JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Rotational Mechanics

Treat the triangle as a uniform lamina and remember that for any family of geometrically similar plane shapes the moment of inertia about the corresponding axis scales as the fourth power of the linear size. The reason is that mass itself follows the area, which goes as the square of the side, and the characteristic distance from the axis contributes another square, so $I\propto m{\ell }^2\propto {\ell }^4$. The inner triangle $DEF$ is formed by joining the midpoints, so its side is exactly half that of $ABC$ and its moment of inertia about the same central axis is ${\left(\frac{1}{2}\right)}^4{I}_0=\frac{1}{16}{I}_0$. Removing it simply subtracts that share, leaving $I=I_0-\frac{1}{16}{I}_0=\frac{15}{16}{I}_0$.