When a ceiling fan is switched off its angular velocity reduces to 50% while it makes 36 rotations. How many more rotation will it make before coming to rest (Assume uniform angular retardation)

The correct option is B 12
By using equation ${\omega }^2$ = ${{\omega }_0}^2$ - 2 $\alpha$$\theta$
(${\frac{{\omega }_0}{2}}^2$) = ${{\omega }_0}^2$ - 2 $\alpha$ (2 $\pi$n) $\Rightarrow$ $\alpha$ = $\frac{3}{4}$ $\frac{{{\omega }_0}^2}{4\pi \times 36}$, (n = 36) ..(i)
Now let fan completes total $n^{{}^{\prime }}$ revolution from the starting to come to rest.
0 = ${w_0}^2$ - 2 $\alpha$ (2 $\pi$ $n^{{}^{\prime }}$) $\Rightarrow$$n^{{}^{\prime }}$ = $\frac{{{\omega }_0}^2}{4\alpha \pi }$
substituting the values of $\alpha$ from equation (i)
$n^{{}^{\prime }}$ = $\frac{{\omega }_{0}2}{4\pi }$ $\frac{4\times 4\pi \times 36}{3{{\omega }_0}^2}$ = 48 revolution
Number of rotation = 48 - 36 = 12