Q 11-03-040NEETJEE MainEasy
A wheel of radius $0.5$ m rotates uniformly at $300$ revolutions per minute. The centripetal acceleration of a point on its rim is
Answer: (A) $50\pi^2\ \text{m/s}^2$
$\omega = 2\pi \times \dfrac{300}{60} = 10\pi$ rad/s.
$$a_c = \omega^2r = 100\pi^2 \times 0.5 = 50\pi^2\ \text{m/s}^2 \approx 493\ \text{m/s}^2$$
Solution by Sreeraj P, M.Sc Physics