JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Rotational Mechanics

Take the axis through one corner and parallel to the diagonal that does not pass through that corner. The mass sitting on the axis contributes nothing. The two masses adjacent to it lie at a perpendicular distance $\frac{l}{\sqrt{2}}$ from the axis, so each contributes $m{\left(\frac{l}{\sqrt{2}}\right)}^2=\frac{m{l}^2}{2}$. The mass at the far corner is at a perpendicular distance $\sqrt{2}l$, contributing $m(\sqrt{2}l{)}^2=2m{l}^2$. The moment of inertia about the axis is therefore $I=2\times \frac{m{l}^2}{2}+2m{l}^2=3m{l}^2$, and the angular momentum is $L=I\omega =3m{l}^2\omega$.