A wheel of moment of inertia `5xx10^(-3) kg m^(2)` is making 20 revolutions per second. If it is stopped in 20 s, then its angular retardation would be

Options

A. `pi rad //s^(2)`
B. `4pi rad//s^(2)`
C. `2pi rad//s^(2)`
D. `8pi rad //s^(2)`

Answer

C

Explanation

`alpha = (omega_(2)-omega_(1))/(t)=(2pi(n_(2)-n_(1)))/(t)=(2pi(0-20))/(t)=2pi`