Find the ratio of radius of gyration of a disc and ring of same radii at their tangential axis in plane.

For disc, moment of inertia about the diameter, $I_d=m{r}^2/4$

and moment of inertia about the tangential axis,

$I_{\mathrm{disc}}=I_d+m{r}^2$

$I_{\mathrm{disc}}=\frac{m{r}^2}{4}+m{r}^2=\frac{5m{r}^2}{4}$

Let the radius of gyration of disc be $K_{{\text{disc }}^{.}}$

$I_{\text{disc }}=m{K}_{\text{disc }}^2$

$\Rightarrow \frac{5m{r}^2}{4}=m{K}_{\mathrm{disc}}^2\Rightarrow K_{\mathrm{disc}}^2=\sqrt{\frac{5}{4}r}$

For ring, moment of inertia about the diameter,

$I_a=\frac{m{r}^2}{2}$

and moment of inertia about the tangential axis,

$I_{\text{ring }}=I_a+m{r}^2=\frac{m{r}^2}{2}+m{r}^2=\frac{3}{2}m{r}^2$

Let the radius of gyration of ring be  $K_{\text{ring }}$

$I_{\text{ring }}=m{K}_{\text{ring }}^2$

$\Rightarrow \frac{3}{2}m{r}^2=m{K}_{\text{ring }}^2\Rightarrow K_{\text{ring }}=\sqrt{\frac{3}{2}r}\text{. }$

$\because \frac{K_{\text{disc }}}{K_{\text{ring }}}=\frac{\sqrt{\frac{5}{4}r}}{\sqrt{\frac{3}{2}r}}=\sqrt{\frac{5}{6}}$