Moment of Inertia and Radius of Gyration

Radius of gyration of a solid sphere: $K_S=\sqrt{\frac{2}{5}}R$ Radius of gyration of a hollow sphere: $K_H=\sqrt{\frac{2}{3}}R$ $\Rightarrow \frac{K_S}{K_H}=\sqrt{\frac{3}{5}}=\frac{\sqrt{3}}{\sqrt{5}}$ Answer key in the source: (None).

For a disc about the axis through its centre, normal to its plane: $I_1 = \frac{mR^2}{2}$ About its diameter: $I_2 = \frac{mR^2}{4}$ Radius of gyration $k = \sqrt{\frac{I}{m}}$ $\frac{k_1}{k_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{mR^2/2}{mR^2/4}} = \sqrt{2}$ So $k_1 : k_2 = \sqrt{2} : 1$