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 $\alpha \theta$
${\frac{{\omega }_0}{2}}^2$ = ${{\omega }_0}^2$ - 2$\alpha \left(2\pi n\right)$ $\Rightarrow$ $\alpha$ = $\frac{3}{4}$ $\frac{{{\omega }_0}^2}{4\pi \times 36}$ , (n = 36) ..(ii)
Now let fan completes total $n^{{}^{\prime }}$ revolution from the starting to come to rest
0 = ${{\omega }_0}^2$ - $2\alpha \left(2\pi n\right)$ $\Rightarrow$ $n^{{}^{\prime }}$ = $\frac{{{\omega }_0}^2}{4\alpha \pi }$
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 revolution
Number of rotation = 48 - 36 = 12