Question

a wheel starts from rest on application of torque which gives it an angular acceleration @=2t^2-t^2 for first two seconds after which @= 0 then find angular velocity of wheel after 4 seconds.

Answer 1

so the angular velocity after four seconds is the same after two seconds, as after two seconds, no more acceleration. 2t^2-t^2 is an odd algebra function, as it reduces quickly to t^2

I don't know your math level, calculus or not.
If you have not had calculus, do this.
Plot angular acceleration vs time from zero to t=2. Now compute the area under that plot: the area is angular velocity (acceleration vs time=velocity). So use a graph paper which lets you count squares. Some graphing calulators will measure the area under the graph for you.
If you are in integral calculus:
velocity=INTegral a(t)dt from zero to 2
= INT (2t^2-t^2)dt=t^3/3 fromzero to 2
= 8/4 rad /sec

Check you acceleration function to verify it is 2t^2-t^2. Even in my Texas, that is odd.