Q 11-06-134JEE MainJEE Main 2022 (26 Jun, Shift 2)Easy
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is
Answer: (C) $\dfrac27$
For rolling, $K_{\text{rot}} = \frac12\cdot\frac25mR^2\omega^2 = \frac15mv^2$ and $K_{\text{total}} = \frac12mv^2 + \frac15mv^2 = \frac7{10}mv^2$.
$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{1/5}{7/10} = \frac27$$
Solution by Sreeraj P, M.Sc Physics