The radius of gyration K of a hollow sphere of mass M and radius R about an axis XY is equal to R as shown in figure. The distance of that axis from the center of the sphere is h. The value of h is

The moment of inertia of a hollow sphere about an axis passing through its center is given by:

$I_{CM} = \dfrac{2}{3}MR^2$

Using the parallel axis theorem, the moment of inertia about the axis $XY$ at a distance $h$ from the center is:

$I_{XY} = I_{CM} + Mh^2$

$I_{XY} = \dfrac{2}{3}MR^2 + Mh^2$

The radius of gyration $K$ about the axis $XY$ is given as $R$. Therefore, the moment of inertia can also be written as:

$I_{XY} = MK^2 = MR^2$

Equating the two expressions for $I_{XY}$:

$MR^2 = \dfrac{2}{3}MR^2 + Mh^2$

$h^2 = R^2 - \dfrac{2}{3}R^2$

$h^2 = \dfrac{1}{3}R^2$

$h = \dfrac{R}{\sqrt{3}}$

Answer: $\dfrac{R}{\sqrt{3}}$