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Physics-

General

Easy

Question

A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a subscript c end subscript$ is varying with time $t$ as, $a subscript c end subscript equals k to the power of 2 end exponent r t to the power of 2 end exponent$ , The power delivered to the particle by the forces acting on it is

$2 π m k^{2} r^{2} t$
$m k^{2} r^{2} t$
$\frac{m k^{4} r^{2} t^{5}}{3}$
Zero

The correct answer is: $m k to the power of 2 end exponent r to the power of 2 end exponent t$

Here the tangential acceleration also exists which requires power
Given that $a subscript C end subscript equals k to the power of 2 end exponent r t to the power of 2 end exponent$ and $a subscript C end subscript equals fraction numerator v to the power of 2 end exponent over denominator r end fraction therefore fraction numerator v to the power of 2 end exponent over denominator r end fraction equals k to the power of 2 end exponent r t to the power of 2 end exponent$
Or $v to the power of 2 end exponent equals k to the power of 2 end exponent r to the power of 2 end exponent t to the power of 2 end exponent$ or $v equals k r t$
Tangential acceleration $a equals fraction numerator d v over denominator d t end fraction equals k r$
Now force $F equals m cross times a equals m k r$
So power $P equals F cross times v equals m k r cross times k r t equals m k to the power of 2 end exponent r to the power of 2 end exponent t$

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