Given, $\mathrm{m}=1 \mathrm{~kg}$ $$ \begin{aligned} & \omega_{\mathrm{i}}=1800 \mathrm{rpm}=1800 \times \frac{2 \pi}{60}=60 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \\ & \omega_{\mathrm{f}}=2100 \mathrm{rpm}=2100 \times \frac{2 \pi}{60}=70 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \\ & \tau_{\text {est }}=25 \pi \mathrm{Nm} \\ & \mathrm{t}=40 \mathrm{sec} \end{aligned} $$ Using equation of motion $$ \begin{aligned} & \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t} \\ & 70 \pi=60 \pi+\alpha(40) \\ & \alpha=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}^2 \end{aligned} $$ Also, $\tau=\mathrm{I} \alpha$ $$ \begin{aligned} & \tau=\frac{\mathrm{mR}^2}{4} \alpha \\ & 25 \pi=\frac{1 \times \mathrm{R}^2}{4} \times \frac{\pi}{4} \\ & \mathrm{R}=20 \mathrm{~m} \end{aligned} $$ Hence, diameter of disk $=2 R=2 \times 20=40 \mathrm{~m}$