$\begin{array}{llll}\text { (A)1 } & \text { (B) } 2 \lg 3 & \text { (C) } \lg 11 & \text { (D) } \lg 15\end{array}$
2. $ABCD$ is a unit square, $P$ is the midpoint of side $AB$. By folding $ABCD$ along $PC$ and $PD$ to form a tetrahedron $PACD$ (where $A$ coincides with $B$), the volume of the tetrahedron $PACD$ is equal to ( ).
(A) $\frac{\sqrt{3}}{30}$
(B) $\frac{\sqrt{3}}{24}$
(C) $\frac{\sqrt{3}}{20}$
(D) $\frac{\sqrt{3}}{12}$
2. B.

After folding, $A$ coincides with $B$.
Since $P A \perp A C, P A \perp A D$, therefore, $P A \perp$ plane $A C D$.
$$
=\frac{1}{3} \times \frac{1}{2} \times 1^{2} \times \frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}}{24} \text {. }
$$