JEE Main 2025 — Rotational Motion Question with Solution

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:

Conservation of Mechanical Energy:

$k_i+U_i=k_f+U_f$

Initially (at the top), we have:

Initial kinetic energy, $k_i=0$ (since they start from rest)

Initial potential energy, $U_i=Mgh$

Finally (at the bottom), we have:

Final potential energy, $U_f=0$

Final kinetic energy, $k_f=\frac{1}{2}m{v}^2\left(1+\frac{k^2}{R^2}\right)$

This gives:

$Mgh=\frac{1}{2}m{v}^2\left(1+\frac{k^2}{R^2}\right)$

Solving for the velocity $V$:

$V=\sqrt{\frac{2gh}{1+\frac{k^2}{R^2}}}$

Ratio of Velocities:

For the circular ring (moment of inertia $I=m{R}^2$), $\frac{k^2}{R^2}=1$.

For the solid sphere (moment of inertia $I=\frac{2}{5}m{R}^2$), $\frac{k^2}{R^2}=\frac{2}{5}$.

The ratio is:

$\frac{V_{\text{Ring}}}{V_{\text{Solid Sphere}}}=\sqrt{\frac{1+\frac{2}{5}}{1+1}}=\sqrt{\frac{7}{10}}$

Thus, $x=3.5$. Rounding off, $x=4$.