Solve this following


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:


  1. $10^{-3}$

  2. $10^{-2}$

  3. $10^{-4}$

  4. $10^{-1}$

Correct Option: 1


$\mathrm{R}=0.1 \mathrm{~m}$

$\omega=\frac{2 \pi}{\mathrm{T}}=\frac{2 \pi}{60}=0.105 \mathrm{rad} / \mathrm{sec}$

$a=\omega^{2} R$



$=1.1 \times 10^{-3}$

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

$\therefore$ correct option is (1)


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