A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution, after the revolution has started, then the tangential acceleration is

Options

A. `(v^2)/(8pir)`
B. `(v^2)/(6pir)`
C. `(v^2)/(4pir)`
D. `(v^2)/(2pir)`

Answer

A

Explanation

According to question, applying equation of motion ,
`v^2 =u^2 +2as`
[Symbols have their usual meanings ]
Given ,u=0
`S=2xx2pir=4pir`
`so , " "v^2 =2axx4pirrArr a =(v^2)/(8pir)`