Q 11-03-144JEE MainJEE Main 2020 (6 Sep, Shift 1)Easy
A clock has a continuously moving second's hand of $0.1$ m length. The average acceleration of the tip of the hand (in units of m s$^{-2}$) is of the order of:
Answer: (A) $10^{-3}$
The tip moves uniformly on a circle of radius $0.1$ m with period $60$ s, so its acceleration has constant magnitude $\omega^2 r$:
$$\omega = \frac{2\pi}{60} \approx 0.105\ \text{rad/s}$$
$$a = \omega^2 r \approx 0.011\times0.1 \approx 1.1\times10^{-3}\ \text{m s}^{-2}$$
So the acceleration is of the order of $10^{-3}$ m s$^{-2}$.
Solution by Sreeraj P, M.Sc Physics