To determine the ratio of velocities for a circular ring and a solid sphere rolling down an inclined plane without slipping, we apply the principle of mechanical energy conservation:

Conservation of Mechanical Energy:

$ k_i + U_i = k_f + U_f $

Initially (at the top), we have:

Initial kinetic energy, $k_i = 0$ (since they start from rest)

Initial potential energy, $U_i = Mgh$

Finally (at the bottom), we have:

Final potential energy, $U_f = 0$

Final kinetic energy, $k_f = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$

This gives:

$ Mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right) $

Solving for the velocity $V$:

$ V = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} $

Ratio of Velocities:

For the circular ring (moment of inertia $I = mR^2$), $\frac{k^2}{R^2} = 1$.

For the solid sphere (moment of inertia $I = \frac{2}{5}mR^2$), $\frac{k^2}{R^2} = \frac{2}{5}$.

The ratio is:

$ \frac{V_{\text{Ring}}}{V_{\text{Solid Sphere}}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}} $

Thus, $x = 3.5$. Rounding off, $x = 4$.