Find ratio of radius of gyration of a disk and ring of same radii...

For disk, moment of inertia about the diameter, $\mathrm{I}_{\mathrm{d}}=\frac{\mathrm{mr}^2}{4}$ and moment of inertia about the tangential axis $$\begin{aligned} & \mathrm{I}_{\text{disc }}=\mathrm{I}_{\mathrm{d}}+\mathrm{mr}^2 \\ & \mathrm{I}_{\text{disc }}=\frac{\mathrm{mr}^2}{4}+\mathrm{mr}^2 \\ & \mathrm{I}_{\text{disc }}=\frac{5 \mathrm{mr}^2}{4}\end{aligned}$$ Let the radius of gyration of disc is $\mathrm{k}_{\text{disc }}$. $$\begin{aligned} \mathrm{I}_{\text{disc }} & =\mathrm{mk}_{\text{disc }}^2 \\ \frac{5 \mathrm{mr}^2}{4} & =\mathrm{mk}_{\text{disc }}^2 \\ \mathrm{k}_{\text{disc }}^2 & =\sqrt{\frac{5}{4} \mathrm{r}}\end{aligned}$$ For ring, moment of inertia about the diameter $\mathrm{I}_{\mathrm{a}}^{\prime}=\mathrm{mr}^2 / 2$ and moment of inertia about the tangential axis $\mathrm{I}_{\text{ring }}=$\mathrm{I}$_{\mathrm{a}}^{\prime}+\mathrm{mr}^2=\frac{\mathrm{mr}^2}{2}+\mathrm{mr}^2=\frac{3}{2} \mathrm{mr}^2$ Let the radius of gyration of ring $\mathrm{k}_{\text{ring }}$. $$\begin{aligned} \mathrm{I}_{\text{ring }} & =\mathrm{mk}^2 \text{ ring } \\ \frac{3}{2} \mathrm{mr}^2 & =\mathrm{mk}^2 \text{ ring } \\ \mathrm{k}_{\text{ring }} & =\sqrt{\frac{3}{2}} \mathrm{r} \\ \frac{\mathrm{k}_{\text{disc }}}{\mathrm{k}_{\text{ring }}} & =\frac{\sqrt{\frac{5}{4} \mathrm{r}}}{\sqrt{\frac{3}{2}} \mathrm{r}}=\sqrt{\frac{5}{6}}\end{aligned}$$