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NEET Rotational Motion MCQs with answers

401 questions on Rotational Motion for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

mediumNEET 2019

1. Moment of inertia of uniform disc about diameter:

  1. MR²/2
  2. MR²/4
  3. MR²
  4. 2MR²
Show answer and explanation

Answer: (B) MR²/4

By perpendicular axis theorem: MR²/2 = 2I_diameter → I = MR²/4.

Quick tip: Disc about center = MR²/2. Divide by 2 for diameter = MR²/4.

NCERT Class 11, Chapter 7, page 152

easy

2. The moment of inertia of a uniform circular disc of mass 6 kg and radius 0.1 m about an axis through its centre, perpendicular to its plane, is:

  1. 0.03 kg m²
  2. 0.06 kg m²
  3. 0.015 kg m²
  4. 0.3 kg m²
Show answer and explanation

Answer: (A) 0.03 kg m²

For a disc about its central perpendicular axis, I = MR²/2 = 6 × 0.1²/2 = 0.03 kg m².

Quick tip: Disc (perpendicular axis): MR²/2. Ring: MR². Disc about a diameter: MR²/4.

NCERT Class 11, Chapter 6

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3. A wheel rotates at 120 rpm. Its angular speed in rad/s is approximately:

  1. 12.566371 rad/s
  2. 2 rad/s
  3. 753.982237 rad/s
  4. 6.283185 rad/s
Show answer and explanation

Answer: (A) 12.566371 rad/s

Unit check: f = 120/60 = 2 s⁻¹. Therefore ω = 2πf = 12.5664 rad/s.

hard

4. A uniform solid cylinder of height H and radius R is joined coaxially with a uniform solid hemisphere of radius R made of the same material, such that the flat circular face of the hemisphere completely covers one circular base of the cylinder. If the centre of mass of the combined body lies on the interface between the cylinder and hemisphere, the height H of the cylinder must be:

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

Answer: (A) R/√2

Let the flat interface lie on the plane z = 0. The cylinder extends from z = 0 to z = -H, so its volume is V₁ = πR²H and its centre of mass is at z₁ = -H/2. The hemisphere extends into z > 0, so its volume is V₂ = (2/3)πR³ and its centre of mass is at z₂ = 3R/8. For the combined centre of mass to lie at z = 0, we must have V₁ z₁ + V₂ z₂ = 0 ⇒ (πR²H)(-H/2) + ((2/3)πR³)(3R/8) = 0. Simplifying gives -(1/2)πR²H² + (1/4)πR⁴ = 0 ⇒ H²/2 = R²/4 ⇒ H² = R²/2 ⇒ H = R/√2.

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5. A thin rod of length L is positioned along the x-axis with one end at the origin x = 0 and the other at x = L. Its linear mass density varies linearly according to the relation λ(x) = λ₀(1 + x/L), where λ₀ is a constant. The x-coordinate of its centre of mass is:

  1. 3L/5
  2. 7L/12
  3. 2L/3
  4. 5L/9
Show answer and explanation

Answer: (D) 5L/9

The total mass is M = ∫₀ᴸ λ₀(1 + x/L)dx = λ₀[x + x²/(2L)]₀ᴸ = (3/2)λ₀L. The first moment of mass is ∫₀ᴸ xλ(x)dx = λ₀∫₀ᴸ (x + x²/L)dx = λ₀[x²/2 + x³/(3L)]₀ᴸ = λ₀(L²/2 + L²/3) = (5/6)λ₀L². Thus, x_cm = [(5/6)λ₀L²] / [(3/2)λ₀L] = 5L/9.

easy

6. A force of 30 N acts at a point 1.5 m from the axis of rotation, at 60° to the position vector. The magnitude of the torque is:

  1. 39 N m
  2. 20 N m
  3. 45 N m
  4. 22.5 N m
Show answer and explanation

Answer: (A) 39 N m

τ = rF sin θ = 1.5 × 30 × sin 60° = 39 N m.

hard

7. A uniform ladder of mass 20 kg and length 5 m rests with its upper end against a smooth vertical wall and its lower end on a rough horizontal floor at a distance of 3 m from the wall. The minimum coefficient of static friction μ between the floor and the ladder to prevent slipping is (take g = 10 m/s²):

  1. 0.250
  2. 0.500
  3. 0.375
  4. 0.750
Show answer and explanation

Answer: (C) 0.375

The vertical height reached on the wall is h = √(5² - 3²) = 4 m. Vertical equilibrium gives normal force from floor N = mg = 200 N. Taking torque about the base on the floor: N_wall × 4 - mg × (1.5) = 0, so N_wall = (200 × 1.5)/4 = 75 N. Horizontal equilibrium gives static friction f = N_wall = 75 N. Thus μ ≥ f / N = 75 / 200 = 0.375.

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8. A uniform hollow right circular cone of height h (without base) has its apex at the origin and its axis along the z-axis. The distance of the centre of mass of this conical surface from the apex is:

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

Answer: (B) 2h/3

For a uniform hollow cone (open at the base), the centre of mass is located at 2/3 of its height measured from the apex along the central axis (or h/3 from the base). For a solid cone, it would be 3h/4 from the apex.

easy

9. A uniform disc has mass 9 kg and radius 0.5 m. Its moment of inertia about the axis through its centre and perpendicular to its face is:

  1. 2.25 kg m²
  2. 0.565 kg m²
  3. 0.563 kg m²
  4. 1.13 kg m²
Show answer and explanation

Answer: (D) 1.13 kg m²

For a disc about its central perpendicular axis, I = MR²/2 = 9 × 0.5²/2 = 1.13 kg m².

Quick tip: Disc (perpendicular axis): MR²/2. Ring: MR². Disc about a diameter: MR²/4.

NCERT Class 11, Chapter 6

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10. A wheel rotates at 900 rpm. Its angular speed in rad/s is approximately:

  1. 5654.866776 rad/s
  2. 15 rad/s
  3. 94.24778 rad/s
  4. 47.12389 rad/s
Show answer and explanation

Answer: (C) 94.24778 rad/s

Unit check: f = 900/60 = 15 s⁻¹. Therefore ω = 2πf = 94.2478 rad/s.

easy

11. The rotational analogue of mass in linear motion is moment of inertia. Which of the following physical quantities is the correct rotational analogue of linear momentum (p = mv)?

  1. Torque (τ)
  2. Angular impulse (J)
  3. Angular momentum (L)
  4. Rotational kinetic energy (K)
Show answer and explanation

Answer: (C) Angular momentum (L)

In rotational motion, moment of inertia (I) corresponds to mass (m), and angular velocity (ω) corresponds to linear velocity (v). Therefore, angular momentum L = Iω is the direct analogue of linear momentum p = mv.

easy

12. Which of the following physical quantities must pass through the centre of mass of a body for the parallel axis theorem, I = I₀ + Md², to be correctly applicable?

  1. The axis about which I₀ is measured
  2. The axis about which I is measured
  3. Both axes simultaneously
  4. Neither axis necessarily
Show answer and explanation

Answer: (A) The axis about which I₀ is measured

The parallel axis theorem states that I = I_cm + Md², where I_cm (here denoted as I₀) must strictly be the moment of inertia about an axis passing through the centre of mass of the body, and the second axis must be parallel to it at distance d.

easy

13. A particle of mass m moves with a constant velocity v parallel to the x-axis along the line y = b. What is the magnitude of its angular momentum about the origin?

  1. Zero
  2. m v / b
  3. m v √(x² + b²)
  4. m v b
Show answer and explanation

Answer: (D) m v b

Angular momentum is given by L = r × p. Its magnitude is L = m v r_⊥, where r_⊥ is the perpendicular distance from the origin to the line of motion. Here, r_⊥ = b, so L = m v b, which remains constant at all times.

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14. A uniform wire of length L is bent into an L-shape with two mutually perpendicular arms of lengths L/3 and 2L/3. The coordinates of the corner are taken as (0, 0), with the arms lying along the positive x-axis and positive y-axis respectively. The position vector of the centre of mass is:

  1. (L/6) î + (L/3) ĵ
  2. (L/18) î + (2L/9) ĵ
  3. (L/9) î + (2L/9) ĵ
  4. (L/3) î + (L/6) ĵ
Show answer and explanation

Answer: (B) (L/18) î + (2L/9) ĵ

Let linear mass density be λ. Mass of horizontal arm is m₁ = λ(L/3) centered at (L/6, 0). Mass of vertical arm is m₂ = λ(2L/3) centered at (0, L/3). Total mass is M = λL. Therefore, x_cm = [λ(L/3)(L/6)] / (λL) = L/18, and y_cm = [λ(2L/3)(L/3)] / (λL) = 2L/9.

easy

15. A skater spins at 4 rad/s with arms outstretched. On folding the arms, the moment of inertia becomes 1/4 of its initial value. The new angular speed is:

  1. 16 rad/s
  2. 4 rad/s
  3. 1 rad/s
  4. 64 rad/s
Show answer and explanation

Answer: (A) 16 rad/s

No external torque, so Iω is conserved: I·4 = (I/4)·ω′, giving ω′ = 4 × 4 = 16 rad/s.

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

NCERT Class 11, Chapter 6

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