Grass-G/jee-main-questions huggingface dataset:
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 — Behaviour of perfect gases — Ideal Gas
Difficulty: Moderate

Question: During an experiment an ideal gas is found to obey condition
gas). The gas is initially at a temperature [IMAGE],
pressure [IMAGE] and density [IMAGE].
The gas expands such that density changes to [IMAGE] then which of following is/are correct?,d iz;ksx

1) The pressure of gas changes to [IMAGE]
2) The temperature of gas changes to [IMAGE].
3) The graph of above process on the P - T diagram is parabola.
4) The graph of above process on the P - T diagram is not rectangular
hyperbola.

Correct option: 2

Solution:
PT = constant
Subject: Physics — Newton's Laws of Motion — Newton's second law for a system
Difficulty: Moderate

Question: Two blocks of masses $M_{1}$ and $M_{2}$are connected to each
other through a light spring as shown in figure. If we push mass $M_{1}$
with force F and cause acceleration $a_{1}$ in mass $M_{1}$, what will
be the magnitude of acceleration in $M_{2}$?

1) $F/M_{2}$
2) $F/\left( M_{1} + M_{2} \right)$
3) $a_{1}$
4) $\left( F - M_{1}a_{1} \right)/M_{2}$

Correct option: 4

Solution:
$kx = M_{2}a_{2}$ [Newton's II law for $M_{2}$]
By adding both equations.
$F = M_{1}a_{1} + M_{2}a_{2} \Rightarrow a_{2} = \frac{F - M_{1}a_{1}}{M_{2}}$
Subject: Physics —  — 

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
velocity[IMAGE]. The emf induced (in volts) is the loop
at t = 2 sec is 6V. Then magnitude of 4B is

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

Correct option: 3

Solution:
[IMAGE] TOPIC:Electromagnetic induction
SUB TOPIC:Faraday's law
LEVEL:Moderate
Subject: Physics — Mechanical Properties of Fluids — Velocity of efflux
Difficulty: Easy

Question: The speed of water flowing out of the orifice before the
cylinder kept inside the tank
csyu dks VSd iwoZ fNnz


Correct option: 1

Solution:
Subject: Physics — Gravitation — Satellite orbiting around earth
Difficulty: Moderate

Question: A satellite is orbiting around the earth in a circular orbit
very close to surface of earth then comparing with other satellites of
same mass moving in circular orbits around the earth.

1) The total energy of earth plus given satellite system is minimum
2) The time period of revolution of given satellite is minimum
3) The orbital speed of given satellite is maximum
4) All of the above

Correct option: 4

Solution:
$v_{0} = \sqrt{\frac{GM}{(R + h)}}$
$T = \frac{2\pi(R + h)}{v_{0}} = \frac{2\pi(R + h)^{3/2}}{\sqrt{GM}}$
$E = \frac{- GMm}{2(R + h)}$
Subject: Physics — Ray Optics — Optical instruments
Difficulty: Moderate

Question: The magnifying power of a telescope with tube length 60 cm is5.
What is the focal length of its eye piece?

1) 30 cm
2) 10 cm
3) 40 cm
4) 50 cm

Correct option: 2

Solution:
Subject: Physics — Capacitance — Dielectric
Difficulty: Moderate

Question: In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K=2. The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
the capacitor have area each. The energy of the capacitor is

1) $\frac{\epsilon_{0}AV^{2}}{3d}$
2) $\frac{2\epsilon_{0}AV^{2}}{d}$
3) $\frac{3}{2}\frac{\epsilon_{0}AV^{2}}{d}$
4) $\frac{2\epsilon_{0}AV^{2}}{3d}$

Correct option: 4

Solution:
$U = \frac{1}{2}C_{eq}V^{2}$
$C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}$
and $C_{2} = \frac{\varepsilon_{0}\ A}{d/2}$
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \frac{(2\frac{\varepsilon_{0}}{d/2}) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}\ A}{d/2}}$
$= \frac{4}{3}\frac{\varepsilon_{0}\ A}{d}$ or
$C_{eq} = \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}$
$U = \frac{1}{2}(\frac{4}{3}\frac{\varepsilon_{0}A}{d}) \vee^{2} = \frac{2}{3}(\frac{\varepsilon_{0}A}{d})v^{2}$
Subject: Physics —  — 
Difficulty: Moderate

Question: The position of a projectile launched from the origin at t = 0 is
given by [IMAGE] at
t = 2s. If the projectile was launched at an angle θ from the
horizontal, then θ is (take g=10 ms-2)

4) [IMAGE] t = 0 t = 2s 
  θ  θ  (g = 10 ms-2 )

Correct option: 1

Solution:
[IMAGE] 40 = Sx = V cos θ × t
50 = Sy = V sin θt - [IMAGE] gt2
t = 2 given
θ = tan-1 [IMAGE] TOPIC: KINEMATIC
SUB TOPIC: 2 D
Subject: Physics —  — 
Difficulty: Moderate

Question: A massless thread passes through a ring (smooth). The ring is in a
circular path with both part of the thread in tension. Which of the
following values of θ and α are possible?

1) θ > α
2) α > θ
3) θ = α
4) All of these
 () 
 θ α ?

Correct option: 2

Solution:
Tension will be same in both parts.
[IMAGE] Tcosθ = mg + Tcosα
⇒ α > θ
Subject: Physics —  — 
Difficulty: Moderate

Question: A particle is projected upwards with velocity v = 30 m/s. There are
two points A and B at heights hA = 25 m and hB = 40m. If the particle
crosses point B twice in a time interval ∆t. Then, it will also cross A
twice in a time interval equal to

1) 2∆t
2) 1.5 ∆t
3) 3∆t
4) 2.5 ∆t
 v = 30 m/s. hA = 25 hB = 40
 A B B ∆t 
 A

Correct option: 1

Solution:
hmax = [IMAGE] = 45m
Time to go from C to B,
5 = [IMAGE] gt2
t = 1s, So ∆t = 2s
Time to go from C to A,
20 =[IMAGE] gt2
t = 2s
∆tA = 2 × 2 = 4s = 2∆t
TOPIC: KINEMATIC
SUB TOPIC: MOTION UNDER GRAVITY
Subject: Physics — AC — LCR
Difficulty: Moderate

Question: In LC circuit the inductance L= 40 mH and
capacitance$C = 100\mu F$. If a voltage V(t)=10 sin (314 t) is applied
to the circuit, the current in the circuit is given as

1) $0.52\ cos314t$
2) $10\ cos314t$
3) $5.2\ cos314t$
4) $0.52\ sin314t$

Correct option: 1

Solution:
$R = 0$
$z = X_{C} - X_{L}$
$= \frac{1}{\omega C} - \omega L$
$= \frac{1}{314 \times 100 \times 10^{- 6}} - 314 \times 40 \times 10^{- 3}$
$= 31.84 - 12.56$
$= 19.28\Omega$
$X_{C} > X_{L}$
Subject: Physics — Mangetism — 
Difficulty: Moderate

Question: In the given diagram aproton is given a speed v at Pin the
horizontal direction then the time after which it comesout of the
magnetic field B. m is the mass of proton.

1) $\frac{m\pi}{eB}$
2) $\frac{2m\pi}{eB}$
3) $\frac{3m\pi}{4eB}$
4) It will never come out of the magnetic field

Correct option: 1

Solution:
Radius of charge particle
$r = \frac{mV}{eB}$
here $r < \frac{d}{2}$, proton will complete a semicircle and comes out
of the magnetic field.
Subject: Physics — Atomic Physics — Bohr's model of hydrogen atom
Difficulty: Moderate

Question: When an electron of hydrogen like atom jumps from a higher energy
level to lower energy level, then

1) angular momentum of the electron about nucleus remains constant
2) kinetic energy increases
3) de-Broglie wavelength, associated with motion of electron
increases
4) none of the above

Correct option: 2

Solution:
Angular momentum [IMAGE] Kinetic energy [IMAGE] De-Broglie wavelength [IMAGE] So as v increases [IMAGE] decreases.
Subject: Physics — Newton's Laws of Motion — Friction
Difficulty: Moderate

Question: In the figure shown a block A of mass m is kept on block B of
mass m. A is given a velocity towards right as shown. The coefficient of
friction between A and B is $\frac{3\mu}{2}$ and that between B and
ground is$\ \mu$.

1) B will accelerate towards right with acceleration$\frac{\mu g}{2}$.
2) B will accelerate towards left with acceleration$\frac{\mu g}{2}$
3) B will not accelerate at all.
4) B will move towards right with acceleration$\ \mu g$.

Correct option: 3

Solution:
as$f_{k^{'}\max} > \frac{3}{2}\mu g$
So it will not move
Subject: Physics —  — 
Difficulty: Moderate

Question: If the mass of block is 1 kg and a force of [IMAGE] N
is applied horizontally on the block as shown in the figure. The
frictional force acting on the block is

1) zero
4) 5 N
 1 kg [IMAGE]

Correct option: 1

Solution:
Net force along the plane
= mg sin 30 - FH cos 30
[IMAGE] So friction force is zero.
TOPIC: NLM
SUB TOPIC: FRICTION
Subject: Physics — Kinetic Theory of Gases — RMS speed
Difficulty: Tough

Question: A container of fixed volume has a mixture of one mole of hydrogen
and one mole of helium in equilibrium at temperature T. Assuming the
gases are ideal, the correct statement(s) is (are)

1) The average energy per mole of the gas mixture is 2RT.
2) The ratio of speed of sound in the gas mixture to that in helium
gas is $\sqrt{6/5}$
3) The ratio of the rms speed of helium atoms to that of hydrogen
molecules is $1/\sqrt{2}$
4) $\ $All of the above

Correct option: 4

Solution:
$\ $For hydrogen,$\ n_{1} = 1,C_{v,} = \frac{5}{2}R$
 For mixture of gases,
$C_{v} = \frac{n_{1}C_{V,} + n_{2}C_{V_{2}}}{n_{1} + n_{2}}$
$= \frac{1 \times \frac{5}{2}R + 1 \times \frac{3}{2}R}{1 + 1} = 2R$
$C_{P} = C_{v} + R = 3R,\gamma_{mix} = \frac{C_{P}}{C_{v}} = \frac{3}{2}$
Also, $\gamma_{mx} = 1 + \frac{2}{f} \Rightarrow \frac{3}{2} = 1 + \frac{2}{f}\therefore f = 4$
$\therefore$ Average energy per mole $= \frac{1}{2}fRT = 2RT$
$M_{\max} = \frac{n_{1}M_{1} + n_{2}M_{2}}{n_{1} + n_{2}} = \frac{1 \times 2 + 1 \times 4}{1 + 1} = 3gm{ol}^{- 1}$
 Speed of sound in a gas, $v = \sqrt{\frac{\gamma RT}{M}}$
For a given value of T,v$\propto \sqrt{\frac{\gamma}{M}}$
 Rms speed of a gas molecule at temperature T is given by 
$T,v \propto \sqrt{\frac{\gamma}{M}}$
 For a given value of T,$\ v_{rms} \propto \frac{1}{\sqrt{M}}$
$\frac{\left( v_{rms\ } \right)He}{\left( v_{rms\ } \right)H_{2}} = \sqrt{\frac{M_{H_{2}}}{M_{He}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$
 So, options (a), (b) and (c) are correct
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: In a Young's double slit experiment 15 fringes are observed on a
small portion of the screen when light of wavelength 500 nm is used. Ten
fringes are observed on the same section of the screen when another
light source of wavelength $\lambda$ is used Then the value of
$\\lambda\$ is (in nm)______.

1) 850 nm
2) 650 nm
3) 750 nm
4) 550 nm

Correct option: 3

Solution:
$15 \times 500 \times \frac{D}{d} = 10 \times \lambda_{2} \times \frac{D}{d}$
$\lambda_{2} = 15 \times 50nm$
$\lambda_{2} = 750nm$
Subject: Physics — Centre of Mass — Impulse momentum equation
Difficulty: Moderate

Question: A particle of mass m is made to move with uniform speed v along
the perimeter of a regular polygon of n sides inscribed in a circle of
radius a. The magnitude of impulse applied at each corner of the polygon
is

1) $2\ mv\ sin\frac{\pi}{n}$
2) $mv\ sin\frac{\pi}{n}$
3) $mv\ sin\frac{2\pi}{n}$
4) $2mv\ sin\frac{n}{\pi}$

Correct option: 1

Solution:
[IMAGE] Magnitude of the impulse at each corner is simply the change in momentum
of the particle.
Magnitude of final momentum = mV
Magnitude of initial momentum = mV
Angle between two momentums =[IMAGE] Hence,
$\left| {\overset{\rightarrow}{P}}_{final\ } - {\overset{\rightarrow}{P}}_{initial\ } \right| = |mpulse = \left| \sqrt{P^{2} + P^{2} - 2P^{2}\cos\left( \frac{2\pi}{n} \right)} \right|$
$\Rightarrow$ Impulse $\left. \ = mV \mid \sqrt{2\left( 1 - cos\left( \frac{2\pi}{n} \right) \right.\ } \right)| = mv|\sqrt{4\sin^{2}\left( \frac{\pi}{n} \right)} \mid = 2mvsin\frac{\pi}{n}$
Subject: Physics — Mechanics — Angular impulse
Difficulty: Moderate

Question: A uniform rod of length L and mass M is placed on a smooth
horizontal surface. The rod is hinged at the end A and is free to rotate
in horizontal plane about a vertical axis passing through A. As shown in
figure there is a nail N at a perpendicular distance $\frac{L}{4}$ from
end A of rod. At t = 0 and impulse P_0 is applied at a distance
$\frac{3L}{4}$ from end A. The impulse is in the horizontal plane and is
perpendicular to the rod. At $t = \frac{8\pi ML}{27P_{0}}$, the rod
returns to its initial position. Find the impulse of the force exerted
by the nail on the rod.

1) $11P_{0}$
2) $12P_{0}$
3) $10P_{0}$
4) $13P_{0}$

Correct option: 2

Solution:
$P_{0}\frac{3L}{4} = \frac{M\mathcal{l}^{2}}{3}\omega_{0}$
$t_{1} = \frac{\pi}{2\omega_{0}}$
$t_{2} = \frac{8\pi ML}{27P_{0}} - t_{1}$
$\omega = \frac{\pi}{2t_{2}}$
$P\frac{L}{4} = \frac{M\mathcal{l}^{2}}{3}(\omega_{0} + \omega)$
$P = 12P_{0}$
Subject: Physics — Motion in 1D — Uniform acceleration
Difficulty: Easy

Question: The velocity time relation of an electron starting from rest is
given by v=K t, where K=2 m/s^2. The distance traversed in 3 sec is 3n
then the find the value of n.



Solution:
$a = \frac{dv}{dt} = k = 2m/s^{2}$
$S = 0 + \frac{1}{2}(2)(3)^{2} = 9m$
Subject: Physics — Simple Harmonic Motion — Amplitude of SHM
Difficulty: Moderate

Question: A horizontal spring-block system of mass 2kg executes S.H.M.
When the block is passing through its equilibrium position, an object of
mass 1kg is put gently on it and the two move together. The new
amplitude of vibration is (A being its initial amplitude)


Correct option: 1

Solution:
Conserving momentum: 2V = 3V′ [IMAGE]
Subject: Physics — Communication system — Modulation
Difficulty: Moderate

Question: The maximum peak to peak voltage of an AM wave is
[IMAGE] and the
minimum peak to peak voltage is[IMAGE]. The modulation factor is 10x. Find x.


Answer: 5

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

Question: A uniform rod of length L and mass M is placed on a smooth
horizontal surface. The rod is hinged at the end A and is free to rotate
in horizontal plane about a vertical axis passing through A. As shown
infigure there is a nail N at a perpendicular distance
[IMAGE] from end A of rod. At t = 0 and impulse
[IMAGE] isapplied at a distance [IMAGE] from end A. The impulse is in the horizontal plane and is perpendicular
to therod. At [IMAGE] the rod returns to its initial
position. If the impulse of the force exerted by the nail on
the rod is $\frac{36\text{P}_{0}}{\text{x}},$then value of x is


Answer: 3

Solution:
Subject: Physics — Rigid Body Dynamics — Pure rolling
Difficulty: Moderate

Question: A disc of radius R rolls on a horizontal surface with linear
velocity V and angular velocity[IMAGE]. There is a point
P on circumference of disc at angle [IMAGE] with upward
vertical diameter measured anticlockwise (see figure), which has a
vertical velocity. Here [IMAGE] is equal to.

1) $\pi \pm \sin^{- 1}\left( \frac{V}{\omega R} \right)$
2) $\pi \pm \cos^{- 1}\left( \frac{V}{\omega R} \right)$
3) $2\pi \pm \cos^{- 1}\left( \frac{V}{\omega R} \right)$
4) $\frac{3\pi}{2} \pm \cos^{- 1}\left( \frac{V}{\omega R} \right)$

Correct option: 2

Solution:
[IMAGE] For vertical velocity of point P
$\omega{Rcos}\theta = - V$
$\Rightarrow$
$\omega{Rcos}(\pi - \theta) = V\ or\ \omega{Rcos}(\pi + \theta) = V$
$\therefore$
$\theta = \pi + \cos^{- 1}\left( \frac{V}{\omega R} \right)$
or $\theta = \pi - \cos^{- 1}\left( \frac{V}{\omega R} \right)$
Subject: Physics — Electrostatics — Electric field due to electric dipole
Difficulty: Tough

Question: For a system of two
 as shown in the
figure (both are at origin and perpendicular toeach other along x and y
axes respectively) (P_1 and P_2denotes magnitudes
 and r is quite large in
comparison to dimensions of dipoles,
[IMAGE] is resultant electric field and QPR is a quarter
of circle whose centre is at O)
fp=kuqlkj nks
fy, ¼nksuksa ewy fcUnq ij gS rFkk,d nwljs ds yEcor~ gS rFkk Øe'k x
o y chp dh nwjh dh rqyuk esa r 
pkSFkkbZ Hkkx

1) Work done in taking electron from P to R on QPR = 0
P ls R rd QPR ij ys tkus
2) $tan\alpha = \frac{\left( P_{1} + P_{2} \right)}{2\left( P_{1} - P_{2} \right)}$
3) $tan\alpha = \frac{\left( P_{1} - P_{2} \right)}{2\left( P_{1} + P_{2} \right)}$

Correct option: 3

Solution:
$E_{1} = \frac{2k\frac{\left( P_{1} + P_{2} \right)}{\sqrt{2}}}{r3}$
$E_{2} = \frac{k\frac{\left( P_{1} - P_{2} \right)}{\sqrt{2}}}{r3}$
$tan\alpha = \frac{E_{2}}{E_{1}}$
$= \frac{P_{1} - P_{2}}{2\left( P_{1} + P_{2} \right)}$
Subject: Physics —  — 

Question: A uniform solid sphere of radius R and mass M purely rolls down
an inclined plane. The coefficient of friction between the sphere and
the inclined plane is[IMAGE]. If the maximum value of
angle of inclination for this to be possible is
[IMAGE] then n is: (Take[IMAGE] )

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

Correct option: 1

Solution:
[IMAGE] put in (1)
[IMAGE] Now [IMAGE] TOPIC:Rotaional dynamics
SUB TOPIC: Rolling down the incline
LEVEL: Moderate
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 — Kinetic Theory of Gases — RMS speed
Difficulty: Tough

Question: A container of fixed volume has a mixture of one mole of
hydrogen and one mole of helium in equilibrium at temperature T.
Assuming the gases are ideal, the correct statement(s) is (are)

1) The average energy per mole of the gas mixture is 2RT.
2) The ratio of speed of sound in the gas mixture to that in helium
gas is $\sqrt{6/5}$
3) The ratio of the rms speed of helium atoms to that of hydrogen
molecules is $1/\sqrt{2}$
4) $\ $All of the above,d ik= esa 1 eksy gkbMªkstu rFkk 1

Correct option: 4

Solution:
$\ $For hydrogen,$\ n_{1} = 1,C_{v,} = \frac{5}{2}R$
gkbMªkstu ds fy, [IMAGE] For mixture of gases,
$C_{v} = \frac{n_{1}C_{V,} + n_{2}C_{V_{2}}}{n_{1} + n_{2}}$
$= \frac{1 \times \frac{5}{2}R + 1 \times \frac{3}{2}R}{1 + 1} = 2R$
[IMAGE] T ds fn, x, eku ds fy,[IMAGE] T ds fn, x, eku ds fy,[IMAGE] Rms speed of a gas molecule at temperature Tis given by
T rkieku ij xSl dhRms pky
[IMAGE] T
Subject: Physics —  — 
Difficulty: Tough

Question: A rotor is rotating with a constant angular velocity ω about its
own axis. At the centre of the rotor a spring is fixed of natural length
(l0) whose other end is connected to a block of mass m. Find the
minimum value of coefficient of friction between block and rotor wall
for which spring remains horizontal. The radius of rotor is R (R >
l0).

4) [IMAGE] ω (l0)
 m 
 
R (R > l0)

Correct option: 2

Solution:
N = mω2R - K(R - l0)... (1)
For the block to remains in the vertical equilibrium,
[IMAGE] f = mg
⇒ N = mg
⇒ µ(mω2R - K (R - l0) = mg 
∴ µ =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS
Subject: Physics — Thermal properties of matter — Latent heat and specific heat
Difficulty: Moderate

Question: M grams of steam at 100°C is mixed with 200g of ice at its
melting point in a thermally insulated container. If it produces liquid
water at 40°C and heat of
fusion of ice is 80 the value of M is______.
100°C rkieku dh M xzke ok"i dks 200g xzke cQZ esa,d crZu
 A ok"i igys cQZ eku vius xyukad ds
cjkcj FkkA izfØ vUr esa 40°C

1) 40
2) 80
3) 120
4) 160

Correct option: 1

Solution:
Subject: Physics —  — 

Question: A projectile is thrown, from the origin along X-Y plane to have
maximum possible horizontal range of 100 m. Then the position vector of
the projectile when its speed is minimum is (take Y on the vertical
axis)


Correct option: 3

Solution:
Subject: Physics —  — 
Difficulty: Moderate

Question: A person projected a small ball with speed U along the floor from
point A. If x = 3R, determine the required speed U so that the ball
return to A after moving the circular surface in the vertical plane from
B to C and becoming projectile at C.

4) [IMAGE] A U 
x = 3R, U B C 
 C A

Correct option: 4

Solution:
V2 = V2 - 4gR
3R = V [IMAGE] V2 = U2 - 2g2R
U2 = 9gR / 4 + 4gR
U2 = 25 gR / 4
U =[IMAGE] TOPIC: WPE
SUB TOPIC: VERTICAL CIRCULAR MOTION
Subject: Physics —  — 

Question: Statement-1: In a dynamo, it is necessary that coil should
be moving and magnet should be stationary and not vice versa.
Statement-2: Magnetic force is perpendicular to the direction of
motion of charges.
Which of the following options is correct?

1) Statement is incorrect.
2) Statement is correct.
3) Both the statements are correct and is the correct reason of
4) Both the statements are correct and is not the reason of

Correct option: 2

Solution:
Subject: Physics —  — 

Question: In a meter bridge, the wire of length 1m has a non-uniform
cross-section such that, the variation [IMAGE] of its
resistance R with length.
Two equal resistances are connected as shown in the figure. The
galvanometer has zero deflection when the jockey is at point P. What is
the eight times the lengthAP ?

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

Answer: 0.25

Solution:
[IMAGE] Acording to Question
[IMAGE] TOPIC:Current and electricity
SUB TOPIC: Potentiometer
LEVEL:Moderate
Subject: Physics — Simple Harmonic Motion — Time period
Difficulty: Easy

Question: A uniform solid hemisphere of mass m and radius R is attached to
the roof with a chord of torsional constant C and performing torsional
SHM. If time period (in seconds) of SHM is x, then find the value of 5x.


Answer: 20

Solution:
Subject: Physics —  — 

Question: Four metal rods all of identical dimensions and made of same
material are welded together at a single point. The configuration is
such that any two rods are oriented at 120°. The far ends of three of
the rods are maintained at 90°C temperature and that of the fourth rod
is maintained at 30°C temperature. The temperature of the junction point
is ( in degree Celsius)


Answer: 75

Solution:
For equilibrium condition, heat flow in and out of junction
should be same. Let k represent some constant representing heat content,
which may depend on dimensions and material of the rods.
k(T - 30) = 3k(90 - T)
4T = 300
T = 75°C
Subject: Physics — Oscillation — Time period
Difficulty: Moderate

Question: A simple pendulum A and a homogeneous rod B hinged at its end
are released from horizontal position as shown in figure

1) The times period of swing of simple pendulum is greater than that
of rod if L = [IMAGE]
2) The time period of swing of simple pendulum is equal to that of
rod if L = [IMAGE]
3) The time period of swing of simple pendulum is greater than that
of rod if [IMAGE]
4) The time period of swing of simple pendulum is equal to that of
rod if L[IMAGE]

Correct option: 1

Solution:
from conservation of energy
for simple pendulum
[IMAGE] for homogeneous rod
[IMAGE] When [IMAGE] at all time
so time periods of swing are equal.
Subject: Physics — Gravitation — Kepler's law
Difficulty: Moderate

Question: India's Mangalyan was sent to the Mars by launching it into a
transfer orbit EOM around the sun. It leaves the earth at E and meets
Mars at M. If the radius of Earth's orbit is a_e = 1.5 × 10^11 m,
that of laws give the
estimate of time for Mangalyan to reach Mars from Earth to be close to
(The time period of revolution of earth around sun is 1 year)

1) $\frac{1}{2}\left( \frac{4}{3} \right)^{3/2}$ year
2) $\frac{1}{2}\left( \frac{5}{3} \right)^{3/2}$ year
3) $\frac{1}{2}\left( \frac{3}{2} \right)^{3/2}$ year
4) $\frac{3}{2}$ year

Correct option: 1

Solution:
$\left( \frac{T_{1}}{T_{2}} \right)^{2} = \left( \frac{a_{1}}{a_{2}} \right)^{3}$
${\left( \frac{1}{T} \right)}^{2} = \left\lbrack \frac{1.5 \times 10^{11}}{2 \times 10^{11}} \right\rbrack^{3}$
(semi major axis for the path from EOM$= \frac{a_{E} + a_{M}}{2}$)
$T = \left( \frac{4}{3} \right)^{3/2}$ years