Question:
A clock has a continuously moving second's hand of $0.1$ $\mathrm{m}$ length. The average acceleration of the tip of the hand (in units of $\mathrm{ms}^{-2}$ ) is of the order of:
Correct Option: 1
Solution:
(1) Here, $R=0.1 \mathrm{~m}$
$\omega=\frac{2 \pi}{T}=\frac{2 \pi}{60}=0.105 \mathrm{rad} / \mathrm{s}$
Acceleration of the tip of the clock second's hand,
$a=\omega^{2} R=(0.105)^{2}(0.1)=0.0011=1.1 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}$
Hence, average acceleration is of the order of $10^{-3}$.
Comments
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.