2. Given point $A(5 \sqrt{3}, 5)$, line $l: x=m y+n$ $(n>0)$ passes through point $A$. If the feasible region
$$
\left\{\begin{array}{l}
x \leqslant m y+n, \\
x-\sqrt{3} y \geqslant 0, \\
y \geqslant 0
\end{array}\right.
$$

has an circumcircle with a diameter of 20, then the value of $n$ is $\qquad$
2. 10 $\sqrt{3}$.

Notice that the line $l^{\prime}: x-\sqrt{3} y=0$ also passes through point $A$, so $A$ is the intersection point of lines $l$ and $l^{\prime}$.

The feasible region is the shaded area (including the boundary) of $\triangle A O B$ as shown in Figure 7.

Let the inclination angle of line $l$ be $\alpha$, then $\angle A B O^{\circ}=\pi-\alpha$. In $\triangle A O B$, $|O A|=10$.

By the Law of Sines, $\frac{10}{\sin (\pi-\alpha)}=20$, then $\alpha=\frac{5 \pi}{6}$.
Therefore, $\frac{1}{m}=\tan \frac{5 \pi}{6}$, which gives $m=-\sqrt{3}$.
Since line $l$ passes through point $A(5 \sqrt{3}, 5)$, we have $5 \sqrt{3}=-\sqrt{3} \times 5+n$. Solving this, we get $n=10 \sqrt{3}$.