Explanation

(b) Rotational kinetic energy $K_{R} = \frac{1}{2}I\omega^{2}$
$K_{R} = \frac{1}{2} \times \frac{MR^{2}}{2} \times \omega^{2} = \frac{1}{4}Mv^{2}\ \ \ \lbrack\because v = R\omega\rbrack$
Translational kinetic energy $K_{T} = \frac{1}{2}Mv^{2}$
Total kinetic energy $= K_{T} + K_{R}$
$= \frac{1}{2}Mv^{2} + \frac{1}{4}Mv^{2} = \frac{3}{4}Mv^{2}$
$\frac{Rotational\ kinetic\ energy}{Total\ kinetic\ energy} = \frac{\frac{1}{4}Mv^{2}}{\frac{3}{4}Mv^{2}} = \frac{1}{3}$