Let the initial angular velocity be $\omega_0$.

Final angular velocity after 24 rotations is $\omega_1 = \frac{\omega_0}{3}$.

Let the angular retardation (deceleration) be $\alpha$ (negative value, since fan is slowing down).

Number of rotations made during retardation is $n$.

Step 1: Use rotational motion equation

The relation between angular velocities, angular acceleration, and angular displacement is:

$$ \omega_1^2 = \omega_0^2 + 2\alpha \theta $$

where $\theta$ is the angle rotated (in radians).

1 rotation = $2\pi$ radians

After 24 rotations:

$$ \theta_1 = 24 \times 2\pi = 48\pi~\text{radians} $$

So,

$$ \left(\frac{\omega_0}{3}\right)^2 = \omega_0^2 + 2\alpha \times 48\pi $$

$$ \frac{\omega_0^2}{9} = \omega_0^2 + 96\pi\alpha $$

Bring like terms together:

$$ \frac{\omega_0^2}{9} - \omega_0^2 = 96\pi\alpha $$

$$ \left( \frac{1}{9} - 1 \right)\omega_0^2 = 96\pi\alpha $$

$$ - \frac{8}{9}\omega_0^2 = 96\pi\alpha $$

$$ \alpha = -\frac{8\omega_0^2}{9 \times 96\pi} $$

$$ \alpha = -\frac{\omega_0^2}{108\pi} $$

Step 2: Find total rotations till rest

Let total number of rotations before coming to rest be $N$.

At rest, final angular velocity $\omega = 0$.

Apply same formula:

$$ \omega^2 = \omega_0^2 + 2\alpha\theta_{\text{total}} $$

$$ 0 = \omega_0^2 + 2\alpha\theta_{\text{total}} $$

$$ 2\alpha\theta_{\text{total}} = -\omega_0^2 $$

$$ \theta_{\text{total}} = \frac{-\omega_0^2}{2\alpha} $$

Substitute $\alpha$ from earlier:

$$ \theta_{\text{total}} = \frac{-\omega_0^2}{2 \times \left(-\frac{\omega_0^2}{108\pi}\right)} $$

$$ \theta_{\text{total}} = \frac{-\omega_0^2}{-\frac{2\omega_0^2}{108\pi}} $$

$$ \theta_{\text{total}} = \frac{-\omega_0^2 \times 108\pi}{-2\omega_0^2} $$

$$ \theta_{\text{total}} = \frac{108\pi}{2} $$

$$ \theta_{\text{total}} = 54\pi~\text{radians} $$

Number of total rotations:

$$ N = \frac{\theta_{\text{total}}}{2\pi} = \frac{54\pi}{2\pi} = 27 $$

Step 3: Find required answer

Rotations still to be made after first $24$ rotations:

$$ N - 24 = 27 - 24 = 3 $$

Final Answer:

$ \boxed{3} $

Option A is correct.

Circular Motion

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