A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $$\frac{x}{5}$$. The value of $$x$$ is ______.

We need to find the ratio of rotational kinetic energy to total kinetic energy for a hollow sphere rolling on a plane surface about its axis of symmetry.

Since the sphere is thin-walled, its moment of inertia about an axis through its centre is $$I = \frac{2}{3}mr^2$$, where $$m$$ is the mass and $$r$$ is the radius.

For rolling without slipping with velocity $$v$$ and angular velocity $$\omega = v/r$$, the translational kinetic energy is $$K_{trans} = \frac{1}{2}mv^2$$ and the rotational kinetic energy is $$K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{3}mv^2$$.

Next, the total kinetic energy is $$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mv^2\bigl(\tfrac{1}{2} + \tfrac{1}{3}\bigr) = mv^2 \cdot \frac{5}{6} = \frac{5}{6}mv^2$$.

From this, the ratio of rotational to total kinetic energy is $$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2} = \frac{\frac{1}{3}}{\frac{5}{6}} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$$. Since this ratio equals $$\frac{x}{5}$$, we obtain $$x = 2$$.

The answer is 2.