Correct Answer - C
We know that `omega=2pin" "therefore omega_(1)=2pin_(1)`
where, `n_(1)=1800"rpm", n_(2)=3000"rpm", Deltat=20s`
`rArr" "omega_(1)=2pixx(1800)/(60)=2pixx30=60pi`
Similarly, `omega_(2)=2pin_(2)=2pixx(3000)/(60)=2pixx50=100pi`
If the angular velocity of a rotating wheel about an axis changes from `omega_(1)` to `omega_(2)` in a time interval `Deltat`, then the angular acceleration of rotation wheel about that axis is given by
`alpha=("change in angular velocity")/("time interval")`
`=(omega_(2)-omega_(1))/(Deltat)=(100pi-60pi)/(20)=(40pi)/(20)=2pi" rad s"^(-2)`