The correct option is D 10 rad/s
Given torque, τ=20 Nm
Moment of inertia of flywheel, I=25 kgm2.
The flywheel turns through 10 revolutions, which is 20π radians.
The work done by the torque,
W=τ(θ2−θ1)=ΔKErot
W=τ(θ2−θ1)=12Iω22−12Iω21
Here, (θ2−θ1)=20π given
ω1=0 (Flywheel initially at rest).
τ(θ2−θ1)=12Iω22−12Iω21
⇒20×20π=12×25×ω22−12×25×0
⇒ω2=10 rad/s
The angular velocity of flywheel after 10 revolutions is 10 rad/s.