Moment of Inertia & Radius of Gyration question from MHT CET 2026 (20 May, Shift 1)

Answer: (c)

1. Two masses lie on the axis (the chosen side), so they contribute nothing.
2. The third mass is at the triangle's height $h = \frac{\sqrt{3}}{2}L$ from that side.
3. $I = 2 \times \left(\frac{\sqrt{3}L}{2}\right)^2 = 2 \times \frac{3L^2}{4} = \frac{3L^2}{2}$
4. So the answer is (c) $\frac{3L^2}{2}$.