# A clock has a continously moving second's hand of  $0.1m$ length. The average acceleration of the tip of the hand (in $m/s^{2}$ is of order of :   Option: 1   Option: 2 Option: 3   Option: 4

\begin{aligned} &\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}\\ &\text { Average acceleration is of the order of } 10^{-3} \end{aligned}

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