Centripetal Acceleration & Force question from JEE Main 2024 (5 Apr, Shift 2)

Answer: (b)

1. Frequency $n = \frac{120}{180\text{ s}} = \frac{2}{3}\text{ Hz}$, so $\omega = 2\pi n = \frac{4\pi}{3}\text{ rad/s}$.
2. The monkey moves with the man on the same circle, so $a = \omega^2 r$.
3. $a = \frac{16\pi^2}{9} \times 9 = 16\pi^2\text{ m/s}^2$
4. So the answer is (b) $16\pi^2\text{ m s}^{-2}$.