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

Options

A. `R // mu_(0)`
B. `mu_(0)R`
C. `(R)/(mu_(0)) (1 - e^(-2 pi))`
D. `(R)/(mu_(0)) e^(-2 pi)`

Answer

C

Explanation

`because` Given `(dv)/(dt) = (v^(2))/(R) " " …….. (i) `
`implies int_(v_(0))^(v) (dv)/(2) = int_(0)^(t) (1)/(R) dt implies R ((1)/(u_(0)) - (1)/(v)) = t " " .....(ii)`
Again from eqns . (i)
` because (dv)/(ds) (ds)/(dt) = (v_(2))/(R) implies v(dv)/(ds) = (v^(2))/(R)`
`int_(u_(0))^(v) (dv)/(v) = int_(0)^(2pi R)(ds)/(R) implies v = u_(0)e^(-2 pi) " " ....(iii)`
From eqns, (ii) and (iii) , `t = (R)/(u_(0)) (1 - e^(-2 pi))`