Medium MCQ +4 / -1 PYQ · JEE Mains 2022

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as

  1. A zero
  2. B mk<sup>2</sup>r<sup>2</sup>t<sup>2</sup>
  3. C mk<sup>2</sup>r<sup>2</sup>t Correct answer
  4. D mk<sup>2</sup>rt

Solution

<p>${a_r} = {k^2}r{t^2} = {{{v^2}} \over r}$</p> <p>$\Rightarrow {v^2} = {k^2}{r^2}{t^2}$ or $v = krt$</p> <p>and ${{d|v|} \over {dt}} = kr$</p> <p>$\Rightarrow {a_t} = kr$</p> <p>$\Rightarrow |\overline F \,.\,\overline v | = (mkr)(krt)$</p> <p>$= m{k^2}{r^2}t =$ power delivered</p>

About this question

Subject: Physics · Chapter: Rotational Motion · Topic: Torque and Angular Momentum

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