A clock has a continuously moving second's hand of $$0.1\,\text{m}$$ length. The average acceleration of the tip of the hand (in units of $$\text{ms}^{-2}$$) is of the order of:

We start by noting that the tip of the second’s hand performs uniform circular motion with radius equal to the length of the hand. Thus, for all instants its speed is constant in magnitude and the only acceleration it possesses is the centripetal acceleration directed toward the centre of the clock face.

The length (radius) of the second’s hand is given as $$r = 0.1\;\text{m}.$$

First we need the angular speed. A second’s hand completes one full revolution in $$T = 60\;\text{s}.$$ The standard formula connecting time-period and angular speed is

$$\omega = \frac{2\pi}{T}.$$

Substituting $$T = 60\;\text{s},$$ we obtain

$$\omega = \frac{2\pi}{60} = \frac{\pi}{30}\;\text{rad s}^{-1}.$$

To find the linear speed $$v$$ of the tip we use the relation

$$v = \omega r.$$

Putting $$\omega = \dfrac{\pi}{30}\;\text{rad s}^{-1}$$ and $$r = 0.1\;\text{m},$$ we get

$$v = \left(\frac{\pi}{30}\right)\!(0.1) = \frac{0.1\pi}{30} = \frac{\pi}{300} \;\text{m s}^{-1}.$$

Taking $$\pi \approx 3.14,$$ this yields

$$v \approx \frac{3.14}{300} = 0.01047\;\text{m s}^{-1}.$$

For uniform circular motion the instantaneous (centripetal) acceleration magnitude is

$$a_c = \frac{v^{2}}{r}.$$

Substituting $$v = 0.01047\;\text{m s}^{-1}$$ and $$r = 0.1\;\text{m},$$ we have

$$a_c = \frac{(0.01047)^{2}}{0.1} = \frac{0.0001097}{0.1} = 0.001097\;\text{m s}^{-2}.$$

This value is approximately $$1.1 \times 10^{-3}\;\text{m s}^{-2}.$$ Since the direction of the centripetal acceleration keeps changing but its magnitude stays the same, the average magnitude over any interval is of the same order. Hence, the required order of the average acceleration is $$10^{-3}\;\text{m s}^{-2}.$$

Hence, the correct answer is Option A.