A particle of mass `m` starts moving in a circular path of canstant radiur `r` , such that iss centripetal acceleration`a_(c)` is varying with time a=`t` as
`(a_(c)=k^(2)r//t)` , where `K` is a contant. What is the power delivered to the particle by the force acting on it ?

Answer

A, B, D

Explanation

As `a_(c)=(v^(2)//r)` so `(v^(2)//r)=k^(2)rt^(2)`
:. Kinetic energy `K=(1)/(2)mv^(2)=(1)/(2)mk^(2)r^(2)t^(2)`
Now, from work-energy theorem
`W=DeltaK=(1)/(2)mk^(2)r^(2)t^(2)-0`
So, `P=(dW)/(dt)=(d)/(dt)((1)/(2)mk^(2)r^(2)t^(2))=mk^(2)r^(2)t`
Alternate solution: Given that `a_(c)=k^(2)rt^(2)` , so that
`F_(c)=ma_(c)=mk^(2)rt^(2)`
Now, as `a_(c)=(v^(2)/r)` , so `(v^(2)//r)=k^(2)rt^(2)`
or `v=krt`
So, that `a_(t)=(dv//dt)=kr`
i.e. `F_(t)=ma_(t)=mkr`
Now, as `F=F_(c)+F_(t)`
So, `P=(dW)/(dt)=F.v=(F_(c)+F_(t))_(v)`
In circular motion, `F_(c)` is perpendicular to `v` while `F_(t)` parallel to it, so
`P=F_(t)v`
`P=mk^(2)r^(2)t`