A ceiling fan rotates about its own axis with some angular veloci… - Vidyadip

MCQ

A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes $(1 / 4)^{\text {th }}$ of the original in time $t$ and $n$ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform)

A
$\frac{4 n}{15}$
B
$\frac{8 n}{15}$
✓
$\frac{16 n}{15}$
D
$\frac{32 n}{15}$

✓

Answer

Correct option: C.

$\frac{16 n}{15}$

(c): The angular velocity is given as
$
\omega^2=\omega_0{ }^2+2 \alpha\left(\theta-\theta_0\right)
$
When the fan is switched off, $\theta=2 \pi n, \theta_0=0, \omega=\frac{\omega_0}{4}$
$
\begin{aligned}
\left(\frac{\omega_0}{4}\right)^2 & =\omega_0^2-2 \alpha(2 \pi n) \Rightarrow 2 \alpha(2 \pi n)=\frac{15}{16} \omega_0^2 \\
2 \pi n^{\prime} & =\left(\frac{\omega_0^2}{2 \alpha}\right)
\end{aligned}
$
When the fan come to rest
$
\begin{aligned}
& 0=\omega_0{ }^2-2 \alpha\left(2 \pi n^{\prime}\right) \\
& 2 \pi n^{\prime}=\left(\frac{\omega_0^2}{2 \alpha}\right) \quad \text { or } n^{\prime}=\frac{16}{15} n
\end{aligned}
$

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