If a constant torque of 5OO N-m turns a wheel of moment of inertia 1OO kg-m2 about an axis passing though its centre, then the gain in angular velocity in 2 sec. is

τ=Iα, where I=moment of inertia of wheel about an axis passing through its centreBut α=dω/dt, hence τ=I(dω/dt)5u6s1i1uting the values, we get500=100dω2  or  dω=10rad/sec