JEE Physics Laws Of Motion MCQs Set 24

Practice MCQs for JEE Physics Laws Of Motion

Access targeted multiple-choice questions for Laws Of Motion designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: A bullet of mass 20 g has an initial speed of 1 m s⁻¹, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of \( 2.5\times10^{-2} \) N, the speed of the bullet after emerging from the other side of the wall is close to
(a) 0.7 m s⁻¹
(b) 0.3 m s⁻¹
(c) 0.4 m s⁻¹
(d) 0.1 m s⁻¹
Show Answer & Explanation

Answer: (a) 0.7 m s⁻¹

Explanation:
1. Retardation \( a=F/m=2.5\times10^{-2}/(20\times10^{-3})=1.25 \) m s⁻².
2. \( v^2=u^2-2as=1-2(1.25)(0.2)=0.5 \).
3. \( v=1/\sqrt2\approx0.7 \) m s⁻¹.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 2:
Assertion (A): It is easier to pull a heavy object than to push it on a level ground.
Reason (R): The magnitude of frictional force depends on the nature of the two surfaces in contact.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Statement I is true: when pulled at an angle upward the normal reaction decreases, and when pushed downward it increases.
2. Statement II is true: friction depends on the nature of the surfaces, but this does not explain Statement I, which depends on the change of the normal reaction.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 3:
Assertion (A): A cloth covers a table with some dishes on it. The cloth can be pulled out without dislodging the dishes from the table.
Reason (R): For every action there is an equal and opposite reaction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Statement I is true: the dishes stay at rest because of inertia (Newton's first law) when the cloth is pulled quickly.
2. Statement II is true (Newton's third law) but it is not the reason for the dishes staying in place.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 4: Arrangement (a): masses m and 2m hang on the two sides of a light pulley, with the rope mass negligible. Arrangement (b): mass m is lifted by pulling the other end of the rope with a constant downward force \( F=2mg \). The ratio of the acceleration of m in (a) to that in (b) is
(a) \( 3:1 \)
(b) \( 1:1 \)
(c) \( 1:2 \)
(d) \( 1:3 \)
Show Answer & Explanation

Answer: (d) \( 1:3 \)

Explanation:
1. (a) \( a_1=\dfrac{2mg-mg}{3m}=\dfrac g3 \).
2. (b) \( a_2=\dfrac{2mg-mg}{m}=g \).
3. \( a_1:a_2=1:3 \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 5: Two blocks of mass 2.9 kg and 1.9 kg are suspended from a rigid support by two inextensible wires each of length 1 m. The upper wire has negligible mass and the lower wire has a uniform mass of 0.2 kg/m. The whole system has an upward acceleration of 0.2 m/s² (g = 9.8 m/s²). The tension at the mid-point of the lower wire is
(a) 20 N
(b) 19.6 N
(c) 10 N
(d) 50 N
Show Answer & Explanation

Answer: (a) 20 N

Explanation:
1. Cut the lower wire at its mid-point. Below the cut are the 1.9 kg block and 0.1 kg of wire, a total of 2.0 kg.
2. \( T-2.0g=2.0a\Rightarrow T=2.0(9.8+0.2)=20 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 6: In the system of the previous question (blocks 2.9 kg and 1.9 kg, lower wire of mass 0.2 kg/m and length 1 m, upper wire massless, upward acceleration 0.2 m/s², g = 9.8 m/s²), the tension at the mid-point of the upper wire is
(a) 20 N
(b) 50 N
(c) 49 N
(d) 52 N
Show Answer & Explanation

Answer: (b) 50 N

Explanation:
1. The upper wire is massless, so its tension is the same everywhere. Below it are 2.9 kg + 0.2 kg + 1.9 kg = 5.0 kg.
2. \( T=5.0(9.8+0.2)=50 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 7: A uniform rope of length L and mass M lying on a smooth table is pulled by a constant force F. The tension in the rope at a distance l from the end where the force is applied is
(a) \( F\,\dfrac lL \)
(b) \( F\left(1+\dfrac lL\right) \)
(c) \( F\left(1-\dfrac lL\right) \)
(d) \( \dfrac{F}{2}\left(1-\dfrac lL\right) \)
Show Answer & Explanation

Answer: (c) \( F\left(1-\dfrac lL\right) \)

Explanation:
1. Acceleration of the whole rope \( a=F/M \).
2. The part of length \( L-l \) farthest from the force has mass \( M(L-l)/L \) and is pulled only by the tension: \( T=\dfrac{M(L-l)}{L}\cdot\dfrac FM=F\left(1-\dfrac lL\right) \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 8: A block of mass 5 kg is (A) pushed along a floor by a force F = 20 N directed 30° below the horizontal, and (B) pulled by the same force directed 30° above the horizontal. The coefficient of friction is 0.2 (g = 10 m s⁻²). The difference between the accelerations of the block in case (B) and case (A) is
(a) 0.4 m s⁻²
(b) 1.2 m s⁻²
(c) 3.2 m s⁻²
(d) 0.8 m s⁻²
Show Answer & Explanation

Answer: (d) 0.8 m s⁻²

Explanation:
1. (A) \( N=mg+F\sin30^\circ=60 \) N, friction \( =12 \) N, \( a_A=\dfrac{20\cos30^\circ-12}{5}\approx1.06 \).
2. (B) \( N=mg-F\sin30^\circ=40 \) N, friction \( =8 \) N, \( a_B=\dfrac{20\cos30^\circ-8}{5}\approx1.86 \).
3. \( a_B-a_A=\dfrac{12-8}{5}=0.8 \) m s⁻².

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 9: A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the mass, the rope deviates at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 m s⁻²)
(a) 100 N
(b) 70 N
(c) 200 N
(d) 140 N
Show Answer & Explanation

Answer: (a) 100 N

Explanation:
1. Equilibrium: \( T\cos45^\circ=mg \) and \( T\sin45^\circ=F \).
2. So \( F=mg\tan45^\circ=100 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 10: A mass \( \sqrt2\,m \) hangs from the junction of two strings. Each string passes over a smooth pulley and carries a mass m at its free end. The strings and pulleys are smooth and massless and the system is in equilibrium. If each string makes an angle \( \theta \) with the vertical at the junction, then \( \theta \) is
(a) \( 0^\circ \)
(b) \( 45^\circ \)
(c) \( 30^\circ \)
(d) \( 60^\circ \)
Show Answer & Explanation

Answer: (b) \( 45^\circ \)

Explanation:
1. Each string has tension \( T=mg \) (set by the hanging mass m).
2. Vertical equilibrium at the junction: \( 2T\cos\theta=\sqrt2\,mg\Rightarrow\cos\theta=\dfrac1{\sqrt2} \).
3. \( \theta=45^\circ \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 11: Two blocks A and B of masses \( m_A=1 \) kg and \( m_B=3 \) kg are placed one over the other on a table. The coefficient of friction between A and B is 0.2 and between B and the table is also 0.2. The maximum horizontal force F that can be applied on B so that A does not slide over B is (g = 10 m s⁻²)
(a) 12 N
(b) 8 N
(c) 16 N
(d) 40 N
Show Answer & Explanation

Answer: (c) 16 N

Explanation:
1. The maximum acceleration of A is \( \mu g=2 \) m s⁻².
2. For the pair: \( F-\mu(m_A+m_B)g=(m_A+m_B)a \), so \( F=0.2(40)+4(2)=16 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 12: A block rests on a rough inclined plane of angle 30°. It stays at rest up to a maximum force of 2 N applied down the incline. The maximum external force up the incline that does not move the block is 10 N. The coefficient of static friction between the block and the plane is
(a) \( \dfrac23 \)
(b) \( \dfrac{\sqrt3}{4} \)
(c) \( \dfrac12 \)
(d) \( \dfrac{\sqrt3}{2} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{\sqrt3}{2} \)

Explanation:
1. Let \( s=mg\sin30^\circ \) and \( f \) be the limiting friction. Down: \( s+2=f \). Up: \( 10=s+f \).
2. Solving, \( s=4 \) N and \( f=6 \) N; \( mg=8 \) N, so \( N=mg\cos30^\circ=4\sqrt3 \) N.
3. \( \mu=\dfrac{6}{4\sqrt3}=\dfrac{\sqrt3}{2} \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 13: A block of mass 10 kg is kept on a rough inclined plane of angle 45°. A force of 3 N is applied on the block down the incline. The coefficient of static friction is 0.6. The minimum value of the force F, applied up the incline, such that the block does not move downward is (g = 10 m s⁻²)
(a) 32 N
(b) 25 N
(c) 23 N
(d) 18 N
Show Answer & Explanation

Answer: (a) 32 N

Explanation:
1. On the point of sliding down: \( F+f=mg\sin45^\circ+3 \), with \( f=\mu mg\cos45^\circ \).
2. \( F=70.7+3-0.6(70.7)=31.3\approx32 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 14: Two masses \( m_1=5 \) kg and \( m_2=10 \) kg are connected by an inextensible string over a frictionless pulley. \( m_2 \) lies on a horizontal surface of coefficient of friction 0.15 and \( m_1 \) hangs freely. The minimum mass m that should be put on top of \( m_2 \) to stop the motion is
(a) 18.3 kg
(b) 23.33 kg
(c) 27.3 kg
(d) 43.3 kg
Show Answer & Explanation

Answer: (b) 23.33 kg

Explanation:
1. To stop the motion, the limiting friction must balance the hanging weight: \( \mu(m_2+m)g=m_1g \).
2. \( 0.15(10+m)=5\Rightarrow m=23.33 \) kg.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 15: A uniform wooden stick of mass 1.6 kg and length l rests on a smooth vertical wall of height \( h\,((a) \( \dfrac hl=\dfrac{\sqrt3}{16},\ f=\dfrac{16\sqrt3}{3}\ \text{N} \)
(b) \( \dfrac hl=\dfrac{3}{16},\ f=\dfrac{16\sqrt3}{3}\ \text{N} \)
(c) \( \dfrac hl=\dfrac{3\sqrt3}{16},\ f=\dfrac{16\sqrt3}{3}\ \text{N} \)
(d) \( \dfrac hl=\dfrac{3\sqrt3}{16},\ f=\dfrac{8\sqrt3}{3}\ \text{N} \)
Show Answer & Explanation

Answer: (c) \( \dfrac hl=\dfrac{3\sqrt3}{16},\ f=\dfrac{16\sqrt3}{3}\ \text{N} \)

Explanation:
1. The wall reaction N is perpendicular to the stick, so it makes 30° with the horizontal. Vertical balance: \( N+N\sin30^\circ=mg=16 \), so \( N=\dfrac{32}{3} \) N.
2. Horizontal balance: \( f=N\cos30^\circ=\dfrac{16\sqrt3}{3} \) N.
3. Torque about the bottom: \( N\,d=mg\cdot\dfrac l2\sin30^\circ \), so \( d=\dfrac{3l}{8} \) along the stick, and \( h=d\cos30^\circ=\dfrac{3\sqrt3}{16}l \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 16: Block A of weight 20 N is pressed against block B of weight 100 N by a horizontal force F, and B is pressed against a vertical wall. The coefficient of friction between the blocks is 0.1 and between B and the wall is 0.15. The blocks are at rest. The frictional force applied by the wall on block B is
(a) 100 N
(b) 80 N
(c) 150 N
(d) 120 N
Show Answer & Explanation

Answer: (d) 120 N

Explanation:
1. Take A and B together: the weight 120 N acting downward must be balanced by the wall friction on B (the wall is the only vertical support).
2. Wall friction \( =20+100=120 \) N, which is below the limiting value as long as F is large enough.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 17: A block of mass m is placed on a surface with a vertical cross-section given by \( y=x^3/6 \). If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
(a) \( \dfrac16\ \text{m} \)
(b) \( \dfrac23\ \text{m} \)
(c) \( \dfrac13\ \text{m} \)
(d) \( \dfrac12\ \text{m} \)
Show Answer & Explanation

Answer: (a) \( \dfrac16\ \text{m} \)

Explanation:
1. The block is on the point of slipping when \( \tan\theta=\mu \), where the slope is \( \tan\theta=\dfrac{dy}{dx}=\dfrac{x^2}2 \).
2. \( \dfrac{x^2}{2}=0.5\Rightarrow x=1 \) m, so \( y=\dfrac16 \) m.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 18: Blocks \( m_1=1 \) kg (smooth contact with an inclined plane) and \( m_2=2 \) kg (coefficient of friction 0.3 with the plane) are placed together on the plane of inclination \( \theta \). Match the angle with the friction on \( m_2 \): P: \( \theta=5^\circ \), Q: \( \theta=10^\circ \), R: \( \theta=15^\circ \), S: \( \theta=20^\circ \); and 1: \( m_2g\sin\theta \), 2: \( (m_1+m_2)g\sin\theta \), 3: \( \mu m_2g\cos\theta \), 4: \( \mu(m_1+m_2)g\cos\theta \). (tan 11.5° ≈ 0.2, tan 16.5° ≈ 0.3)
(a) P-1, Q-1, R-1, S-3
(b) P-2, Q-2, R-3, S-3
(c) P-2, Q-2, R-2, S-4
(d) P-2, Q-2, R-2, S-3
Show Answer & Explanation

Answer: (b) P-2, Q-2, R-3, S-3

Explanation:
1. The pair slides when \( (m_1+m_2)g\sin\theta>\mu m_2g\cos\theta \), i.e. \( \tan\theta>0.3\times\tfrac23=0.2 \), so \( \theta>11.5^\circ \).
2. For 5° and 10° the pair is at rest; static friction on \( m_2 \) must hold both blocks: \( (m_1+m_2)g\sin\theta \) (item 2).
3. For 15° and 20° the block slides, so friction is kinetic: \( \mu m_2g\cos\theta \) (item 3). Hence P-2, Q-2, R-3, S-3.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 19: What is the maximum value of the force F such that a block of mass \( \sqrt3 \) kg does not move on a horizontal surface when F pushes it at 60° below the horizontal? The coefficient of friction is \( \dfrac1{2\sqrt3} \) (g = 10 m s⁻²).
(a) 10 N
(b) 12 N
(c) 20 N
(d) 15 N
Show Answer & Explanation

Answer: (c) 20 N

Explanation:
1. At the limit: \( F\cos60^\circ=\mu(mg+F\sin60^\circ) \).
2. \( \dfrac F2=\dfrac1{2\sqrt3}\left(10\sqrt3+\dfrac{\sqrt3}{2}F\right)=5+\dfrac F4 \), so \( F=20 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 20: An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle \( \alpha \) with the vertical, the maximum possible value of \( \alpha \) is given by
(a) \( \tan\alpha=3 \)
(b) \( \sec\alpha=3 \)
(c) \( \operatorname{cosec}\alpha=3 \)
(d) \( \cot\alpha=3 \)
Show Answer & Explanation

Answer: (d) \( \cot\alpha=3 \)

Explanation:
1. At the limit, \( \tan\alpha=\mu=\tfrac13 \) (the slope angle equals the angle of friction).
2. So \( \cot\alpha=3 \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Question 21: A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is (g = 9.8 m s⁻²)
(a) 0.98 N
(b) 2.5 N
(c) 4.9 N
(d) 0.49 N
Show Answer & Explanation

Answer: (a) 0.98 N

Explanation:
1. Limiting friction \( =0.5\times5=2.5 \) N, but the block is at rest, so friction equals the weight.
2. \( f=mg=0.1\times9.8=0.98 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

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