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Conservation of angular momentum and applications: NEET MCQs with answers

86 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. A skater spins at 3 rad/s with arms outstretched. On folding the arms, the moment of inertia becomes 1/3 of its initial value. The new angular speed is:

  1. 1 rad/s
  2. 3 rad/s
  3. 27 rad/s
  4. 9 rad/s
Show answer and explanation

Answer: (D) 9 rad/s

No external torque, so Iω is conserved: I·3 = (I/3)·ω′, giving ω′ = 3 × 3 = 9 rad/s.

Quick tip: L = Iω is constant: smaller I means proportionally larger ω.

NCERT Class 11, Chapter 6

medium

2. A celestial body of mass M collapses gravitationally to form a neutron star. If its original radius R shrinks to R/100 while losing no mass, the ratio of its final rotational kinetic energy to its initial rotational kinetic energy will be:

  1. 10⁻⁴
  2. 10²
  3. 10⁴
  4. 10⁻²
Show answer and explanation

Answer: (C) 10⁴

Since no external torque acts, angular momentum L is conserved. The rotational kinetic energy is given by K = L² / (2I). Since I is proportional to R², when R becomes R/100, I becomes I/10000. Therefore, K_final / K_initial = I_initial / I_final = (R / (R/100))² = 10⁴.

hard

3. A uniform rod of mass M and length L lies at rest on a smooth horizontal table. A particle of mass m moving with velocity v perpendicular to the rod hits one of its ends and sticks to it. The angular velocity of the compound system immediately after the collision is:

  1. 12mv / [(M + 3m)L]
  2. 6mv / [(M + 4m)L]
  3. 3mv / [(M + m)L]
  4. 6mv / [(2M + 3m)L]
Show answer and explanation

Answer: (B) 6mv / [(M + 4m)L]

Let the center of mass of the system be at distance x from the center of the rod towards the struck end: x = m(L/2) / (M + m). The initial angular momentum about the center of mass is L_i = m v [L/2 - x] = m v [M(L/2) / (M + m)] = MmLv / [2(M + m)]. The moment of inertia about the center of mass is I = [ML²/12 + Mx²] + m(L/2 - x)² = (ML²/12) + [Mm/(M + m)](L/2)² = ML²(M + 4m) / [12(M + m)]. Equating L_i = Iω gives ω = 6mv / [(M + 4m)L].

medium

4. A disc of moment of inertia I rotates with an angular speed ω about a vertical axis passing through its centre. A second disc of moment of inertia 2I, initially at rest, is dropped coaxially onto the first disc. The two discs eventually rotate together. The fractional loss of rotational kinetic energy of the system is:

  1. 1/3
  2. 1/2
  3. 2/3
  4. 3/4
Show answer and explanation

Answer: (C) 2/3

By conservation of angular momentum, Iω = (I + 2I)ω_f, which yields ω_f = ω/3. Initial kinetic energy is K_i = (1/2)Iω². Final kinetic energy is K_f = (1/2)(3I)(ω/3)² = (1/6)Iω² = K_i / 3. Therefore, fractional loss = (K_i - K_f) / K_i = 1 - 1/3 = 2/3.

easy

5. A person on a frictionless turntable rotates at 10 rad/s. Drawing in the arms reduces the moment of inertia to 1/2 of its value. The new angular speed is:

  1. 20 rad/s
  2. 40 rad/s
  3. 5 rad/s
  4. 10 rad/s
Show answer and explanation

Answer: (A) 20 rad/s

No external torque, so Iω is conserved: I·10 = (I/2)·ω′, giving ω′ = 2 × 10 = 20 rad/s.

Quick tip: L = Iω is constant: smaller I means proportionally larger ω.

NCERT Class 11, Chapter 6

hard

6. A uniform disc of mass M and radius R rotates about its central perpendicular axis with angular velocity ω. A concentric circular portion of radius R/2 is scooped out and discarded from the disc while it is rotating (without applying any tangential force). Assuming angular momentum of the remaining portion is conserved during the cut, what is the new angular velocity of the remaining annular disc?

  1. (16/15) ω
  2. ω
  3. (15/16) ω
  4. (4/3) ω
Show answer and explanation

Answer: (B) ω

Since every concentric ring element of the disc rotates with the same angular speed ω, removing any concentric portion without tangential forces does not alter the angular velocity of the remaining part. Each remaining mass element continues to rotate with its original angular speed ω.

medium

7. When a ballet dancer pulls her arms inward during a pirouette, her moment of inertia decreases by 20%. By what percentage does her rotational kinetic energy change?

  1. Remains unchanged
  2. Decreases by 20%
  3. Increases by 20%
  4. Increases by 25%
Show answer and explanation

Answer: (D) Increases by 25%

Since net external torque is zero, angular momentum L is conserved. Rotational kinetic energy is K = L² / (2I). Since I_final = 0.8 I_initial, K_final = K_initial / 0.8 = 1.25 K_initial. Thus, kinetic energy increases by 25% due to the internal muscular work done by the dancer.

easy

8. A skater spins at 6 rad/s with arms outstretched. On folding the arms, the moment of inertia becomes 1/3 of its initial value. The new angular speed is:

  1. 6 rad/s
  2. 54 rad/s
  3. 18 rad/s
  4. 2 rad/s
Show answer and explanation

Answer: (C) 18 rad/s

No external torque, so Iω is conserved: I·6 = (I/3)·ω′, giving ω′ = 3 × 6 = 18 rad/s.

Quick tip: L = Iω is constant: smaller I means proportionally larger ω.

NCERT Class 11, Chapter 6

easy

9. A light rigid rod of length 2L has two point masses, each of mass m, attached at its two ends. The rod is rotating horizontally in a plane about a fixed vertical axis passing through its centre with angular speed ω. If an internal mechanism suddenly shifts both masses symmetrically to a distance L/2 from the centre, the new angular speed becomes:

  1. 2ω
  2. ω/4
  3. 8ω
  4. 4ω
Show answer and explanation

Answer: (D) 4ω

Initial moment of inertia I₁ = 2 × m × L² = 2mL². Final moment of inertia when both are at distance L/2 is I₂ = 2 × m × (L/2)² = mL²/2. Since no external torque acts, I₁ω = I₂ω′, which gives (2mL²)ω = (mL²/2)ω′, so ω′ = 4ω.

easy

10. A point mass m is moving with a constant velocity v in a straight line on a smooth horizontal table. With respect to an arbitrary point P lying at a perpendicular distance d from this line of motion, the angular momentum of the particle:

  1. increases linearly with time
  2. remains constant with magnitude mvd
  3. decreases linearly with time
  4. is zero at all times
Show answer and explanation

Answer: (B) remains constant with magnitude mvd

Angular momentum is given by L = r × p. The magnitude is L = mvd, where d is the perpendicular distance between the fixed point P and the straight line of motion. Since m, v, and d are all constant and the direction of L remains unchanged, the angular momentum about point P is constant in time.

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11. A small particle of mass m is tied to one end of a light string which passes through a small smooth hole in a horizontal table. The particle revolves in a circle of radius r₁ with angular speed ω₁. If the string is slowly pulled down through the hole so that the radius of the circle decreases to r₂ = r₁ / 2, the ratio of the final kinetic energy to the initial kinetic energy of the particle is:

  1. 8 : 1
  2. 2 : 1
  3. 1 : 4
  4. 4 : 1
Show answer and explanation

Answer: (D) 4 : 1

The tension in the string exerts no torque about the central hole, so the angular momentum L of the particle is conserved. The rotational kinetic energy can be expressed as K = L² / (2I). Since I = m r², we have K ∝ 1/r². When r is halved, the moment of inertia becomes 1/4 of its initial value, and therefore kinetic energy increases by a factor of 4.

medium

12. A star of radius R rotates with period T. If it collapses gravitationally under internal forces to a uniform sphere of radius R/n without losing any mass, its new period of rotation will be:

  1. T / n
  2. T √n
  3. n² T
  4. T / n²
Show answer and explanation

Answer: (D) T / n²

Since the collapse is driven purely by internal gravitational forces, no external torque acts, and angular momentum is conserved: I₁ω₁ = I₂ω₂. For a uniform sphere, I ∝ R², so I₂ = I₁ / n². Therefore, ω₂ = n² ω₁. Since period T = 2π/ω, the new period is T₂ = T₁ / n² = T / n².

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