Moment of inertia of an equilateral triangular lamina $ABC$, about the axis passing through its centre $O$ and perpendicular to its plane is $I_0$ as shown in the figure. A cavity $DEF$ is cut out from the lamina, where $D, E, F$ are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is:
Answer: (B) $\dfrac{15}{16}I_0$
For a uniform equilateral triangle of mass $M$ and side $a$, the moment of inertia about the perpendicular axis through its centroid is $I = kMa^2$ (with $k$ a fixed number).
Triangle $DEF$ is equilateral with side $a/2$, so its area (and mass) is $M/4$, and its centroid is also $O$:
$$I_{DEF} = k\cdot\frac M4\cdot\frac{a^2}{4} = \frac{I_0}{16}$$
$$I_{\text{remaining}} = I_0 - \frac{I_0}{16} = \frac{15}{16}I_0$$
Solution by Sreeraj P, M.Sc Physics