JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(10^{-3}\)
- B \(10^{-2}\)
- C \(10^{-4}\)
- D \(10^{-1}\)
Answer & Solution
Correct Answer
(A) \(10^{-3}\)
Step-by-step Solution
Detailed explanation
\(R=0.1 m\) \(\omega=\frac{2 \pi}{ T }=\frac{2 \pi}{60}=0.105 rad / sec\) \(a =\omega^{2} R\) \(=(0.105)^{2}(0.1)\) \(=0.0011\) \(=1.1 \times 10^{-3}\) Average acceleration is of the order of \(10^{-3}\)
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