Physics · Rotational Dynamics

JEE Main 2025 — 24 January, Evening Shift — Question 50

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 :

Option A:
2/5
Option B:
5/2
Correct
Option C:
3/4
Option D:
4/3

Answer: B

Step-by-step solution

$\frac{\text { Linear KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}}{\frac{1}{2} \mathrm{I} \omega^{2}}$ $\frac{\mathrm{mv}_{\mathrm{cm}}^{2}}{\frac{2}{5} \mathrm{mR}^{2} \omega^{2}}=\frac{5}{2} \quad(\mathrm{~V}=\omega \mathrm{R})$

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Rolling Motion

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the…

Answer: B

Step-by-step solution

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