt} = 0$for both forces
(D)
$\Delta\overset{\rightarrow}{L} = ( - m2v2a\overset{\hat{}}{k}) - ( - mva\overset{\hat{}}{k})$
$|\Delta\overset{\rightarrow}{L}| = 3mva$
Subject: Physics — AC circuit — LCR circuit
Difficulty: Moderate

Question: A LCR circuit behaves like a damped harmonic oscillator.
Comparing it with a physical spring-mass damped oscillator having
damping constant the correct equivalence would be

1) $L \leftrightarrow \frac{1}{b},C \leftrightarrow \frac{1}{m},R \leftrightarrow \frac{1}{k}$
2) $L \leftrightarrow k,C \leftrightarrow b,R \leftrightarrow m$
3) $L \leftrightarrow m,C \leftrightarrow \frac{1}{k},R \leftrightarrow b$
4) $L \leftrightarrow m,C \leftrightarrow k,R \leftrightarrow b$

Correct option: 3

Solution:
In damped oscillation
$ma + bv + kx = 0$
[IMAGE] $m\frac{d^{2}x}{{dt}^{2}} + b\frac{dx}{dt} + kx = 0$....(i)
In the circuit
$- iR - L\frac{di}{dt} - \frac{q}{c} = 0$
$L\frac{d^{2}q}{{dt}^{2}} + R\frac{dq}{dt} + \frac{1}{c} \cdot q = 0$....(ii)
Comparing equation (i) and (ii)
$m = L,b = R,k = \frac{1}{c}$
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

Question: A non-viscous fluid of density ρ is flowing in a tube as shown
in figure. Area of section

1) is double that of section
2) . Centre
of mass of section
3) Work done by gravitational force per unit volume from section
4) Work done by elastic forces (pressure) per unit volume from
section

Correct option: 2

Solution:
Applying Bernoulli is equation from section-(1) and (2)
[k.M-(1).M-(2) esa cjukWyh lehdj.k yxkus ij
[IMAGE] andrFkkP_1- P_2= $\rho$g(2h)
Solving we get, gy djus ij çkIr gksxk
$V = \sqrt{\frac{2gh}{3}}$
(C) Work done by gravitation force per unit value
[IMAGE] decrease in gravitational
PT_value = $\rho{gh}_{1} - \rho{gh}_{2}$
 
PT_value = $\rho{gh}_{1}\ - \rho{gh}_{2}$
$W_{gr} = 0 - \rho gh$
(D) Work done by elastic force volume, çR;kLFk cyksa }kjk çfr bdkbZ
W_e=decrease in elastic P.E._vol = decrease in pressure
W_e=çR;kLFk fLFkfrt P.E._vol= nkc
Subject: Physics — Center of mass — Collision
Difficulty: Moderate

Question: A small smooth disc of mass m and radius r moving with an initial
velocity along the positive x-axis collided with a big disc of mass
2 m and radius 2 r which was initially at rest with its centre at origin
as show in figure.
[IMAGE] If the coefficient of restitution is 0 then velocity of larger disc
after collision is

1) $\frac{8v}{27}\overset{\hat{}}{i} - \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$
2) $\frac{8v}{27}\overset{\hat{}}{i} + \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$
3) $\frac{v}{3}\overset{\hat{}}{i}$
4) $\frac{2\sqrt{2}}{27}v\overset{\hat{}}{i} - \frac{8v}{27}\overset{\hat{}}{j}$

Correct option: 1

Solution:
[IMAGE] The larger sphere will move along line of impact.
e=0, magnitude of velocity of larger sphere
$v^{'} = \frac{mvcos\theta}{m + 2m} = \frac{vcos\theta}{3}$
Cos$\theta$= $2\sqrt{2}/3$
Sin $\theta$= 1/3
Velocity of larger sphere
$= v^{'}cos\theta\overset{\hat{}}{i} - v^{'}sin\theta\overset{\hat{}}{j}$
$= \frac{v}{3}\cos^{2}\theta\overset{\hat{}}{i} - \frac{v}{3}sin\theta cos\theta\overset{\hat{}}{j} = \frac{8}{27}v\overset{\hat{}}{i} - \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$
Subject: Physics — Newton's Laws of Motion — Friction
Difficulty: Moderate

Question: Initially spring is in its natural length and blocks A and B are
at rest. Find maximum value of constant force F that can be applied on B
such that block A remains at rest. (g = 10 ms-2).

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

Correct option: 2

Solution:
Value of $\mu mg$ for A and B are 2N and 2N.
Work energy theorem on B gives
Fx - 2x - (k/2) x^2 = 0
F = 2 + kx/2....................(i)
For equilibrium of A, kx = $\mu mg = 2$
So, F = 2 + 1 = 3N
Subject: Physics —  — 
Difficulty: Moderate

Question: In the system shown in the figure the friction and mass of rope is
negligible, then acceleration of 2 m is

3) 0
4) [IMAGE], 2 m

Correct option: 3

Solution:
2mg - 2T = 2ma
[IMAGE] 2 [T - mg = m (2a)]
0 = 6 ma ∴ a = 0
TOPIC: NLM
SUB TOPIC: EQUILIBRIUM
Subject: Physics — Mechanical Properties of Fluids — Surface tension
Difficulty: Moderate

Question: In a cylinder-piston arrangement, air is under a pressure P_1.
A soap bubble of radius r lies inside the cylinder, soap bubble has
surface tension T. The radius of soap bubble is to be reduced to half,
The new pressure P_2 to which air should be compressed isothermally.
(Assume r is very small as compared to height of cylinder)

1) $\ P_{1} + \frac{4T}{r}$
2) $\ 4P_{1} + \frac{12T}{r}$
3) $\ 8P_{1} + \frac{24T}{r}$
4) $\ P_{1} + \frac{2T}{r}$

Correct option: 3

Solution:
Isothermal process.
$\left( P_{1} + \frac{4T}{r} \right)\left( \frac{4}{3}\pi r^{3} \right) = \left( P_{2} + \frac{4T}{r/2} \right)\left( \frac{4}{3}\pi(r/2)^{3} \right)$
$P_{2} = 8P_{1} + \frac{24T}{r}$
Subject: Physics —  — 

Question: A man sitting inside a train moving with uniform acceleration
[IMAGE]. Man throws a ball from point O (on the floor of
train) with relative velocity [IMAGE] (where
[IMAGE] = velocity of ball with respect to train
[IMAGE] Ball falls on the train at point A (where OA = R) after time T.
Choose correct option


Correct option: 1

Solution:
[IMAGE] Time of flight = T = $\frac{2\ (20\sqrt{}2)sin45{^\circ}}{10} = 4s$
Solving we get[IMAGE] TOPIC:Projectile motion
SUB TOPIC:Relative motion
LEVEL:Moderate
Subject: Physics —  — 

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

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

Correct option: 1

Solution:
From (1) and (2)
[IMAGE] TOPIC:Structure of Atom
SUB TOPIC:Bohr's model of hydrogen
LEVEL:Moderate
Subject: Physics — Work Power & Energy — Average power
Difficulty: Tough

Question: A uniform rope of linear mass density $\lambda$ and length 1 is
coiled on a smooth horizontal surface. One end is pulled up with
constant velocity v. Then the average power applied by the external
agent in pulling the entire rope just off the ground is

1) $\frac{1}{2}\lambda\mathcal{l}v^{2} + \frac{\lambda\mathcal{l}^{2}g}{2}$
2) $\lambda\mathcal{l}gv$
3) $\frac{1}{2}\lambda v^{3} + \frac{\lambda\mathcal{l}vg}{2}$
4) $\lambda\mathcal{l}gv + \frac{1}{2}\lambda v^{3}$

Correct option: 3

Solution:
Initial K.E. = 0
Initial P.E. = 0
When the rope is just pulled off the table.
Final K.E. = $\frac{1}{2}(\lambda l)v^{2}$
Final P.E$= \frac{\left( \lambda\mathcal{l} \right)g\mathcal{l}}{2}$
Time taken $t = 1/v$
Average power = $\frac{\text{ net change in energy }}{\text{ time }}$
$= \frac{\frac{1}{2}\lambda\mathcal{l}v^{2} + \lambda\mathcal{l}g\mathcal{l/}2}{\mathcal{l/}v} = \frac{1}{2}\lambda v^{3} + \frac{\lambda\mathcal{l}vg}{2}$
Subject: Physics — Motion in 1D — Uniform accelerated motion
Difficulty: Moderate

Question: A particle moves along x-axis with initial position x=0. Its
velocity varies with x-coordinate as shown in graph. The acceleration
'a" of this particle varies with x as


Correct option: 1

Solution:
The linear relationship between V and x is $V = - mx + C$
where m and C are positive constants.
Acceleration
$a = v\frac{dV}{dx} = - m( - mx + C)$
Hence the graph relating a tox is.
Subject: Physics —  — 

Question: A particle is projected vertically upward from the surface of
the earth with a speed of [IMAGE], R being the radius
of the earth and g is the acceleration due to gravity on the surface of
the earth. Then the maximum height ascended is nR,then value of n
is(neglect cosmic dust resistance)


Answer: 03

Solution:
[IMAGE] and g = [IMAGE] Therefore [IMAGE] ×[IMAGE] × gR - Rg =
- [IMAGE] We get h = 3R