If a constant torque of `500 N-m` turns a wheel of moment of inertia `100 kg m^(2)` about an axis passing through its centre, find the gain in angular velocity in `2 s`.

Explanation

Here, `tau = 500N-m, I = 100 kg m^(2)`,
`(omega_(2) - omega_(1)) = ?, t = 2 s`
`tau = I alpha = (I(omega_(2)- omega_(1)))/(t)`
`(omega_(2) - omega_(1)) = (tau xx1)/(I) = (500 xx 2)/(100) = 10 rad//s`