Q70 / 75 · JEE Main 2025 (4 Apr Shift 1) · Physics
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: 4
Solution
Applying Mechanical Energy conservation :
$\mathrm{k}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}$
$\begin{aligned}<br/>& \Rightarrow 0+\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)+0 \\ & \Rightarrow \mathrm{~V}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}<br/>\end{aligned}$
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 $\mathrm{x}=4$
$\mathrm{k}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}$
$\begin{aligned}<br/>& \Rightarrow 0+\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)+0 \\ & \Rightarrow \mathrm{~V}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}<br/>\end{aligned}$
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 $\mathrm{x}=4$
Hence, the correct answer is 4.