From a disc of mass ' m 'and radius ' r ' a circular portion of radius $\frac{r}{3}$ is removed from the edges. Find the M.I of the remaining system about a normal through centre of original disc

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$\frac{9}{4} {mr}^{2}$

$\frac{4 {mr}^{2}}{9}$

$\frac{3}{2} {mr}^{2}$

$\frac{2}{3} {mr}^{2}$

answer is C.

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Detailed Solution

Let I₂ be the M.I of remaining portion, M.I of the removed portion about a normal axis through centre of the original disc is
$\begin{array}{l} I_{1} = \frac{\frac{m}{9} {(\frac{r}{3})}^{2}}{2} + \frac{m}{9} {(\frac{2 r}{3})}^{2} \\ I_{1} = \frac{{mr}^{2}}{18} \\ I_{2} = I - I_{1} = \frac{{mr}^{2}}{2} - \frac{{mr}^{2}}{18} \\ I_{2} = \frac{4 {mr}^{2}}{9} \end{array}$