In a circular motion of a particle of mass m in orbit of constant radius R, centripetal acceleration varies as square of time given by `(g = pi^(2)ms^(-2))` . The force acting on the particle provides a power given by

Options

A. `m beta Rt`
B. `m beta Rt^(2)`
C. `m beta R^(2)t`
D. `m beta R^(2)t^(2)`

Answer

C

Explanation

Here tangential acceleration on `(a_(r ))` which is equal to `(dv)/(dt)` also exists .
Hence power is required .
Acceleration = `betaRt^(2)=(v^(2))/R :. (v^(2))/R = beta Rt^(2) " or " v = sqrt(beta) . Rt`
`a_(r ) = (dv)/(dt) =d/(dt) (sqrt(beta).R.t) = sqrt(beta).R :. " Force " = m . sqrt(beta).R`
Power = force `xx` velocity = `m . sqrt(beta).R xx sqrt(beta).R.t = mbeta .R^(2)t`