JEE Physics 2025 Question with Solution

Given: A solid sphere and a hollow sphere have the same mass and the same radius, and both roll down the same inclined plane.

Find: Compare the times $t_1$ and $t_2$ taken to reach the bottom.

For rolling motion, the acceleration depends on the moment of inertia. The object with smaller moment of inertia accelerates more and reaches the bottom sooner.

The relevant energy relation is

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For a solid sphere,

$$I_{\text{solid}} = \frac{2}{5}mr^2$$

For a hollow sphere,

$$I_{\text{hollow}} = \frac{2}{3}mr^2$$

Since

$$\frac{2}{5}mr^2 < \frac{2}{3}mr^2$$

the solid sphere has smaller moment of inertia, so it accelerates faster on the incline and takes less time to reach the bottom.

Therefore, $t_1 < t_2$. The correct option is D.

Given: A solid sphere and a hollow sphere of equal mass and radius are rolled on an inclined plane.

Find: Whether $t_1$ is greater than, equal to, or less than $t_2$.

When an object rolls down an incline, its motion includes both translational and rotational kinetic energy. The total mechanical energy is written as

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

where $m$ is mass, $g$ is acceleration due to gravity, $h$ is height, $v$ is speed of the center of mass, $I$ is moment of inertia, and $\omega$ is angular velocity.

For the two bodies,

$$I_{\text{solid}} = \frac{2}{5}mr^2$$

and

$$I_{\text{hollow}} = \frac{2}{3}mr^2$$

The body with the larger moment of inertia stores more of the gravitational energy in rotation, so less is available for translational acceleration. Hence it moves down the incline more slowly.

Because the hollow sphere has greater moment of inertia than the solid sphere, the solid sphere reaches the bottom first.

Thus, $t_1 < t_2$, so the correct option is D.