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A particle is moving in a circle of radi...

A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of the acceleration are equal. If its speed at `t = 0` is `u_(0)` the time taken to complete the first revolution is :

A

`R//u_(0)`

B

`R/u_(0)(1-e^(-2pi))`

C

`R/u_(0)(1-e^(2pi))`

D

`R/u_(0)e^(-2pi)`

Text Solution

Verified by Experts

The correct Answer is:
C
`:'Given (dv)/(dt)=v^(2)/R`...(1)
`rArroverset(v)underset(u_(0))(int)(dv)/v^(2)=overset(t)underset(0)(int)1/RdtrArrR(1/u_(0)-1/v)=t`...(2)
Again from eqn (1)
`:'(dv)/(ds).(ds)/(dt)=v^(2)/RrArrv(dv)/(ds)=v^(2)/R`
`overset(v)underset(u_(0))(int)(dv)/v=overset(2piR)underset(0)(int)(ds)/R rArr v=u_(0)e^(-2pi)`...(3)
from (2) and (3), `t=R/u_(0)(1-e^(-2pi))`
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