A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of \(25 \pi \mathrm{Nm}\) for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ________ m.

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