Question

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

• A. ${10}^{-3}$
• B. ${10}^{-1}$
• C. ${10}^{-2}$
• D. ${10}^{-4}$

Answer 1

The correct option is A

${10}^{-3}$

Step 1 : Given data

$R$ $=0.1m$ (It is the radius )

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

$T=60$

Therefore

$$\begin{gathered} \omega =\frac{2\pi }{60} \\ =0.105rad/sec \end{gathered}$$

Step 2 : To find the acceleration

$$\begin{gathered} a={\omega }^{2}R \\ ={\left(0.105\right)}^2\times \left(0.1\right) \\ =0.0011 \\ a=1.1\times {10}^{-3} \end{gathered}$$

Hence the average acceleration is of the order of ${10}^{-3}$

Therefore the correct answer is Option A