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Q.
A clock has a continuously moving second's hand of $0.1 m$ length. The average acceleration of the tip of the hand (in units of $ms ^{-2}$ ) is of the order of :
$R=0.1 m$
$\omega=\frac{2 \pi}{ T }=\frac{2 \pi}{60}=0.105 rad / sec$
$a=\omega^{2} R$
$=(0.105)^{2}(0.1)$
$=0.0011$
$=1.1 \times 10^{-3}$
Average acceleration is of the order of $10^{-3}$