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 equals 1, then when the volume of the regular tetrahedron is minimized, the height of the regular tetrahedron equals ( ).
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) $\sqrt{6}$
(D) $2 \sqrt{3}$
$10 \mathrm{~B}$ Prompt: As shown in the figure, $O$ is the center of the base of a regular triangular pyramid and also the center of a hemisphere. $CD$ is the height of the base of the regular triangular pyramid. The side $PAB$ is tangent to the hemisphere at point $E$. Connect $OE, OE \perp PD, OE=1, PO=$ $h$.

Let $\angle PDO=\alpha\left(0^{\circ}<\alpha<90^{\circ}\right)$. Since $OE \perp PD$, we have $OE \perp PE$. Therefore, $OE$ is the radius of the hemisphere, and $OE=1$. Thus, $PE=1$. In the right triangle $POE$, we have $PO^2=PE^2+OE^2$, which means $h^2=1^2+1^2=2$, so $h=\sqrt{2}$. However, this is not the final answer, as we need to find the minimum value of the volume $V$ of the regular triangular pyramid.

The volume $V$ of the regular triangular pyramid is given by
$$
V=\frac{1}{3} \times \text{base area} \times \text{height}=\frac{1}{3} \times \frac{\sqrt{3}}{4} a^2 \times h=\frac{\sqrt{3}}{12} a^2 h,
$$
where $a$ is the side length of the base. Since $OE \perp PD$, we have $\sin \alpha=\frac{OE}{PO}=\frac{1}{h}$, and $\cos \alpha=\frac{PE}{PO}=\frac{1}{h}$. Therefore, $h=\frac{1}{\sin \alpha}=\frac{1}{\cos \alpha}$. Since $a=2 \sqrt{3} \cos \alpha$, we have
$$
V=\frac{\sqrt{3}}{12} \times (2 \sqrt{3} \cos \alpha)^2 \times \frac{1}{\cos \alpha}=\frac{\sqrt{3}}{12} \times 12 \cos^2 \alpha \times \frac{1}{\cos \alpha}=\frac{\sqrt{3}}{\sin^2 \alpha \cos \alpha}.
$$

Since $\sin^2 \alpha + \cos^2 \alpha = 1$, we have $\sin^2 \alpha = 1 - \cos^2 \alpha$. Therefore, $\sin^2 \alpha \cos \alpha = (1 - \cos^2 \alpha) \cos \alpha = \cos \alpha - \cos^3 \alpha$. Let $f(\alpha) = \cos \alpha - \cos^3 \alpha$. Then $f'(\alpha) = -\sin \alpha + 3 \cos^2 \alpha \sin \alpha = \sin \alpha (3 \cos^2 \alpha - 1)$. Since $0 < \alpha < 90^{\circ}$, we have $\sin \alpha > 0$. Therefore, $f'(\alpha) > 0 \Leftrightarrow 3 \cos^2 \alpha - 1 > 0 \Leftrightarrow \cos^2 \alpha > \frac{1}{3} \Leftrightarrow \cos \alpha > \frac{\sqrt{3}}{3}$, so by
$$
\begin{aligned}
\sin^4 \alpha \cos^2 \alpha & =\frac{1}{2} \sin^2 \alpha \cdot \sin^2 \alpha \cdot 2 \cos^2 \alpha \\
& \leqslant \frac{1}{2}\left(\frac{\sin^2 \alpha+\sin^2 \alpha+2 \cos^2 \alpha}{3}\right)^3 \\
& =\frac{4}{27},
\end{aligned}
$$
we know $0<\sin^2 \alpha \cos \alpha \leqslant \frac{2 \sqrt{3}}{9}$, so
$$
V=\frac{\sqrt{3}}{\sin^2 \alpha \cos \alpha} \geqslant \frac{\sqrt{3}}{\frac{2 \sqrt{3}}{9}}=\frac{9}{2}.
$$

When and only when $\sin^2 \alpha=2 \cos^2 \alpha$, i.e., $\cos \alpha=\frac{\sqrt{3}}{3}$, the above equation takes the equal sign, i.e., $V$ takes the minimum value $\frac{9}{2}$, and the corresponding $h=\frac{1}{\cos \alpha}=\sqrt{3}$. Therefore, the answer is B.