Q 11-06-071JEE MainJEE Main 2025 (24 Jan, Shift 2)Easy
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
Answer: (C) $\dfrac{5}{2}$
$K_{lin} = \tfrac{1}{2}mv^2$ and, with $I = \tfrac{2}{5}mR^2$ and $\omega = v/R$, $K_{rot} = \tfrac{1}{2}\cdot\tfrac{2}{5}mR^2\cdot\dfrac{v^2}{R^2} = \tfrac{1}{5}mv^2$.
$$\frac{K_{lin}}{K_{rot}} = \frac{1/2}{1/5} = \frac{5}{2}$$
Solution by Sreeraj P, M.Sc Physics