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 $a = k^{2} r t^{2}$ , where k is a constant. The power delivered to the particle by the force acting on it is given as

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$m k^{2} r^{2} t^{2}$

$m k^{2} r^{2} t$

$m k^{2} r t$

answer is C.

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Detailed Solution

$\begin{array}{rcl} Centripetal acceleration \\ \frac{v^{2}}{r} & = & k^{2} {rt}^{2} \\ v & = & krt \\ tangential acceleration \\ a_{t} & = & \frac{dv}{dt} = kr \\ Power delivered \\ P & = & {ma}_{t} v = {mk}^{2} r^{2} t \end{array}$

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