Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis AB as shown in figure is $$\sqrt{8/x}$$. The value of $$x$$ is :

For a hollow sphere of mass $$M$$ and radius $$R$$, the moment of inertia about its diameter is:

$$I_1 = \frac{2}{3}MR^2$$

The radius of gyration is $$k_1 = \sqrt{\frac{I_1}{M}} = \sqrt{\frac{2}{3}R^2}$$

For the solid cylinder of mass $$M$$, radius $$R$$, and length $$L = 4R$$, the axis AB passes through one end. Using the Parallel Axis Theorem:

$$I_2 = I_{cm} + M\left(\frac{L}{2}\right)^2$$

$$I_2 = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) + M\frac{L^2}{4} = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right)$$

$$I_2 = M\left(\frac{R^2}{4} + \frac{(4R)^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{16R^2}{3}\right)$$

$$I_2 = MR^2 \left(\frac{3 + 64}{12}\right) = \frac{67}{12}MR^2$$

The radius of gyration is $$k_2 = \sqrt{\frac{I_2}{M}} = \sqrt{\frac{67}{12}R^2}$$

$$\text{Ratio} = \frac{k_1}{k_2} = \frac{\sqrt{\frac{2}{3}R^2}}{\sqrt{\frac{67}{12}R^2}}$$ = $$\sqrt{\frac{8}{67}}$$

$$x\ =\ 67$$