JEE Physics 2023 Question with Solution

Given: A ring and a solid sphere have the same radius of gyration.

Find: The value of $x$ in the relation $\dfrac{R_{\text{ring}}}{R_{\text{sphere}}} = \sqrt{\dfrac{2}{x}}$.

For a solid sphere,

$$I = \frac{2}{5}mR_1^2 = mK_1^2$$

So,

$$K_1 = \sqrt{\frac{2}{5}}\,R_1$$

For a ring,

$$K_2 = R_2$$

Since the radii of gyration are the same,

$$K_1 = K_2$$

Therefore,

$$\sqrt{\frac{2}{5}}\,R_1 = R_2$$

Hence,

$$\frac{R_2}{R_1} = \sqrt{\frac{2}{5}}$$

Comparing with $\sqrt{\frac{2}{x}}$, we get

$$x = 5$$

Therefore, the value of $x$ is $5$.

Given: The radius of gyration of the ring and the solid sphere are equal.

Find: The value of $x$.

Radius of gyration of a ring about its central axis is

$$K_{\text{ring}} = R_{\text{ring}}$$

Radius of gyration of a solid sphere about its diameter is

$$K_{\text{sphere}} = \sqrt{\frac{2}{5}}\,R_{\text{sphere}}$$

Since they are equal,

$$R_{\text{ring}} = \sqrt{\frac{2}{5}}\,R_{\text{sphere}}$$

Thus,

$$\frac{R_{\text{ring}}}{R_{\text{sphere}}} = \sqrt{\frac{2}{5}}$$

Comparing with $\sqrt{\frac{2}{x}}$, we obtain

$$x = 5$$

Therefore, the value of $x$ is $5$.