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MHT CET 2021 (24 Sep, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2021 (24 Sep, Shift 2)

Q35MHT CET 2021 · 24 SEP · SHIFT 2Medium

Centripetal Acceleration & Force

A body of mass 'm' is moving with speed 'V' along a circular path of radius 'r'. Now the speed is reduced to V2\frac{V}{2} and radius is increased to '3r'. For this change, initial centripetal force needs to be

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Answer: (c)

  1. F=mV2rF = \frac{mV^2}{r}
  2. F′=m(V/2)23r=112mV2r=F12F' = \frac{m(V/2)^2}{3r} = \frac{1}{12}\frac{mV^2}{r} = \frac{F}{12}
  3. Change =F−F12=1112F= F - \frac{F}{12} = \frac{11}{12}F, a decrease.
  4. So the answer is (c) decreased by 1112\frac{11}{12}.
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