Solid sphere is rotating about an axis as shown in figure. If the radius of the sphere is 5 cm, then radius of gyration about that axis will be $\sqrt{x}$ cm. The value of x is

The moment of inertia of a solid sphere about an axis passing through its center of mass is given by $I_{cm} = \dfrac{2}{5}MR^2$.

Using the parallel axis theorem, the moment of inertia about the given axis of rotation is:

$I = I_{cm} + Md^2$

where $d$ is the distance from the axis of rotation to the center of mass of the sphere.

Given $R = 5$ cm and $d = 10$ cm, we have:

$I = \dfrac{2}{5}M(5)^2 + M(10)^2$

$I = 10M + 100M = 110M$

The radius of gyration $k$ is defined by $I = Mk^2$.

$Mk^2 = 110M$

$k^2 = 110$

$k = \sqrt{110}$ cm

Comparing this with $\sqrt{x}$ cm, we get $x = 110$.