Physics - Rotation Question with Solution | TestHub

Let $M$ and $R$ be the mass and radius of ring and disc then M.I. of ring about an axis passing through its centre and perpendicular to its plane is $$I_{\text{ring }}=M{R}^2$$ And M.I. of disc about an axis passing through centre is $\begin{array}{rl}& I_{\text{disc }}=\frac{M{R}^2}{2}\\ & \text{ i.e. }\begin{array}{rl}I& =M{K}^2\\ \text{ where }K=\text{ radius of }& \text{ gyration }\\ \therefore I_{\text{ring }}& =M{K}_{\text{ring }}^2\\ & =M{R}^2\\ I_{\text{fring }}& =R\\ I_{\text{disc }}& =M{K}_{\text{disc }}^2\\ & =\frac{M{K}^2}{2}\\ K_{\text{disc }}& =\frac{R}{\sqrt{2}}\end{array}\end{array}$ $\begin{array}{rl}\frac{{\mathrm{K}}_{\text{ring }}}{{\mathrm{K}}_{\text{desc }}}& =\frac{R}{\frac{R}{\sqrt{2}}}=\frac{\sqrt{2}}{1}\\ & =\sqrt{2}:1\end{array}$