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

Question: On a smooth pulley of radius R hangs a homogeneous flexible rope of
mass / length λ and length  (Fig.). Find the maximum tension in the
rope.

4) [IMAGE]  λ / R 
 ()

Correct option: 4

Solution:
[IMAGE] Tmax - T0 = λgR [1 - cos θ]
Tmax = λgR + λg[IMAGE] = λ 
TOPIC: NLM
SUB TOPIC: EQUILIBRIUM
Subject: Physics — Simple Harmonic Motion — Time period of oscillation
Difficulty: Moderate

Question: The minimum time in which the potential energy of a particle
executing SHM changes from maximum to minimum is 5 s. The time period of
SHM in seconds is 4T. Find T.


Answer: 5

Solution:
P.E. is maximum at extreme position and minimum at mean
position.
Time to go from extreme position to mean position is,
 where T is time period of SHM.
Subject: Physics — Electrostatics — Capacitance
Difficulty: Moderate

Question: Three metallic plates out of which middle is given a charge Q,
as shown in the figure. The outer plates can be earthed with switches S1
and S2. The area of plates is same.The charge appearing on outer surface
of external plates when S_1 and S_2 are open and external platesare
uncharged.
Ckgjh ifV~Vdkvksa dh ckgjh lrgks ij S_1 rFkk S_2

1) $- \left( \frac{Q}{2} \right)$
2) $\left( \frac{Q}{2} \right)$
3) Q
4) - Q

Correct option: 2

Solution:
[IMAGE] (conductor pkyd)
[IMAGE] So vr -x = x + Q + y - y
2x = - Q
$x = \frac{- Q}{2}$
$- x = \frac{Q}{2}$
Subject: Physics — Newton's Laws of Motion — Constraint motion
Difficulty: Tough

Question: System is shown in the figure. Assume that cylinder remains in
contact with the two wedges. The velocity of cylinder is

1) $\sqrt{19 - 4\sqrt{3}}\frac{u}{2}\ m/s$
2) $\frac{\sqrt{13}u}{2}\ m/s$
3) $\sqrt{3}u\ m/s$
4) $\sqrt{7}u\ m/s$

Correct option: 4

Solution:
[IMAGE] As cylinder will remains in contact with wedge A
Vx = 2u
As it also remain in contact with wedge B
u sin 30° = Vy cos30° - Vxsin30°
Vy = Vx$\frac{Sin30{^\circ}}{Cos30{^\circ}}$ +
$u\frac{Sin30{^\circ}}{Cos30{^\circ}}$
Vy = Vx tan30° + u tan 30°
Vy = 3utan30° = $\sqrt{3}$u
V = $\sqrt{7}$u
Subject: Physics — Thermodynamics — Thermodynamic cycle
Difficulty: Moderate

Question: Which of the following is an equivalent cyclic process
corresponding to the thermodynamic cyclic givenin the figure ? where, 1
→ 2 is adiabatic. (Graphs are schematic and are not to scale)


Correct option: 1

Solution:
For process A - B izØe ds fy,; Volume is constant 
PV = nRT T increases cM+rk gS
For process B - C izØe 
$\Rightarrow {TV}^{\gamma - 1}$ = Constant 
For process C - A izØe ds fy,; pressure is constant
Subject: Physics — Relative motion — closest distance of approach
Difficulty: Moderate

Question: Two stones are in the same horizontal line and are$50\sqrt{5}m$
away. They are projected with a velocity of 10 
respectively and with an angle of 37º and 53º with horizontal
respectively. The minimum distance between them in meters is 5n. Find s.
Suppose they don't strike the ground during the motion.
(take$g = 10m/s^{2}$)


Answer: 4

Solution:
$d_{\min} = 50\sqrt{5}\ \ \ \ \ \ \ \ sin\theta = 50\sqrt{5} \times \frac{2}{5\sqrt{5}} = 20m$
Subject: Physics — Mechanical Properties of Fluids — Buoyancy
Difficulty: Moderate

Question: A leak proof cylinder of length 1 m, made of a metal which has
very low coefficient of expansion is floating vertically in water at 0°C
such that its height above the water surface is 20 cm. When the
temperature of water is increased to 4°C, the height of the cylinder
above the water surface becomes 21 cm. The density of water at T = 4°C,
relative to the density at T = 0°C is close to

1) 1.03
2) 1.04
3) 1.26
4) 1.01

Correct option: 4

Solution:
$mg = A(79)\rho_{4^{\circ}C}g$
$\frac{\rho_{4^{\circ}C}}{\rho_{0^{\circ}C}} = \frac{80}{79} = 1.01$
Subject: Physics — Circular motion — Vertical circular motion
Difficulty: Moderate

Question: A particle of mass $m = \frac{5}{42}kg$ is tied to an end of an
inextensible string and is whirled in a vertical circle as shown below.
Then, the difference in tension of string at A and B i.e.
$T_{A} - T_{B}$ in Newton is: (particle completes the circle)

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

Correct option: 4

Solution:
$\Delta T = \frac{3mg\Delta h}{I}$
$h_{B} - h_{A} = I(\frac{3}{5} + \frac{4}{5}) = \frac{7I}{5}$
$T_{A} - B_{B} = \frac{3mg}{I}(h_{B} - h_{A})$
$= \frac{3mg}{I} \times \frac{7I}{5}$
$= \frac{21mg}{5} = \frac{21}{5} \times \frac{5}{42} \times 10$
$= 5N$
Subject: Physics — Mechanical Properties of Fluids — Equation of continuity
Difficulty: Moderate

Question: An ideal fluid flows (laminar flow) through a pipe of
non-uniform diameter. The maximum and minimum diameters of the pipes are
6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the
maximum velocities of fluid in this pipe is

1) $\frac{9}{16}$
2) $\frac{\sqrt{3}}{2}$
3) $\frac{81}{256}$
4) $\frac{8}{4}$

Correct option: 1

Solution:
Using equation of continuity
$\frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{2}} = \left( \frac{4.8}{6.4} \right)^{2} = \frac{9}{16}$
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

Question: Consider a uniform rod of mass [IMAGE] and length
[IMAGE] pivoted about its centre. A mass m moving
withvelocity v making angle [IMAGE] to the rod's long
axis collides with one end of the rod and sticks to it. The angular
speed of the rod-mass system just after the collision is


Correct option: 2

Solution:
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) 3
2) 1
3) 7
4) 8

Correct option: 2

Solution:
[IMAGE] TOPIC:Electromagnetic induction
SUB TOPIC:Faraday's law
LEVEL: Moderate
Subject: Physics — Current Electricity — KVL
Difficulty: Moderate

Question: In the figure shown $C_{1} = 11\mu F$ and $C_{2} = 5\mu F$ then
at steady state

1) The potential difference across C_1 is 5V
2) The potential difference between terminals of 15V battery is 9V
3) The potential difference V_a - V_b = - 4V
4) All of the above

Correct option: 4

Solution:
[IMAGE] at steady state
$I(3) + 1(2) = 15$
I = 3
$kvL$
$C \rightarrow D \rightarrow E \rightarrow a \rightarrow b \rightarrow C$
$V_{c} - I(3) + \frac{q}{11} - 7 + \frac{q}{5} = V_{c}$
$\frac{q}{11} + \frac{q}{5} = 7 + 3 \times 3 = 16$
$q = 55\mu C$
Now KVL for a$\rightarrow$b
$V_{a} - 7 + \frac{q}{5} = V_{b}$
$\Rightarrow V_{a} - V_{b} = 7 - \frac{q}{5}$
$= 7 - \frac{55}{5}$ $= - \frac{20}{5} = - 4V$