Explanation

(d) Linear kinetic energy $= \frac{1}{2}mv^{2}$
Rotational kinetic energy $= \frac{1}{2}I\omega^{2}$
$= \frac{1}{2}\left( \frac{2}{5}mr^{2} \right)\frac{v^{2}}{r^{2}}\ $
$= \frac{1}{5}mv^{2}$
Total KE $= \frac{1}{2}mv^{2} + \frac{1}{5}mv^{2}$
$= \frac{7}{10}mv^{2}$
Required fraction $= \frac{\frac{1}{5}mv^{2}}{\frac{7}{10}mv^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$