For a body rolling without slipping, $K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}}$

Translational kinetic energy is $K_{\text{trans}} = \frac{1}{2}Mv^2$

Rotational kinetic energy is $K_{\text{rot}} = \frac{1}{2}I\omega^2$

For a spherical shell about its center, $I = \frac{2}{3}MR^2$

Since it is rolling without slipping, $v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R}$

Substitute into rotational kinetic energy:

Total kinetic energy:

Therefore, the required ratio is $\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3}Mv^2}{\frac{5}{6}Mv^2} = \frac{2}{5}$

Given that this ratio is $\frac{x}{5}$, we compare: $\frac{x}{5} = \frac{2}{5} \Rightarrow x = 2$