JEE Main · 2008 · 27 Apr 2008 · Offline · Rotational Mechanics

For a uniform square plate the moment of inertia about an axis through its centre and perpendicular to its plane is $\frac{m{a}^2}{6}$, which follows from the perpendicular-axis theorem applied to the two in-plane central axes, each contributing $\frac{m{a}^2}{12}$. To move the axis to a corner, the parallel-axis theorem adds $m{d}^2$, where $d$ is the centre-to-corner distance; for a square of side $a$ that distance is $\frac{a}{\sqrt{2}}$, so $d^2=\frac{a^2}{2}$. Adding, $I=\frac{m{a}^2}{6}+\frac{m{a}^2}{2}=\frac{m{a}^2+3m{a}^2}{6}=\frac{2}{3}m{a}^2$.