Question

A lamina is made by removing a small disc of diameter $2R$ from a bigger disc of uniform mass density and radius $2R$, as shown in the figure. The moment of inertia of this lamina about axes passing through $O$ and $P$ is $I_0$ and $I_P$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_P}{I_0}$

• A. $13/37$
• B. $37/13$
• C. $73/31$
• D. $8/13$

Answer 1

The Correct Option is B

Let $M$ be mass of the whole disc. Then, the mass of the removed disc $= \frac{M}{\pi\left(2R\right)^{2}}\pi R^{2} = \frac{M}{4} $ So, moment of inertia of the remaining disc about an axis passing through $O$ ,br> $I_{0} = \frac{1}{2}M \left(2R\right)^{2} - \left[\frac{1}{2}\left(\frac{M}{4}\right)R^{2} +\frac{M}{4}R^{2}\right]$ $ = 2MR^{2} -\left[\frac{MR^{2}+2MR^{2}}{8}\right] $ $ = MR^{2}\left[2-\frac{3}{8}\right] = \frac{13}{8}MR^{2}$ Moment of the inertia of the remaining disc about an axis passing through $P$ is $I_{p} = \left[\frac{1}{2}M\left(2R^{2}\right) +M\left(2R\right)^{2}\right] -\left[\frac{1}{2}\left(\frac{M}{4}\right)R^{2} + \frac{M}{4}\left(\sqrt{5}R\right)^{2}\right] $ $ = \left[2MR^{2} +4MR^{2}\right] -\left[\frac{MR^{2}}{8} +\frac{5MR^{2}}{4}\right] $ $= 6MR^{2} -\frac{11}{8}MR^{2} $ $=\frac{37}{8}MR^{2} $ $ \therefore\frac{I_{P}}{I_{O}} = \frac{37}{8}\times\frac{8}{13} = \frac{37}{13}$