Class XI · Chapter 06
System of Particles and Rotational Motion — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering torque, moment of inertia, and angular momentum.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why does a diver curl their body while performing a somersault in mid-air?
Answer
Curling the body pulls mass closer to the axis of rotation, reducing moment of inertia. Since angular momentum (L = Iω) is conserved in the air (no external torque), a smaller I means a larger ω — the diver spins faster.
Curling the body pulls mass closer to the axis of rotation, reducing moment of inertia. Since angular momentum (L = Iω) is conserved in the air (no external torque), a smaller I means a larger ω — the diver spins faster.
Derive the relation between torque and angular acceleration for a rigid body rotating about a fixed axis.
Answer
For a particle of mass mi at distance ri from the axis, tangential force Fi = miai = mi(riα). Torque due to this particle: τi = riFi = miri²α.
For a particle of mass mi at distance ri from the axis, tangential force Fi = miai = mi(riα). Torque due to this particle: τi = riFi = miri²α.
Summing over all particles: τ = (Σmiri²)α = Iα, where I = Σmiri² is the moment of inertia.
Q3. (a) State and derive the parallel axis theorem. (b) A uniform rod of mass 2 kg and length 1 m has moment of inertia ML²/12 about a central axis perpendicular to its length. Find its moment of inertia about a parallel axis through one end.
Answer
(a) The moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass, plus Md² (d = distance between the axes): I = Icm + Md².
(a) The moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass, plus Md² (d = distance between the axes): I = Icm + Md².
(b) Icm = ML²/12 = (2×1)/12 ≈ 0.1667 kg·m². The end axis is at distance d = L/2 = 0.5 m from the centre.
Iend = Icm + Md² = 0.1667 + 2×(0.5)² = 0.1667 + 0.5 = 0.667 kg·m² (equivalently ML²/3).
2NEET Style (Single-Correct MCQ)
Q4. The moment of inertia of a uniform ring of mass M and radius R about an axis through its centre, perpendicular to its plane, is:
Solution
Every point on a ring lies at exactly the same distance R from the central axis, so I = MR² directly — the largest moment of inertia (for given M, R) among common shapes.
Every point on a ring lies at exactly the same distance R from the central axis, so I = MR² directly — the largest moment of inertia (for given M, R) among common shapes.
Q5. Angular momentum of a rotating body is conserved when:
Solution
τ = dL/dt. With zero net external torque, L stays exactly constant, even if I and ω individually change (as in the diver example).
τ = dL/dt. With zero net external torque, L stays exactly constant, even if I and ω individually change (as in the diver example).
Q6. The SI unit of torque is:
Solution
Torque = force × perpendicular distance, giving N·m — numerically the same unit as work/energy (joule), though torque is a vector and treated distinctly.
Torque = force × perpendicular distance, giving N·m — numerically the same unit as work/energy (joule), though torque is a vector and treated distinctly.
Q7. The rotational kinetic energy of a rigid body rotating with angular velocity ω and moment of inertia I is:
Solution
KErot = ½Iω² — the direct rotational analogue of ½mv² for linear motion, with I replacing m and ω replacing v.
KErot = ½Iω² — the direct rotational analogue of ½mv² for linear motion, with I replacing m and ω replacing v.
3JEE Main Style
Q8. A disc of mass 1 kg and radius 0.5 m spins about its central axis at ω = 10 rad/s. Find its rotational kinetic energy.
Solution
I = ½MR² = ½×1×(0.5)² = 0.125 kg·m².
I = ½MR² = ½×1×(0.5)² = 0.125 kg·m².
KE = ½Iω² = ½×0.125×(10)² = ½×0.125×100 = 6.25 J.
Q9. A solid sphere and a hollow spherical shell, of equal mass and radius, are released from rest at the top of the same incline and roll without slipping. Which reaches the bottom first?
Solution
The solid sphere has a smaller moment of inertia (2/5 MR² vs. 2/3 MR² for the shell), so less of its gravitational PE goes into rotational KE and more into translational KE — giving it greater linear acceleration down the incline.
The solid sphere has a smaller moment of inertia (2/5 MR² vs. 2/3 MR² for the shell), so less of its gravitational PE goes into rotational KE and more into translational KE — giving it greater linear acceleration down the incline.
Q10. A torque of 10 N·m is applied to a wheel of moment of inertia 2 kg·m². Find its angular acceleration.
Solution
α = τ/I = 10/2 = 5 rad/s².
α = τ/I = 10/2 = 5 rad/s².
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