A flywheel is in the form of a uniform circular disc of radius `1 m` and mass `2 kg`. The work which must be done on it to increase its frequency of rotation from `5` to `10 rev//s` is approximately

Options

A. `1.5xx10^(2)J`
B. `3.0xx10^(2)J`
C. `1.5xx10^(3)J`
D. `3.0xx10^(3)J`

Answer

C

Explanation

Work done `W=(1)/(2)l(omega^(2)-omega_(0)^(2))`
`thereforeW=(1)/(2)((1)/(2)xx2xx1^(2))" "[(20pi)^(2)-(10pi)^(2)]`
`W=(pi^(2))/(2)(400-100)`
Taking `pi^(2)=10`, we get
`W=(10)/(2)xx300=5xx300`
`W=1500J=1.5xx10^(3)J`