Q 11-06-083JEE MainJEE Main 2024 (5 Apr, Shift 2)Easy
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of its rotational kinetic energy to its total kinetic energy is $\dfrac{x}{5}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 2
For a hollow sphere $I = \dfrac23mR^2$, so $K_{rot} = \dfrac12\cdot\dfrac23mR^2\omega^2 = \dfrac13mv^2$ and $K_{trans} = \dfrac12mv^2$.
$$\frac{K_{rot}}{K_{total}} = \frac{1/3}{1/2 + 1/3} = \frac{2}{5} \Rightarrow x = 2$$
Solution by Sreeraj P, M.Sc Physics