2. The axial section of a cone with vertex $P$ is an isosceles right triangle, $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, the foot of the perpendicular is $B, O H \perp P B$, the foot of the perpendicular is $H$, and $P A=4, C$ is the midpoint of $P A$. Then, when the volume of the tetrahedron $O-H P C$ is maximized, the length of $O B$ is
A. $\frac{\sqrt{5}}{3}$
B. $\frac{2 \sqrt{3}}{3}$
C. $\frac{\sqrt{6}}{3}$
D. $\frac{2 \sqrt{6}}{3}$
2. D As shown in the figure, $A B \perp O B, A B \perp O P$, so $A B \perp$ plane $P O B$, thus plane $P A B \perp$ plane $P O B$. Since $O H \perp P B, O H \subseteq$ plane $P O B$, it follows that $O H \perp$ plane $P A B$. Therefore, $O H \perp H C, O H \perp P A$. In the isosceles right $\triangle P O A$, $C$ is the midpoint of $P A$, so $O C \perp P A$.
$$
S_{\triangle H C C}=\frac{1}{2} O H \cdot H C \leqslant \frac{1}{4}\left(O H^{2}+H C^{2}\right)=\frac{1}{4} O C^{2},
$$

Since $O C=2$ is a constant, $S_{\triangle H C C}$ is maximized when $O H=H C$, and $V_{O-H P C}=V_{P-H C O}$ is maximized. At this time, $O H=H C=\sqrt{2}, O P=2 \sqrt{2}$, hence $H O=\frac{1}{2} O P$, so $\angle H P O=30^{\circ}$, thus $O B=O P \cdot \tan 30^{\circ}=\frac{2 \sqrt{6}}{3}$.