A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where $x=$ ________.

<p>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:</p> <p><strong>Conservation of Mechanical Energy:</strong></p> <p>$ k_i + U_i = k_f + U_f $</p> <p>Initially (at the top), we have:</p> <p><p>Initial kinetic energy, $k_i = 0$ (since they start from rest)</p></p> <p><p>Initial potential energy, $U_i = Mgh$</p></p> <p>Finally (at the bottom), we have:</p> <p><p>Final potential energy, $U_f = 0$</p></p> <p><p>Final kinetic energy, $k_f = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$</p></p> <p>This gives:</p> <p>$ Mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right) $</p> <p>Solving for the velocity $V$:</p> <p>$ V = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} $</p> <p><strong>Ratio of Velocities:</strong></p> <p>For the circular ring (moment of inertia $I = mR^2$), $\frac{k^2}{R^2} = 1$.</p> <p>For the solid sphere (moment of inertia $I = \frac{2}{5}mR^2$), $\frac{k^2}{R^2} = \frac{2}{5}$.</p> <p>The ratio is:</p> <p>$ \frac{V_{\text{Ring}}}{V_{\text{Solid Sphere}}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + 1}} = \sqrt{\frac{7}{10}} $</p> <p>Thus, $x = 3.5$. Rounding off, $x = 4$.</p>