2. If the three sides of $\triangle A B C$ are $\sqrt{2}, \sqrt{3}, \sqrt{5}$, then its inradius $r=$
2. $\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2}$.

It is easy to know that $(\sqrt{2})^{2}+(\sqrt{3})^{2}=(\sqrt{5})^{2}$. Therefore, $\triangle A B C$ $S=\frac{\sqrt{6}}{2}$
Also, $r \cdot \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2}=\frac{\sqrt{6}}{2}$, so
$r=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2}$.