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 \(\mathrm{x}=\) _____

Applying Mechanical Energy conservation :
\(\mathrm{k}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}\)
\(\begin{aligned}
& \Rightarrow 0+\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)+0 \\
& \Rightarrow \mathrm{~V}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}
\end{aligned}\)
So Ratio of velocities
\(\frac{\mathrm{V}_{\text {Ring }}}{\mathrm{V}_{\text {solids sphere }}}=\sqrt{\frac{1+\frac{2}{5}}{1+1}}=\sqrt{\frac{7}{10}}\)
\(\mathrm{x}=3.5\) Rounding off \(\mathrm{x}=4\)