A particluar moves along a circle of radius P with constant tangential accelertion . It the velocity of the particles is V at the end of third revolution , after the revolution has started then the tangenital accleration is

Options

A. `(v^(2))/(12pir)`
B. `(v^(2))/(10 pir)`
C. `(v^(2))/(14 pi r)`
D. `(v^(2))/(9 pir)`

Answer

A

Explanation

Given , Initial velocity u =0
`because ` Radius of circle r .
Velocity particle V = V
`therefore ` Distance moved in 3 revolution , `S = 3 xx 2pi r = 6pir `
we know that `v^(2) = u^(2) + 2as`
On putting the value of all variable es , we get
`V^(2) = 2a xx 6pi`
`rArr " "a = (v^(2))/(12 pir)`