From a circular disc of radius R and mass $9 M$ , a small disc of radius $\frac{R}{3}$ is removed from the edge of the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

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$10 M R^{2}$

$\frac{40}{9} M R^{2}$

$\frac{37}{9} M R^{2}$

$4 M R^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} {}^{_{I_{r e m a i n i n g}} =_{I_{w h o l e} - I_{r e m o v e d}}} \\ o r I = \frac{1}{2} (9 m) {(R)}^{2} - [\frac{1}{2} m {(\frac{R}{3})}^{2} + m {(\frac{2 R}{3})}^{2}] \\ H e r e, m = \frac{9 M}{π R^{2}} x π {(\frac{R}{3})}^{2} = M \\ S u b s t i t u t i n g I n E q . (1), w e h a v e \\ I = 4 M R^{2} \end{array}$