JEE Main 2023 — Rotational Motion Question with Solution

Given that the radii of gyration for the ring and the solid sphere are equal, we have:

$$K_1=K_2$$

For the ring, the moment of inertia is:

$$I_{ring}=m{R}_1^2=m{K}_1^2$$

Thus, the radius of gyration for the ring is:

$$K_1=R_1$$

For the solid sphere, the moment of inertia is:

$$I_{sphere}=\frac{2}{5}{m}^{\prime }{R}_2^2=m^{\prime }{K}_2^2$$

Hence, the radius of gyration for the solid sphere is:

$$K_2=\sqrt{\frac{2}{5}}{R}_2$$

Since the radii of gyration are equal:

$$R_1=\sqrt{\frac{2}{5}}{R}_2$$

Therefore, the ratio of the radius of the ring to that of the sphere is:

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

So, the value of $$x$$ is:

$$x=5$$