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$\text{Hint: } \alpha = \frac{d\omega}{dt}$ $\text{Step 1: Find the time when the direction of motion is reversed.}$ $\text{The torque due to force } F \text{ at a distance } R \text{ is given by;} \Rightarrow F \times R = I \alpha$ $\text{where } I \text{ is the moment of inertia, } \alpha \text{ is the angular acceleration.}$ $\text{The angular acceleration is defined as;} \Rightarrow \alpha = \frac{d\omega}{dt} \Rightarrow \omega = \frac{R}{I} \int_0^t (12t - 3t^2) \, dt$ $\Rightarrow \omega = 2t^2 - \frac{t^3}{3}$ $\text{For } \omega = 0,$ $\Rightarrow 0 = 2t^2 - \frac{t^3}{3} \Rightarrow t^2 \left(2 - \frac{t}{3}\right) = 0$ $\Rightarrow t = 0, 6.$ $\text{Step 2: Find the number of revolutions.}$ $\text{The angular velocity is given by;} \Rightarrow \frac{d\theta}{dt} = \omega \Rightarrow \frac{d\theta}{dt} = 2t^2 - \frac{t^3}{3}$ $\Rightarrow \theta = \int_0^6 \left(2t^2 - \frac{t^3}{3}\right) dt$ $\Rightarrow \theta = \left[\frac{2t^3}{3} - \frac{t^4}{12}\right]_0^6$ $\Rightarrow \theta = 6^3 \left(\frac{2}{3} - \frac{6}{12}\right) = 6^3 \left(\frac{8 - 6}{12}\right)$ $\Rightarrow \theta = \frac{6^3}{6} = 36$ $\text{Thus, the total number of revolutions is given by;} \Rightarrow \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$ $\Rightarrow \frac{K}{\pi} = \frac{18}{\pi} \Rightarrow K = 18$ $\text{Hence, option (4) is the correct answer.}$