1. For a body rolling without slipping,

v = omega R

where $v$ is the speed of the center of mass.

2. Total kinetic energy of a rolling body is the sum of translational and rotational kinetic energies:

$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}}$

$K_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$

3. For a hollow sphere (thin spherical shell) about its symmetry axis,

$I = \frac{2}{3}MR^2$

4. Rotational kinetic energy:

$K_{\text{rot}} = \frac{1}{2}\left(\frac{2}{3}MR^2\right)\omega^2 = \frac{1}{3}MR^2\omega^2$

Using $v = \omega R$,

$K_{\text{rot}} = \frac{1}{3}Mv^2$

5. Translational kinetic energy:

$K_{\text{trans}} = \frac{1}{2}Mv^2$

So total kinetic energy is

$K_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{3}Mv^2 = \left(\frac{3}{6}+\frac{2}{6}\right)Mv^2 = \frac{5}{6}Mv^2$

6. Therefore, the required ratio is

$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3}Mv^2}{\frac{5}{6}Mv^2} = \frac{1}{3}\cdot\frac{6}{5} = \frac{2}{5}$

Given that this ratio is $\frac{x}{5}$,

$\frac{x}{5} = \frac{2}{5}$

Hence,

$x = 2$