Rotational Motion is a Class 11 Physics chapter in the NEET (UG) syllabus. NEET720 has 1,018 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.
207
easy
629
medium
182
hard
Topics covered
Work and energy in rotation · Combined translation and rotation · Centre of mass · Torque and rigid-body equilibrium · Moment of inertia · Angular momentum and its conservation · Rotational dynamics and rolling motion · Rotational kinetic energy and work · COM · TORQ · MOI · ANGM · ROLL · RKE · Angular momentum and torque · Angular kinematics · Rotational power · Radius of gyration · Rotational work-energy theorem · Parallel axis theorem · Perpendicular axis theorem · Rotational work and energy · Conservation of angular momentum · Rotational energy · Angular velocity · Rigid body motion · Rolling with slipping · Rotational work · Equilibrium of rigid bodies · Angular impulse · Equilibrium · Rotational Dynamics · Rotational Motion · Center of Mass · Centre of mass & torque · Angular momentum · Rolling motion · Rotational kinetic energy · Torque & equilibrium · Moment of inertia & rolling
8 free Rotational Motion practice questions with answers
Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.
Question 1 · easy · Centre of mass
For which one of the following uniform bodies does the centre of mass lie at a point where the body has NO mass?
- A.A circular ring
- B.A solid sphere
- C.A straight uniform rod
- D.A solid cube
Show answer and explanation
Answer: A. A circular ring
A ring's centre of mass is at its geometric centre, but all the ring's mass lies on the circumference, so there is no material at the COM.
The centre of mass is a mathematical average position of the mass distribution; it need not coincide with any material particle of the body. For a uniform ring, symmetry places the COM at the centre of the circle, yet every particle of the ring lies on the circumference — the centre is empty space. For the solid sphere, rod and cube, the COM (geometric centre or midpoint) is a point occupied by material. Remember: for hollow or bent bodies (ring, hollow sphere, L-shaped lamina, horseshoe), the COM can lie outside the material entirely.
Common mistake: Believing the centre of mass must always be a physical point inside the body.
Question 2 · easy · Rotational dynamics and rolling motion
A solid ball rolling without slipping on a horizontal floor gradually slows down and stops because of
- A.kinetic friction at the contact point continuously opposing the motion
- B.rolling friction / resistance, arising from deformation of the ball and/or the surface, and air resistance
- C.the gradual conversion of translational kinetic energy into rotational kinetic energy
- D.gravity acting to decelerate all horizontal motion
Show answer and explanation
Answer: B. rolling friction / resistance, arising from deformation of the ball and/or the surface, and air resistance
Real rolling on a horizontal floor is not perfectly rigid rolling; small deformation of the ball/surface and air drag provide the actual retarding effect, not kinetic friction, since ideal rolling has zero sliding at the contact.
In the idealized rigid-body picture, a ball rolling without slipping on a horizontal floor experiences zero net friction (as established for uniform rolling) and would roll forever. In reality, balls and surfaces are not perfectly rigid: the contact region deforms slightly, creating an asymmetric pressure distribution that generates a small retarding torque  this is called rolling friction or rolling resistance, and is much smaller than typical kinetic friction. Air resistance also contributes at higher speeds. This is why a bowling ball rolls a long way before stopping, unlike a sliding block, which is stopped comparatively quickly by ordinary kinetic friction. NEET tests this distinction between kinetic friction (irrelevant here) and rolling resistance (the real cause).
Common mistake: Attributing the slow-down of a rolling ball to ordinary kinetic friction
Question 3 · easy · Torque
A force of 15 N is applied perpendicular to a spanner at a distance of 0.4 m from the bolt it is turning. What is the magnitude of the torque produced about the bolt?
- A.6 N·m
- B.3.75 N·m
- C.15.4 N·m
- D.0.6 N·m
Show answer and explanation
Answer: A. 6 N·m
τ = F × r (since F ⊥ r) = 15 × 0.4 = 6 N·m.
Torque is τ = r F sinθ, where θ is the angle between the position vector and the force. Here the force is perpendicular to the spanner, so θ = 90° and sinθ = 1. Thus τ = (0.4 m)(15 N)(1) = 6 N·m.
Common mistake: Forgetting that sinθ = 1 only when the force is perpendicular; some students divide instead of multiply.
Question 4 · easy · Moment of inertia
A uniform ring and a uniform disc have the same mass M and the same radius R. Both rotate about an axis through their centre, perpendicular to their plane. What is the ratio of the moment of inertia of the ring to that of the disc?
- A.2 : 1
- B.1 : 2
- C.1 : 1
- D.4 : 1
Show answer and explanation
Answer: A. 2 : 1
I_ring = MR², I_disc = ½MR², so the ratio is 2 : 1.
For a ring, all mass lies at distance R from the axis, giving I_ring = MR². For a disc, mass is distributed from 0 to R, giving I_disc = ½MR². The ratio I_ring : I_disc = MR² : ½MR² = 2 : 1. This reflects that mass concentrated farther from the axis gives a larger moment of inertia.
Common mistake: Assuming equal mass and radius automatically means equal moment of inertia, ignoring the mass distribution.
Question 5 · medium · Rolling motion
A thin hollow spherical shell (I = (2/3)MR²) rolls without slipping down a rough incline of angle 30°. Taking g = 10 m/s², find its linear acceleration.
- A.3 m/s²
- B.5 m/s²
- C.3.57 m/s²
- D.2 m/s²
Show answer and explanation
Answer: A. 3 m/s²
a = g sinθ /(1 + I/MR²) = 10 × 0.5 /(1 + 2/3) = 5/(5/3) = 3 m/s².
For a body of moment of inertia I = kMR² rolling without slipping down an incline, a = g sinθ/(1+k). Here k = 2/3 for a thin hollow sphere. With θ = 30°, sinθ = 0.5: a = (10 × 0.5)/(1 + 2/3) = 5/(5/3) = 5 × 3/5 = 3 m/s².
Common mistake: Using the wrong fractional constant for the body's moment of inertia, or omitting the (1+k) factor entirely.
Question 6 · medium · Angular Momentum Conservation
A person sitting on a frictionless rotating stool with arms outstretched has a moment of inertia of 5 kg·m² and spins at 2 rad/s. He pulls his arms in, reducing his moment of inertia to 2 kg·m². What is his new angular velocity?
- A.5 rad/s
- B.0.8 rad/s
- C.2 rad/s
- D.10 rad/s
Show answer and explanation
Answer: A. 5 rad/s
By conservation of angular momentum, I₁ω₁ = I₂ω₂ ⇒ ω₂ = (5×2)/2 = 5 rad/s.
No external torque acts on the person-stool system about the vertical axis, so angular momentum is conserved: I₁ω₁ = I₂ω₂. Substituting: 5 × 2 = 2 × ω₂, so ω₂ = 10/2 = 5 rad/s.
Common mistake: Forgetting that a decrease in I must correspond to an increase in ω to keep L constant.
Question 7 · easy · Angular momentum
A rotating disc has a moment of inertia of 0.4 kg·m² and an angular velocity of 10 rad/s. What is its angular momentum?
- A.4 kg·m²/s
- B.0.04 kg·m²/s
- C.25 kg·m²/s
- D.10.4 kg·m²/s
Show answer and explanation
Answer: A. 4 kg·m²/s
L = Iω = 0.4 × 10 = 4 kg·m²/s.
Angular momentum of a rigid body rotating about a fixed axis is given by L = Iω. Substituting the given values: L = 0.4 × 10 = 4 kg·m²/s.
Common mistake: Confusing the multiplication with division of I and ω.
Question 8 · medium · Rolling motion
A solid sphere of mass 2 kg rolls without slipping on a horizontal surface with a centre-of-mass speed of 4 m/s. What fraction of its total kinetic energy is rotational?
- A.2/7
- B.1/2
- C.2/5
- D.5/7
Show answer and explanation
Answer: A. 2/7
For any rolling body with I = kMR², rotational KE fraction = k/(1+k). For a solid sphere k = 2/5, so fraction = (2/5)/(7/5) = 2/7.
Total KE = ½Mv² + ½Iω² with I = (2/5)MR² and ω = v/R. KE_trans = ½(2)(4²) = 16 J. KE_rot = ½ × (2/5)(2)R² × (v/R)² = (1/5)Mv² = (1/5)(2)(16) = 6.4 J. Total KE = 22.4 J. Fraction rotational = 6.4/22.4 = 2/7 ≈ 0.286.
Common mistake: Confusing the moment-of-inertia coefficient k with the KE fraction k/(1+k), or swapping the rotational and translational fractions.
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Questions about Rotational Motion for NEET
How many NEET questions does NEET720 have on Rotational Motion?＋
NEET720 has 1,018 reviewed practice questions on Rotational Motion (Physics): 207 easy, 629 medium and 182 hard. 8 of them are free on this page with full explanations; the rest are available in the app.
Is Rotational Motion a Class 11 or Class 12 chapter for NEET?＋
Rotational Motion is a Class 11 Physics chapter in the NEET (UG) syllabus. Read the NCERT chapter first, then practise chapter-wise MCQs and previous-year questions.
How should I practise Rotational Motion for NEET?＋
Attempt the questions below without looking at the options for more than a few seconds, mark your answer, then read the explanation even when you were right. Record every mistake and revisit it after a gap. On NEET720 this happens automatically: wrong answers go to your Mistake Book and are scheduled for spaced revision.
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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (1,018 active practice questions in this chapter today).