Free Online Motion In Plane PYQ 001

JEE Mains Previous Year Questions on Laws of Motion

Welcome to this interactive blog post featuring previous year JEE Mains questions from the Laws of Motion chapter. Practice these questions to strengthen your understanding of key physics concepts. Click the "Show Answer" button to reveal detailed solutions for numerical questions, or select an option for multiple-choice questions to check if you're correct!

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Question 1

A body of mass 1000 kg is moving horizontally with a velocity 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: [27-Jan-2024 Shift 1]

Solution:

This problem involves conservation of momentum since no external forces are mentioned.

Initial mass, \( m_1 = 1000 \) kg, initial velocity, \( v_1 = 6 \) m/s.

Extra mass added, \( m_2 = 200 \) kg (assumed at rest), so final mass, \( m_f = 1000 + 200 = 1200 \) kg.

Initial momentum: \( m_1 \times v_1 = 1000 \times 6 = 6000 \) kg·m/s.

Final momentum: \( m_f \times v_f = 1200 \times v_f \).

By conservation of momentum: \( 1200 v_f = 6000 \).

\[ v_f = \frac{6000}{1200} = 5 \] m/s.

Final velocity: 5 m/s

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Question 2

A train is moving with a speed of 12 m/s on rails which are 1.5 m apart. To negotiate a curve of radius 400 m, the height by which the outer rail should be raised with respect to the inner rail is (Given, \( g = 10 \) m/s²): [27-Jan-2024 Shift 1]

Solution:

For a train to negotiate a curve, the outer rail is raised to provide centripetal force. The height \( h \) is given by:

\[ h = \frac{v^2 w}{g r} \]

Where \( v = 12 \) m/s, \( w = 1.5 \) m, \( r = 400 \) m, \( g = 10 \) m/s².

\[ h = \frac{(12)^2 \times 1.5}{10 \times 400} = \frac{144 \times 1.5}{4000} = \frac{216}{4000} = 0.054 \] m.

Height: 0.054 m

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Question 3

A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60° with the horizontal, the weight experienced by the man is: [27-Jan-2024 Shift 2]

Solution:

Assuming "weight 12 kg" means a mass of 12 kg, the rod’s weight is \( W = 12g \) N (where \( g \) is typically 9.8 m/s² unless specified). The rod is uniform, with one end on the ground and the other supported vertically by the man’s force \( F \).

Taking torques about the ground end: Torque due to weight (at midpoint) = \( W \times \frac{L}{2} \cos 60^\circ \), torque due to \( F \) = \( F \times L \cos 60^\circ \).

\[ F \times L \cos 60^\circ = W \times \frac{L}{2} \cos 60^\circ \]

\[ F = \frac{W}{2} = \frac{12g}{2} = 6g \] N, equivalent to 6 kg weight.

Weight experienced: 6 kg

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Question 4

Given below are two statements:

Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.

In the light of the above statements, choose the most appropriate answer from the options given below: [27-Jan-2024 Shift 2]

A. Statement I is correct but Statement II is incorrect
B. Statement I is incorrect but Statement II is correct
C. Both Statement I and Statement II are incorrect
D. Both Statement I and Statement II are correct

Solution:

Statement (I): Static friction \( f_s \leq \mu_s N \) depends on materials via \( \mu_s \), not area. So, it’s incorrect.

Statement (II): Kinetic friction \( f_k = \mu_k N \) is independent of area and depends on materials via \( \mu_k \). So, it’s correct.

Correct Answer: B

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Question 5

A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string when the stone is at the lowest point is (if \( \pi^2 = 9.8 \) and \( g = 9.8 \) m/s²) [29-Jan-2024 Shift 2]

Solution:

At the lowest point: \( T = mg + \frac{m v^2}{r} \).

\( m = 0.9 \) kg, \( r = 1 \) m, \( \omega = 10 \, \text{rpm} = \frac{10}{60} \times 2\pi = \frac{\pi}{3} \) rad/s.

\( v = \omega r = \frac{\pi}{3} \) m/s, \( \frac{v^2}{r} = \left( \frac{\pi}{3} \right)^2 = \frac{\pi^2}{9} = \frac{9.8}{9} \).

\[ T = 0.9 \times 9.8 + 0.9 \times \frac{9.8}{9} = 9.8 \] N.

Tension: 9.8 N

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Question 6

A block of mass \( m \) is placed on a surface with vertical cross-section given by \( y = \frac{x^2}{4} \). If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is: [30-Jan-2024 Shift 2]

Solution:

Slope: \( \frac{dy}{dx} = \frac{x}{2} = \tan \theta \), \( \mu = 0.5 \).

For no slipping: \( \tan \theta \leq \mu \), so \( \frac{x}{2} \leq 0.5 \), \( |x| \leq 1 \).

Height: \( y = \frac{x^2}{4} \), max at \( x = 1 \), \( y = \frac{1}{4} = 0.25 \) m.

Maximum height: 0.25 m

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Question 7

A coin is placed on a disc. The coefficient of friction between the coin and the disc is \( \mu \). If the distance of the coin from the center of the disc is \( r \), the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is: [31-Jan-2024 Shift 1]

Solution:

Friction provides centripetal force: \( m \omega^2 r \leq \mu m g \).

\[ \omega^2 \leq \frac{\mu g}{r} \], \( \omega_{\text{max}} = \sqrt{\frac{\mu g}{r}} \).

Maximum angular velocity: \( \sqrt{\frac{\mu g}{r}} \)

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Question 8

A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s, then the magnitude of force exerted by the ball on the hand of player will be (in SI unit): [1-Feb-2024 Shift 2]

Solution:

Force = rate of change of momentum.

\( m = 0.12 \) kg, \( \Delta v = 25 - 0 = 25 \) m/s, \( \Delta t = 0.1 \) s.

\[ F = \frac{m \Delta v}{\Delta t} = \frac{0.12 \times 25}{0.1} = 30 \] N.

Force: 30 N

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Question 9

Given below are two statements:

Statement-I: An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable.

Statement-II: Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed.

In the light of the above statements, choose the correct answer from the options given below: [24-Jan-2023 Shift 1]

A. Both Statement I and Statement II are false
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are true
D. Statement I is false but Statement II is true

Solution:

Statement-I: If \( T = mg \), net force is zero, so elevator moves with uniform speed. True.

Statement-II: Elevator accelerating downward (\( a \downarrow \)), \( N = mg - ma < mg \). False.

Correct Answer: B

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Question 10

A block of mass \( m \) slides down a plane inclined at angle 30° with an acceleration \( \frac{g}{4} \). The value of coefficient of kinetic friction will be: [29-Jan-2023 Shift 1]

Solution:

Net force: \( mg \sin 30^\circ - \mu mg \cos 30^\circ = ma \).

\[ g \cdot \frac{1}{2} - \mu g \cdot \frac{\sqrt{3}}{2} = \frac{g}{4} \]

\[ \frac{1}{2} - \mu \frac{\sqrt{3}}{2} = \frac{1}{4} \], \( \mu \frac{\sqrt{3}}{2} = \frac{1}{4} \).

\[ \mu = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6} \].

Coefficient of kinetic friction: \( \frac{\sqrt{3}}{6} \)

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