10. In a regular tetrahedron $P-ABC$, there is a hemisphere whose base coincides with the base of the regular tetrahedron, and the three lateral faces of the regular tetrahedron are all tangent to the hemisphere. If the radius of the hemisphere is 1, then when the volume of the regular tetrahedron is minimized, the height of the regular tetrahedron is ( ).
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) $\sqrt{6}$
(D) $2 \sqrt{3}$
10. B.

As shown in Figure 3, $O$ is the center of the base of the regular triangular pyramid and also the center of the hemisphere. $CD$ is the height of the base of the regular triangular pyramid. The side face $PAB$ is tangent to the hemisphere at point $E$, and $OE$ is connected. Then $OE \perp PD, OE=1, PO=h$. Let $\angle PDO=\alpha\left(00 \Leftrightarrow \frac{\sqrt{3}}{3}<t<1$.
Therefore, when $t=\frac{\sqrt{3}}{3}$, $V$ reaches its minimum value $\frac{\sqrt{3}}{\frac{\sqrt{3}}{3}-\left(\frac{\sqrt{3}}{3}\right)^{3}}=\frac{9}{2}$.
The corresponding $h=\frac{1}{\cos \alpha}=\frac{1}{t}=\sqrt{3}$.