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MHT CET 2020 (16 Oct, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2020 (16 Oct, Shift 2)

Q41MHT CET 2020 · 16 OCT · SHIFT 2Medium

Centripetal Acceleration & Force

A particle of mass mm is performing UCM along a circle of radius rr. The relation between centripetal acceleration aa and kinetic energy EE is given by

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Answer: (a)

  1. E=12mv2⇒v2=2EmE = \frac{1}{2}mv^2 \Rightarrow v^2 = \frac{2E}{m}
  2. a=v2r=2Emra = \frac{v^2}{r} = \frac{2E}{mr}
  3. So the answer is (a) a=2Emra = \frac{2E}{mr}.
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