For rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is x 5 . The…

The rotational kinetic energy is given by 

$K_{rot}=\frac{I{\omega }^2}{2}=\frac{1}{3}M{R}^2{\omega }^2$.

The total kinetic energy is given by 

$$\begin{gathered} K_{tot}=\frac{1}{2}m{v}^2+\frac{1}{3}M{R}^2{\omega }^2 \\ =\frac{m{R}^2{\omega }^2}{2}+\frac{1}{3}M{R}^2{\omega }^2 \\ =\frac{5}{6}M{R}^2{\omega }^2 \end{gathered}$$

Thus, the ratio is 

$\frac{K_{rot}}{K_{tot}}=\frac{{\displaystyle \frac{1}{3}M{R}^2{\omega }^2}}{{\displaystyle \frac{5}{6}M{R}^2{\omega }^2}}=\frac{2}{5}$

Comparing with the given value,

$$\begin{gathered} \frac{x}{5}=\frac{2}{5} \\ \Rightarrow x=2 \end{gathered}$$