(d) $\theta = 2\pi n = \omega_{0}t + \frac{1}{2}\alpha t^{2}$
$2\pi \times 10 = \frac{1}{2}\ \alpha 4^{2}$ or $\alpha = \frac{40\pi}{16}$
Let it make $N$ rotations in the first 8 s
Then, $2\pi N = \frac{1}{2}\alpha\ 8^{2}$
or $N = \frac{1}{2\pi} \times \frac{64}{2} \times \frac{40\pi}{16} = 40$
$\therefore$ The required number of rotations
$= 40 - 10 = 30$