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 lt brgt `v^2=u^2+2as` lt brgt Given, u=0
`S=2xx2pir=4pir`
So, `v^2=2axx4pir Rightarrow a= v^2/(8pir)`