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

The correct option is B 12
By using equation
${\omega }^2={\omega }_0^2-2\alpha \theta$
(for $n$ rotations, $\theta =2n\pi$)
${\left(\frac{{\omega }_0}{2}\right)}^2={\omega }_0^2-2\alpha \left(2n\pi \right)$
$\Rightarrow \alpha =\frac{3}{4}\frac{{\omega }_0^2}{4\pi \times 36}..\left(i\right)$
(given $n=36$)
Now let fan complete total $n$ revolutions from the starting to the time it comes to rest
$0={\omega }_0^2-2\alpha \left(2\pi n^{\prime }\right)\Rightarrow n^{\prime }=\frac{{\omega }_0^2}{4\alpha \pi }$
Substituting the value of $\alpha$ from equation (i)
$n^{\prime }=\frac{{\omega }_0^2}{4\pi }\frac{4\times 4\pi \times 36}{3{\omega }_0^2}=48$ revolutions
Therefore, number of rotation it'll make $=48-36=12$