To solve this problem, we will use the kinematic equations for rotational motion with uniform angular acceleration.$\\$Given:$\\$• Initial angular speed $\omega_i = 1200$ rpm$\\$• Final angular speed $\omega_f = 3120$ rpm$\\$• Time interval $t = 16$ seconds$\\$First, convert the angular speeds from rpm to rad/s:$\\ \omega_i = 1200 \times \frac{2\pi}{60} = 40\pi$ rad/s$\\ \omega_f = 3120 \times \frac{2\pi}{60} = 104\pi$ rad/s$\\$We use the equation for angular acceleration $\alpha$ given by:$\\ \omega_f = \omega_i + \alpha t \\$Substitute the known values:$\\ 104\pi = 40\pi + \alpha \times 16 \\$Rearrange to solve for $\alpha: \\ \alpha \times 16 = 104\pi - 40\pi \\ \alpha \times 16 = 64\pi \\ \alpha = \frac{64\pi}{16} \\ \alpha = 4\pi$ rad/s$^2 \\$Therefore, the angular acceleration is $4\pi$ rad/s$^2$, which corresponds to Option 2.