Moment of Inertia & Radius of Gyration question from MHT CET 2021 (24 Sep, Shift 1)

Answer: (c)

1. Each particle is at $\frac{d}{2}$ from the axis: $I = 2m\left(\frac{d}{2}\right)^2 = \frac{md^2}{2}$.
2. $K = \frac{1}{2}I\omega^2 = \frac{md^2\omega^2}{4}$
3. $\omega^2 = \frac{4K}{md^2} \Rightarrow \omega = \frac{2}{d}\sqrt{\frac{K}{m}}$
4. So the answer is (c) $\frac{2}{d}\sqrt{\frac{K}{m}}$.