System of Particles and Rotational Motion question for NEET / JEE Main, with solution

Answer: (B) $\dfrac{1}{3}$

$K_{rot} = \dfrac{1}{2}\left(\dfrac{1}{2}MR^2\right)\omega^2 = \dfrac{1}{4}Mv^2$ and $K_{trans} = \dfrac{1}{2}Mv^2$.
$$\frac{K_{rot}}{K_{total}} = \frac{1/4}{3/4} = \frac{1}{3}$$

Solution by Sreeraj P, M.Sc Physics