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Rigid-body equilibrium, rotational equations and comparison with linear motion: NEET MCQs with answers

52 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 circular disc rotates about its central axis with a constant angular acceleration α. If the ratio of the angle rotated during the 5th second to the angle rotated during the 3rd second from rest is given by r, then r equals:

  1. 7 / 5
  2. 5 / 3
  3. 25 / 9
  4. 9 / 5
Show answer and explanation

Answer: (D) 9 / 5

The angle traversed in the nth second starting from rest is θₙ = (1/2)α(2n - 1). Therefore, θ₅ / θ₃ = (2(5) - 1) / (2(3) - 1) = 9 / 5.

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2. A flywheel rotating at 240 rpm slows down to 120 rpm in 6 s under a constant retarding torque. The number of revolutions made by the flywheel during this time interval is:

  1. 12
  2. 24
  3. 36
  4. 18
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Answer: (D) 18

Initial frequency f₀ = 240/60 = 4 rev/s and final frequency f = 120/60 = 2 rev/s. Since angular deceleration is uniform, the average rotational speed is f_avg = (f₀ + f)/2 = (4 + 2)/2 = 3 rev/s. The total number of revolutions is N = f_avg × t = 3 rev/s × 6 s = 18 revolutions.

hard

3. A heavy uniform rod of mass 6 kg and length 2 m is pivoted at one end to a vertical wall and held in a horizontal position by a wire attached to its other end making an angle of 30° with the horizontal rod. Taking g = 10 m/s², the tension in the wire is:

  1. 30 N
  2. 120 N
  3. 60 N
  4. 45 N
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Answer: (C) 60 N

Taking torque about the pivot on the wall: the weight of the rod acts downwards at its midpoint (1 m from the pivot) with force W = mg = 6 × 10 = 60 N. The vertical component of the tension acts upwards at the outer end (2 m from the pivot) with magnitude T sin 30°. For equilibrium: W × (L/2) = (T sin 30°) × L. Thus, 60 × 1 = T × (1/2) × 2, which gives T = 60 N.

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4. A wheel rotates at 180 rpm. Its angular speed in rad/s is approximately:

  1. 3 rad/s
  2. 18.849556 rad/s
  3. 1130.973355 rad/s
  4. 9.424778 rad/s
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Answer: (B) 18.849556 rad/s

Unit check: f = 180/60 = 3 s⁻¹. Therefore ω = 2πf = 18.8496 rad/s.

easy

5. Which of the following physical quantities is the rotational analogue of force in linear motion?

  1. Angular momentum
  2. Torque
  3. Moment of inertia
  4. Angular impulse
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Answer: (B) Torque

In linear translational motion, force F causes linear acceleration (F = dp/dt). In rotational motion about a fixed axis, torque τ causes angular acceleration (τ = dL/dt), making torque the exact rotational analogue of force.

hard

6. A uniform ladder of mass M and length L rests against a frictionless vertical wall, with its lower end on a rough horizontal floor. If the ladder makes an angle θ with the horizontal floor, what is the magnitude of the normal reaction exerted by the vertical wall on the ladder?

  1. (1/2) M g tan θ
  2. M g tan θ
  3. M g cot θ
  4. (1/2) M g cot θ
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Answer: (D) (1/2) M g cot θ

Let N_w be the normal reaction from the wall. For translational equilibrium in the vertical direction, the normal force from the floor is N_f = Mg. Taking torque about the base on the floor: N_w × (L sin θ) - Mg × ((L/2) cos θ) = 0. Solving for N_w gives N_w = (1/2) Mg (cos θ / sin θ) = (1/2) Mg cot θ.

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7. An electric fan is switched off. Its angular speed decreases uniformly from 50 rad/s to 10 rad/s while making 30 complete revolutions. The uniform angular acceleration of the fan is:

  1. - 2.0 rad/s²
  2. - 4.0 rad/s²
  3. - 6.4 rad/s²
  4. - 8.0 rad/s²
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Answer: (C) - 6.4 rad/s²

Using the rotational kinematic equation ω² = ω₀² + 2αθ: Here ω₀ = 50 rad/s, ω = 10 rad/s, and θ = 30 × 2π = 60π rad. Then 10² - 50² = 2 × α × (60π), which gives -2400 = 120π α, so α = -20/π rad/s² ≈ -6.37 rad/s² ≈ -6.4 rad/s².

medium

8. A wheel rotating at an initial angular speed ω₀ decelerates uniformly to an angular speed ω₀/3 after completing N full revolutions. How many further revolutions will it make before coming completely to rest?

  1. N/4
  2. N/8
  3. N/9
  4. N/3
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Answer: (B) N/8

Using ω² = ω₀² - 2αθ: (ω₀/3)² = ω₀² - 2α(2πN), which gives 2α(2πN) = 8ω₀²/9. For coming to rest from ω₀/3: 0 = (ω₀/3)² - 2α(2πN'), so 2α(2πN') = ω₀²/9. Dividing the two relations gives N/N' = 8, so N' = N/8.

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9. Consider a pair of equal and opposite forces acting along two different, parallel lines of action on a rigid body (a couple). Which of the following statements is correct?

  1. The resultant force is zero, and the torque is independent of the choice of reference point
  2. The couple produces a torque that depends on the choice of the origin
  3. The couple produces translational acceleration but no rotation
  4. The couple cannot change the angular momentum of the body
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Answer: (A) The resultant force is zero, and the torque is independent of the choice of reference point

A couple consists of two equal and opposite forces whose vector sum is identically zero, so there is no net translational force. The torque produced by a couple is equal to the product of one force and the perpendicular distance between them, and its value is completely independent of the reference origin.

medium

10. A light horizontal beam of length 6 m is supported by two vertical pillars at its ends A and B. A point load of 600 N is placed on the beam at a distance of 2 m from end A. Neglecting the weight of the beam, the normal reaction force exerted by the pillar at end B is:

  1. 150 N
  2. 400 N
  3. 300 N
  4. 200 N
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Answer: (D) 200 N

Taking torque about end A for rotational equilibrium: Στ_A = 0 implies (600 N)(2 m) - R_B(6 m) = 0. Solving gives R_B = 1200 / 6 = 200 N.

easy

11. In the context of rigid-body dynamics, which of the following pairs does NOT represent correctly matched analogous linear and rotational physical quantities?

  1. Linear velocity — Angular acceleration
  2. Force — Torque
  3. Mass — Moment of inertia
  4. Linear momentum — Angular momentum
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Answer: (A) Linear velocity — Angular acceleration

The rotational analogue of linear velocity v is angular velocity ω, not angular acceleration α. Angular acceleration α is analogous to linear acceleration a. All other pairs correctly map translational quantities to their rotational counterparts.

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12. Statement I: A body in rotational equilibrium may not be in translational equilibrium. Statement II: The moment of a couple is independent of the point about which moments are taken.

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct.

Statement I is correct because rotational equilibrium only requires Στ = 0; if a net external force acts through the center of mass, translational acceleration exists while torque remains zero. Statement II is correct because the moment of a couple (pair of equal and opposite forces) has the magnitude F × d regardless of the reference origin chosen.

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