From a circular disc of radius $R$ and mass $9M$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its center is: (2010 Mains)
Solution:
Initial moment of inertia $I_{original} = \frac{1}{2}(9M)R^{2} = \frac{9}{2}MR^{2}$. Moment of inertia of the removed part is $I_{removed} = \frac{1}{2}M(\frac{R}{3})^{2} = \frac{1}{18}MR^{2}$. The remaining moment of inertia is $I = I_{original} - I_{removed} = \frac{9}{2}MR^{2} - \frac{1}{18}MR^{2} = \frac{40}{9}MR^{2}$.
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