Physics · Rotational Dynamics

JEE Main 2024 — 5 April, Shift 2 — Question 54

A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$ . The value of $x$ is $\qquad$ .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\frac{\frac{1}{2} I \omega^{2}}{\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} m^{2}\right) \omega^{2}}{\left(\frac{1}{2}\right)\left(\frac{2}{3} m R^{2}\right) \omega^{2}+\frac{1}{2} m(R \omega)^{2}}$

$=\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5}$ $\mathrm{x}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Work-Energy Theorem in General Motion