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:
$\mathrm{R}=0.1 \mathrm{~m}$
$\omega=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{60}=0.105 \mathrm{rad} / \mathrm{sec}$
$a=\omega^{2} R$
$=(0.105)^{2}(0.1)$
$=0.0011$
$=1.1 \times 10^{-3}$
Average acceleration is of the order of $10^{-3}$
$\therefore$ correct option is (1)
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