A fan makes 10 revolutions in 3 second which is just switched on. Considering uniform acceleration the number of revolution made by the fan in next 3 seconds is

Solutions:

we have, $\theta = 2 \pi n$ where $\theta$ is the angular displacement and $n$ is the number of revolutions

from $t=0$ to $t=3s$ , the fan made $n =10$ revolutions.

$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$

Initial angular velocity $\omega_0 = 0$ So,

$2 \pi n = 0 + \frac{1}{2}\alpha t^2$

$2 \pi .10 = \frac{1}{2} \alpha 3^2$

$40 \pi = 9 \alpha$

$\boxed{\alpha = \frac{40\pi}{9}}$

Now lets find the number of revolutions made from $t=0$ to $t=6s$ . Applying the same equations of rotational motions as before:

$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$

using $\theta = 2 \pi n$ and $\omega_0 = 0$ We get,

$2 \pi n = \frac{1}{2} \alpha t^2$

$2 \pi n = \frac{1}{2}. \frac{40\pi}{9}. 6^2$

$n = \frac{1}{2}.\frac{20}{9}.36$

$n = 40$

number of revolution from $t=0$ to $t=3s$ is 10

number of revolution from $t=0$ to $t=6s$ is 40

hence, number of revolution made from $t=3s$ to $t=6s$ is (40-10) = 30

Ans: 30

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