Correct Answer - D
From third equation of angular motion, `omega^(2)=omega_(0)^(2)-2alphatheta`
[Here, `omega=(omega_(0))/(2),theta=36xx2pi]`
`:.((omega_(0))/(2))^(2)=omega_(0)^(2),theta=36xx2pi`
or `4xx36pialpha=omega_(0)^(2)-(omega_(0)^(2))/(4)`
or `4xx36pialpha=(3omega_(0)^(2))/(4)`
or `alpha=(omega_(0)^(2))/(16xx12pi)" ".......(i)`
According to qestion again applying the third equation of angular motion
`omega^(2)=omega_(0)^(2)-2alphatheta" "["Here",omega=0]`
`:.0=((omega_(0))/(2))-2xx(omega_(0)^(2).theta)/(16xx12pi)`
or `theta=24piortheta=12xx2pi`
But `2pi1` cycle
So, `theta=12` cycle