Q84 / 90 · JEE Main 2024 (05 Apr Shift 2) · Physics
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 _____ .
Correct answer: 2
Solution
$\begin{aligned} & \frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{mv}^2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2}{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^2} \\ & =\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5} \\ & \mathrm{x}=2\end{aligned}$
Hence, the correct answer is 2.