Moment of Inertia & Radius of Gyration question from JEE Main 2021 (24 Feb, Shift 2)

Answer: 8

1. Each side: $m = 1\text{ kg}$, $a = 0.4\text{ m}$, at distance $\frac{\sqrt{3}a}{2}$ from the centre.
2. One side: $\frac{ma^2}{12} + m\frac{3a^2}{4} = \frac{5}{6}ma^2$
3. Hexagon: $I = 6 \times \frac{5}{6}ma^2 = 5ma^2 = 5 \times 1 \times 0.16 = 0.8\text{ kg m}^2$
4. So the answer is $8 \times 10^{-1}\text{ kg m}^2$, i.e. $8$.