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MHT CET 2021 (21 Sep, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2021 (21 Sep, Shift 1)

Q36MHT CET 2021 · 21 SEP · SHIFT 1Medium

Centripetal Acceleration & Force

A particle moves in a circular orbit of radius 'rr' under a central attractive force, F=−krF = -\frac{k}{r}, where kk is a constant. The periodic time of its motion is proportional to

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Answer: (c)

  1. Central force gives the centripetal force: mv2r=kr\frac{mv^2}{r} = \frac{k}{r}.
  2. v=kmv = \sqrt{\frac{k}{m}}, independent of rr.
  3. T=2πrv∝rT = \frac{2\pi r}{v} \propto r
  4. So the answer is (c) rr.
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