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A. 2370 $ms^{-2}$

B. 5055 $ms^{-2}$

C. 1000 $ms^{-2}$

D. 4740 $ms^{-2}$

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Acceleration of a rotating body has a magnitude given by:

$a = \omega^2 r$

We are given distance from the axis, i.e., r= 30 cm or 0.3m.

Also the fan is making a rotation with frequency 1200 rpm (or rotations per minute):

$\nu = 1200$ rpm.

In terms of S.I. units:

$\nu= \dfrac{1200}{60} =20$ rps or rotations per second.

The angular frequency (in radians) is the total angle that the fan covers in a second. So, when rotating at a rate of 20 rotations in a second (cover $2\pi$ in one rotation):

$\omega = 2 \pi \times \nu$

We get

$\omega = 2 \pi 20$ radians per second.

The formula for acceleration is just:

$a = \omega^2 r$

So, keeping the values in it, we get:

$a = (40 \pi)^2 \times 0.3 ms^{-2}$

Therefore, we get:

$a = 4741.2 ms^{-2}$

after keeping the value of pi to be 22/7.

This is pretty close to option D.

The formula for velocity is $\omega r$ for a rotating body. The closer the body is to the axis, more will be the velocity. The formula for force is $mv^2/r$, equating this with ma, we get the magnitude of the acceleration as $v^2/r$. Thus, even if one doesn't remember the formula, this trick can be used to get the acceleration.

One might know the formula for acceleration right but here, one might substitute 1200 rpm in place of $\omega$ directly, without much bothering about the rpm part. Therefore, one must remember the unit of $\omega$ at such time. Rpm is clearly not radians per second but it is rotations per second.