Solid sphere $$A$$ is rotating about an axis $$PQ$$. If the radius of the sphere is 5 cm, then its radius of gyration about $$PQ$$ will be $$\sqrt{x}$$ cm. The value of $$x$$ is _____.

For a uniform solid sphere of mass $$M$$ and radius $$R$$: $$I_{cm} = \frac{2}{5}MR^2$$

$$I_{PQ} = I_{cm} + Md^2 = \frac{2}{5}MR^2 + Md^2$$ (Parallel axis theorem)

The radius of gyration ($$k$$) is the distance from the axis at which the entire mass of the body could be concentrated to result in the same moment of inertia. It is defined by the relation $$I = Mk^2$$.

$$Mk^2 = M \left( \frac{2}{5}R^2 + d^2 \right)$$

$$k^2 = \frac{2}{5}R^2 + d^2$$

$$k^2 = \frac{2}{5}(5)^2 + (10)^2 = 10 + 100 = 110$$

We have $$k = \sqrt{x}$$, which implies $$x = k^2$$

The value of $$x$$ is 110.