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 $$\sqrt{\dfrac{x}{5}}$$ where x = ________.

For any rigid body rolling without slipping down an inclined plane, conservation of mechanical energy gives

$$mgh \;=\;\frac12\,m v^{2}\;+\;\frac12\,I\omega^{2}$$
Since the body rolls without slipping, $$\omega=\dfrac{v}{R}$$. Thus

$$mgh \;=\;\frac12\,m v^{2}\;+\;\frac12\,I\dfrac{v^{2}}{R^{2}} =\frac12\,m v^{2}\Bigl(1+\frac{I}{mR^{2}}\Bigr)$$

Therefore the speed attained at the bottom is

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

For a thin ring about its symmetry axis, $$I_{ring}=mR^{2}$$, so $$\dfrac{I}{mR^{2}}=1$$.

$$v_{ring}=\sqrt{\frac{2gh}{1+1}} =\sqrt{\frac{2gh}{2}} =\sqrt{gh}$$

For a solid sphere about a diameter, $$I_{sphere}=\tfrac{3}{5}mR^{2}$$, hence $$\dfrac{I}{mR^{2}}=\tfrac{3}{5}$$.

$$v_{sphere}=\sqrt{\frac{2gh}{1+\tfrac{3}{5}}} =\sqrt{\frac{2gh}{\tfrac{8}{5}}} =\sqrt{\frac{5gh}{4}}$$

Required ratio:

$$\frac{v_{ring}}{v_{sphere}} =\sqrt{\frac{gh}{\tfrac{5gh}{4}}} =\sqrt{\frac{4}{5}} =\sqrt{\frac{x}{5}}$$

Hence $$x=4$$.

Thus, the ratio of their velocities at the bottom of the plane is $$\sqrt{\dfrac{4}{5}}$$, giving $$x = 4$$.