If a circular concentric hole is made in a disc, then about its own geometrical axis

If M be the mass of solid disc and m be the mass of remaining disc, Then

$m=\frac{M}{\pi b^2}\times \pi \left(b^2-a^2\right)=\frac{M\left(b^2-a^2\right)}{b^2}$

Now for the solid disc  $\mathrm{I}=\frac{1}{2}{\mathrm{Mb}}^2$
and for the hollow disc  ${\mathrm{I}}^{\text{'}}=\frac{1}{2}\mathrm{m}\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)=\frac{\mathrm{M}\left({\mathrm{b}}^2-{\mathrm{a}}^2\right)\left({\mathrm{b}}^2+{\mathrm{a}}^2\right)}{2{\mathrm{b}}^2}$
                                             $\therefore {\mathrm{I}}^{\text{'}}<\mathrm{I}$
K = Radius of gyration of solid disc =  $\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}$ and $\mathrm{K}\text{'}=$ Radius of gyration of hollow disc $=\sqrt{\frac{\mathrm{I}\text{'}}{\mathrm{m}}}$

$\therefore \mathrm{K}=\sqrt{\frac{\frac{1}{2}{\mathrm{Mb}}^2}{\mathrm{M}}}=\frac{\mathrm{b}}{\sqrt{2}}\mathrm{and}\mathrm{K}\text{'}=\sqrt{\frac{{\mathrm{b}}^2+{\mathrm{a}}^2}{2}}>\mathrm{K}$