Explanation

(d) From third equation of angular motion,
$\omega^{2} = \omega_{0}^{2} = 2\alpha\ \ (Here,\omega = \frac{\omega_{0}}{2},\theta = 36 \times 2\pi)$
$\therefore$ $\left( \frac{\omega_{0}}{2} \right)^{2} = \omega_{0}^{2} - 2\alpha \times 36 \times 2\pi$
or $4 \times 36\pi\alpha = \omega_{0}^{2} - \frac{\omega_{0}^{2}}{4}$
or $4 \times 36\pi\alpha = \frac{3\omega_{0}^{2}}{4}$
or $\alpha = \frac{\omega_{0}^{2}}{16 \times 12\pi}$ …(i)
According to question again applying the third equation of angular motion
$\omega^{2} = \omega_{0}^{2} - 2\alpha\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Here,\ \omega = 0)$
$\therefore$ $0 = \left( \frac{\omega_{0}}{2} \right)^{2} - 2 \times \frac{\omega_{0}^{2}.\theta}{16 \times 12\pi}$
or $\theta = 24\pi\ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \theta = 12 \times 2\pi$
But $2\pi = 1$ cycle
So, $\theta = 12\ cycle$