Mass of the disk, $ m = 1 \, \text{kg} $.

Initial angular velocity, $ \omega_i = 1800 \, \text{rpm} $.

Final angular velocity, $ \omega_f = 2100 \, \text{rpm} $.

External torque, $ \tau_{\text{ext}} = 25\pi \, \text{Nm} $.

Time, $ t = 40 \, \text{seconds} $.

First, convert the rotational speeds from revolutions per minute (rpm) to radians per second (rad/s):

$ \omega_i = 1800 \times \frac{2\pi}{60} = 60\pi \, \text{rad/s} $

$ \omega_f = 2100 \times \frac{2\pi}{60} = 70\pi \, \text{rad/s} $

Next, use the equation of motion for rotation to find the angular acceleration $\alpha$:

$ \omega_f = \omega_i + \alpha t $

$ 70\pi = 60\pi + \alpha(40) $

Solving for $\alpha$:

$ \alpha = \frac{70\pi - 60\pi}{40} = \frac{10\pi}{40} = \frac{\pi}{4} \, \text{rad/s}^2 $

The torque and moment of inertia relationship is given by:

$ \tau = I \alpha $

For a thin solid disk rotating along its diameter, the moment of inertia $ I $ is $ \frac{mR^2}{4} $. Thus:

$ 25\pi = \frac{1 \times R^2}{4} \times \frac{\pi}{4} $

Solving for $ R $:

$ 25\pi = \frac{\pi R^2}{16} $

$ R^2 = \frac{25\pi \times 16}{\pi} = 400 $

$ R = 20 \, \text{m} $

The diameter of the disk is:

$ \text{Diameter} = 2R = 2 \times 20 = 40 \, \text{m} $