1. Moment of inertia of a square plate about its center

For a square lamina of side $l$ and mass $M$, the moment of inertia about an axis through its center and perpendicular to its plane is

$I_C = \frac{1}{12}M(l^2+l^2)=\frac{1}{6}Ml^2$

since for a rectangular plate,

$I_C=\frac{1}{12}M(a^2+b^2)$

and here $a=b=l$.

2. Distance from center to a corner

The center of the square is at equal distance from all four corners.

Half the diagonal of the square is:

$d=\frac{l\sqrt{2}}{2}=\frac{l}{\sqrt{2}}$

So,

$d^2=\frac{l^2}{2}$

3. Apply parallel axis theorem

The required axis passes through a corner and is perpendicular to the plane of the square. By the parallel axis theorem,

$I_{\text{corner}} = I_C + Md^2$

Substitute the values:

$I_{\text{corner}} = \frac{1}{6}Ml^2 + M\cdot \frac{l^2}{2}$

$I_{\text{corner}} = Ml^2\left(\frac{1}{6}+\frac{1}{2}\right)$

$I_{\text{corner}} = Ml^2\left(\frac{1+3}{6}\right)=\frac{4}{6}Ml^2$

$I_{\text{corner}}=\frac{2}{3}Ml^2$

4. Check options

• A: $\frac{Ml^2}{6}$ ❌
• B: $Ml^2$ ❌
• C: $\frac{Ml^2}{12}$ ❌
• D: $\frac{2}{3}Ml^2$ ✅

Hence, the correct option is D.