Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $$\frac{\sqrt{x}}{2}$$. Then, the value of $$x$$ is ___.

For a body rolling without slipping down an inclined plane, energy conservation gives:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Since rolling without slipping: $$\omega = \frac{v}{R}$$, so:

$$mgh = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$

$$v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}$$

For a ring: $$I_{\text{ring}} = mR^2$$, so $$\frac{I}{mR^2} = 1$$, giving:

$$v_{\text{ring}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh}$$

For a solid cylinder: $$I_{\text{cylinder}} = \frac{1}{2}mR^2$$, so $$\frac{I}{mR^2} = \frac{1}{2}$$, giving:

$$v_{\text{cylinder}} = \sqrt{\frac{2gh}{\frac{3}{2}}} = \sqrt{\frac{4gh}{3}}$$

The ratio of velocities:

$$\frac{v_{\text{ring}}}{v_{\text{cylinder}}} = \frac{\sqrt{gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$

Given that this ratio equals $$\frac{\sqrt{x}}{2}$$, we have $$\sqrt{x} = \sqrt{3}$$, so $$x = 3$$.