Exp. (a)Given force, F=20t−5t2α=lFR​=10(20t−5t2)2​=4t−t2⇒dtdω​=4t−t2 ⇒∫0∞​dω=∫0t​(4t−t2)dt⇒ω=2t2−3t3​When direction is reversed, ω=0, i.e., t=0,6 sNow, dθ=ωdt⇒∫0θ​dθ=∫06​(2t2−3t3​)dt⇒θ=[32t3​−12t4​]06​⇒θ=144−108=36rad∴ Number of rotations, n=2πθ​=2π36​<6