A circular ring and a solid sphere having same radius roll down on...

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}=

_____

Correct answer

Step-by-step solution

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$

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