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JEE Main 2020 (6 Sep, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2020 (6 Sep, Shift 1)

Q40JEE Main 2020 · 6 SEP · SHIFT 1Easy

Centripetal Acceleration & Force

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:

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Answer: (a)

  1. Second hand: T=60 sT = 60\text{ s}, so ω=2π60≈0.105 rad/s\omega = \frac{2\pi}{60} \approx 0.105\text{ rad/s}.
  2. Tip moves uniformly, so its acceleration is centripetal: a=ω2ra = \omega^2 r.
  3. a=(0.105)2×0.1≈1.1×10−3 m/s2a = (0.105)^2 \times 0.1 \approx 1.1 \times 10^{-3}\text{ m/s}^2
  4. So the answer is (a) of the order 10−310^{-3}.
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