QJEEM23Ph708e: JEE Main 2023 Physics - Class 11 JEEM23Ph708e Solution: A ring and a solid sphere rotating about an axis passing…

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:

$I_{ring} = mR_1^2 = mK_1^2$

Thus, the radius of gyration for the ring is:

$K_1 = R_1$

For the solid sphere, the moment of inertia is:

$I_{sphere} = \frac{2}{5}m'R_2^2 = m'K_2^2$

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:

$x = 5$