JEE Main · 2006 · 30 Apr 2006 · Offline · Rotational Mechanics

Place the square with $A$ at the origin and the sides along the axes: $A(0,0)$, $B(\ell ,0)$, $C(\ell ,\ell )$, $D(0,\ell )$. The diagonal $BD$ runs along the direction $(-1,1)$, and the axis is the parallel line through $A$. The perpendicular distance of a corner $(x,y)$ from that line is $\frac{|x+y|}{\sqrt{2}}$, giving squared distances $0$ for $A$, $\frac{{\ell }^2}{2}$ for $B$, $2{\ell }^2$ for $C$ and $\frac{{\ell }^2}{2}$ for $D$. Summing $m{r}^2$ over the four masses, $I=m\left(0+\frac{{\ell }^2}{2}+2{\ell }^2+\frac{{\ell }^2}{2}\right)=3m{\ell }^2$.