JEE Physics 2025 Question with Solution

Given: A solid sphere rolls without slipping down an incline of length $L$. In the first case, the incline angle is $30^\circ$ and the speed at the bottom is $v_1$. In the second case, the incline angle is $45^\circ$ and the speed at the bottom is $v_2$.

Find: The ratio $v_1^2 : v_2^2$.

For a solid sphere rolling without slipping, the speed at the bottom is obtained from conservation of energy:

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

For a solid sphere,

$$I = \frac{2}{5}mr^2, \qquad \omega = \frac{v}{r}$$

Substituting,

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\frac{v^2}{r^2}$$ $$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$

Hence,

$$v^2 = \frac{10}{7}gh$$

Now the vertical height is

$$h = L\sin\theta$$

So,

$$v^2 = \frac{10}{7}gL\sin\theta$$

For $\theta = 30^\circ$,

$$v_1^2 = \frac{10}{7}gL\sin 30^\circ = \frac{10}{7}gL \cdot \frac{1}{2} = \frac{5gL}{7}$$

For $\theta = 45^\circ$,

$$v_2^2 = \frac{10}{7}gL\sin 45^\circ = \frac{10}{7}gL \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}gL}{7}$$

Therefore,

$$v_1^2 : v_2^2 = \frac{5gL}{7} : \frac{5\sqrt{2}gL}{7} = 1 : \sqrt{2}$$

Therefore, the correct option is A.

Given: The sphere rolls without slipping down inclines of the same length $L$, with angles $30^\circ$ and $45^\circ$.

Find: The ratio $v_1^2 : v_2^2$.

Since for a given rolling body,

$$v^2 \propto h$$

and

$$h = L\sin\theta$$

with $L$ constant, we get

$$v^2 \propto \sin\theta$$

Thus,

$$v_1^2 : v_2^2 = \sin 30^\circ : \sin 45^\circ = \frac{1}{2} : \frac{\sqrt{2}}{2} = 1 : \sqrt{2}$$

This works because the rotational factor is the same in both cases and cancels in the ratio. Therefore, the correct option is A.