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Moment of force, torque and angular momentum: NEET MCQs with answers

75 questions on this topic of Rotational Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. If the direction of the position vector r⃗ is reversed while keeping the force F⃗ unchanged, the torque τ⃗ = r⃗ × F⃗ will:

  1. Reverse its direction
  2. Double in magnitude
  3. Remain unchanged in magnitude and direction
  4. Become zero
Show answer and explanation

Answer: (A) Reverse its direction

Replacing r⃗ with -r⃗ yields τ⃗_new = (-r⃗) × F⃗ = -(r⃗ × F⃗) = -τ⃗. Thus, its magnitude remains the same, but its direction is reversed.

medium

2. A particle of mass 2 kg is moving in the xy-plane with a constant velocity v⃗ = 3 î m s⁻¹. At an instant, its position vector is r⃗ = (2 î + 4 ĵ) m. What is the angular momentum of the particle about the origin at this instant?

  1. 24 k̂ kg m² s⁻¹
  2. -24 k̂ kg m² s⁻¹
  3. -12 k̂ kg m² s⁻¹
  4. 12 k̂ kg m² s⁻¹
Show answer and explanation

Answer: (B) -24 k̂ kg m² s⁻¹

Angular momentum is given by L⃗ = r⃗ × p⃗ = m(r⃗ × v⃗). Here, r⃗ × v⃗ = (2 î + 4 ĵ) × (3 î) = 4 × 3 (ĵ × î) = -12 k̂ m² s⁻¹. Therefore, L⃗ = 2 × (-12 k̂) = -24 k̂ kg m² s⁻¹.

hard

3. A thin uniform rod AB of mass M and length L is free to rotate in a vertical plane about a smooth horizontal axis passing through its end A. It is released from rest from a horizontal position. The initial angular acceleration of the rod immediately after release is:

  1. 2g / (3L)
  2. 3g / (2L)
  3. g / L
  4. 3g / L
Show answer and explanation

Answer: (B) 3g / (2L)

The torque about hinge A immediately upon release is due to gravity acting at the center of mass (distance L/2 from A): τ = Mg(L/2). The moment of inertia of the rod about one end is I = ML²/3. Using τ = I α, we get Mg(L/2) = (ML²/3) α, which gives α = 3g / (2L).

medium

4. A light string is wound around the rim of a uniform solid cylinder of mass 4 kg and radius 0.2 m, which is mounted on a horizontal frictionless axle. A constant pulling force of 30 N is applied to the free end of the string. The angular acceleration produced in the cylinder is:

  1. 150 rad s⁻²
  2. 37.5 rad s⁻²
  3. 75 rad s⁻²
  4. 50 rad s⁻²
Show answer and explanation

Answer: (C) 75 rad s⁻²

The torque applied by the pulling force is τ = F × R = 30 × 0.2 = 6 N m. The moment of inertia of the solid cylinder about its central axis is I = (1/2) M R² = 0.5 × 4 × (0.2)² = 0.08 kg m². The angular acceleration is α = τ / I = 6 / 0.08 = 75 rad s⁻².

easy

5. Which of the following statements is INCORRECT regarding torque and angular momentum?

  1. Torque is the rotational analogue of force.
  2. The rate of change of total angular momentum of a system of particles about a point equals the net external torque about that point.
  3. If net external torque on a system is zero, its total angular momentum must be conserved.
  4. A pair of equal and opposite forces acting along the same line of action constitutes a couple and produces maximum torque.
Show answer and explanation

Answer: (D) A pair of equal and opposite forces acting along the same line of action constitutes a couple and produces maximum torque.

A couple consists of two equal and opposite forces whose lines of action do NOT coincide. When two equal and opposite forces act along the same line of action, their perpendicular distance is zero, so the net torque is zero, not maximum.

hard

6. A uniform rod of mass M and length L lies at rest on a frictionless horizontal plane. An impulse J is applied perpendicular to the rod at one of its ends. The angular velocity acquired by the rod immediately after the impact is:

  1. 12J / (M L)
  2. 3J / (M L)
  3. 6J / (M L)
  4. 2J / (M L)
Show answer and explanation

Answer: (C) 6J / (M L)

The angular impulse about the center of mass of the rod is equal to the change in angular momentum: J × (L/2) = I_cm ω. The moment of inertia of the rod about its center of mass is I_cm = (1/12) M L². Therefore, J (L/2) = (1/12) M L² ω, which yields ω = 6J / (M L).

easy

7. A force of 20 N acts at a point 0.25 m from the axis of rotation, at 90° to the position vector. The magnitude of the torque is:

  1. 80 N m
  2. 5 N m
  3. 10 N m
  4. 0 N m
Show answer and explanation

Answer: (B) 5 N m

τ = rF sin θ = 0.25 × 20 × sin 90° = 5 N m.

easy

8. A force of 60 N acts at a point 1 m from the axis of rotation, at 60° to the position vector. The magnitude of the torque is:

  1. 30 N m
  2. 104 N m
  3. 52 N m
  4. 60 N m
Show answer and explanation

Answer: (C) 52 N m

τ = rF sin θ = 1 × 60 × sin 60° = 52 N m.

easy

9. A force F⃗ = (3î + 4ĵ) N acts on a particle whose position vector relative to the origin is r⃗ = (2î - 5ĵ) m. The torque acting on the particle about the origin is:

  1. 7 k̂ N m
  2. -23 k̂ N m
  3. 23 k̂ N m
  4. -7 k̂ N m
Show answer and explanation

Answer: (C) 23 k̂ N m

Torque is given by τ⃗ = r⃗ × F⃗. Here, r⃗ = 2î - 5ĵ and F⃗ = 3î + 4ĵ. Therefore, τ⃗ = (2î - 5ĵ) × (3î + 4ĵ) = [2(4) - (-5)(3)] k̂ = (8 + 15) k̂ = 23 k̂ N m.

easy

10. A force of 40 N acts at a point 0.5 m from the axis of rotation, at 60° to the position vector. The magnitude of the torque is:

  1. 10 N m
  2. 80 N m
  3. 17.3 N m
  4. 20 N m
Show answer and explanation

Answer: (C) 17.3 N m

τ = rF sin θ = 0.5 × 40 × sin 60° = 17.3 N m.

easy

11. A force of 100 N acts at a point 0.25 m from the axis of rotation, at 30° to the position vector. The magnitude of the torque is:

  1. 400 N m
  2. 21.7 N m
  3. 12.5 N m
  4. 25 N m
Show answer and explanation

Answer: (C) 12.5 N m

τ = rF sin θ = 0.25 × 100 × sin 30° = 12.5 N m.

easy

12. A force F⃗ = (î - 2ĵ + 3k̂) N acts on a particle located at position vector r⃗ = (2î + ĵ + k̂) m with respect to the origin. What is the torque acting on the particle about the origin?

  1. (-5î + 5ĵ + 5k̂) N m
  2. (5î + 5ĵ - 5k̂) N m
  3. (5î - 5ĵ - 5k̂) N m
  4. (î - 5ĵ - 5k̂) N m
Show answer and explanation

Answer: (C) (5î - 5ĵ - 5k̂) N m

Torque is given by the cross product τ⃗ = r⃗ × F⃗. Computing the determinant: î[(1)(3) - (1)(-2)] - ĵ[(2)(3) - (1)(1)] + k̂[(2)(-2) - (1)(1)] = 5î - 5ĵ - 5k̂ N m.

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