JEE Main 2022 · 28 Jun · Shift 1 · Physics Q32

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PhysicsSingle correct+4 / -1Question 32 of 90

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

zero

mk^2r^2t^2

mk^2r^2t

mk^2rt

Answer and worked solution

Official answer: C

Speed. $\frac{v^2}r=k^2rt^2$ , so $v=krt$ and the tangential acceleration is $kr$ .

Power. Only the tangential force does work: $P=m(kr)(krt)=mk^2r^2t$ .

(C)

Topics:Power Uniform circular motion

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JEE Main 2022 · 28 Jun · Shift 1, Physics question 32