JEE Physics 2025 Question with Solution

Given: A circular ring and a solid sphere of the same radius roll down an inclined plane from rest without slipping.

Find: The value of $x$ in

$$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \sqrt{\frac{x}{5}}$$

Using mechanical energy conservation for rolling motion,

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

With the rolling condition $v = R\omega$,

$$mgh = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$

So,

$$v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}$$

For the circular ring,

$$I_{\text{ring}} = mR^2$$

Hence,

$$v_{\text{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$$

For the solid sphere,

$$I_{\text{sphere}} = \frac{2}{5}mR^2$$

Hence,

$$v_{\text{sphere}} = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{10gh}{7}}$$

Now the ratio is

$$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \frac{\sqrt{gh}}{\sqrt{\frac{10gh}{7}}} = \sqrt{\frac{7}{10}}$$

Comparing with

$$\sqrt{\frac{x}{5}}$$

we get

$$\frac{x}{5} = \frac{7}{10}$$

Therefore,

$$x = \frac{7}{2} = 3.5$$

So, the value of $x$ is $3.5$.

Given: Both bodies start from rest and roll without slipping.

Find: The numerical value of $x$.

For a rolling body,

$$v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}$$

For the ring,

$$\frac{I}{mR^2} = 1$$

so

$$v_{\text{ring}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh}$$

For the solid sphere,

$$\frac{I}{mR^2} = \frac{2}{5}$$

so

$$v_{\text{sphere}} = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{2gh}{\frac{7}{5}}} = \sqrt{\frac{10gh}{7}}$$

Then,

$$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \sqrt{\frac{gh}{10gh/7}} = \sqrt{\frac{7}{10}}$$

Given that

$$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \sqrt{\frac{x}{5}}$$

Equating the terms inside the square roots,

$$\frac{x}{5} = \frac{7}{10}$$

Thus,

$$x = \frac{7}{2} = 3.5$$

the solution also mentions an approximate value of $4$ in one approach, but the worked derivation gives the exact value $3.5$, so that is the correct numerical answer.