Question

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

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

Answer 1

The correct option is A ${10}^{-3}$

Here, $R=0.1\text{m}$
$\omega =\frac{2\pi }{T}=\frac{2\pi }{60}=0.105\text{rad/s}$
Accelaration of the tip of the clock second's hand will be,
$a={\omega }^{2}R=\left(0.105{)}^2\right(0.1)=0.0011=1.1\times {10}^{-3}{\text{m/s}}^2$
Hence, average accelaration is of the order of ${10}^{-3}$.