Moment of Inertia & Radius of Gyration question from MHT CET 2025 (25 Apr, Shift 1)

Answer: (b)

1. Each corner is at half the diagonal from the centre: $r = \frac{\sqrt{2}L}{2} = \frac{L}{\sqrt{2}}$.
2. $I = 4M\left(\frac{L}{\sqrt{2}}\right)^2 = 2ML^2$
3. $K^2 = \frac{I}{4M} = \frac{L^2}{2} \Rightarrow K = \frac{L}{\sqrt{2}}$
4. So the answer is (b) $\frac{L}{\sqrt{2}}$.