JEE Main 2025 · 24 January · evening · Single correct · +4 / −1

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

  1. (a) 3/4
  2. (b) 4/3
  3. (c) 5/2
  4. (d) 2/5
Answer and solution

Answer (c) 5/2

Solution

Physics sir · 3 steps

Write both energies in terms of v, using v = ωR, and the ratio drops out without any algebra fuss.

  1. Rolling condition. Rolling without slipping means v = ω R, so ω = v/R. For a solid sphere about its centre, I = 2/5MR².
  2. Rotational energy. Kᵣₒₜ = 1/2Iω² = 1/2×2/5MR² × v²/R² = 1/5Mv².

    Kₗᵢₙ/Kᵣₒₜ = (1/2Mv²)/(1/5Mv²) = 5/2

  3. So. The linear kinetic energy is 5/2 times the rotational kinetic energy, which is option (c).

The idea to keep. For any rolling body with I = kMR², the ratio of linear to rotational kinetic energy is 1/k. Read k off the moment of inertia and you are done.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Rotational Motion

Asked about 180 times in JEE Main since 2020. This question is from 2025.