Moment of Inertia & Radius of Gyration question from JEE Main 2015 (10 Apr, Shift 1)

Answer: (c)

1. About the axis through the centre perpendicular to the sheet: $I_z = \frac{m(a^2+a^2)}{12} = \frac{ma^2}{6}$.
2. Perpendicular axis theorem with the two diagonals (equal by symmetry): $I_z = I_{d1} + I_{d2} = 2I$.
3. $I = \frac{ma^2}{12}$
4. So the answer is (c) $I = \frac{ma^2}{12}$.