6. As shown in the figure, in a regular tetrahedron $A B C D$ with edge length 1, $G$ is the centroid of $\triangle B C D$, and $M$ is the midpoint of line segment $A G$. Then the surface area of the circumscribed sphere of the tetrahedron $M-B C D$ is
A. $\pi$
B. $\frac{3}{2} \pi$
C. $\frac{\sqrt{6}}{4} \pi$
D. $\frac{\sqrt{6}}{8} \pi$
Answer
As shown in the figure, it is evident that the center $O$ of the circumscribed sphere of the tetrahedron $M-BCD$ lies on the line $AG$, with $GC=\frac{\sqrt{3}}{3}$ and $GM=\frac{\sqrt{6}}{6}$. Let $GO=x$, then we have $\left(x+\frac{\sqrt{6}}{6}\right)^{2}=\frac{1}{3}+x^{2}$ $\Rightarrow \frac{\sqrt{6}}{3} x=\frac{1}{6} \Rightarrow x=\frac{\sqrt{6}}{12} \Rightarrow R=\frac{\sqrt{6}}{4}$,
Thus, the surface area of the circumscribed sphere is $S=\frac{3}{2} \pi$, so the answer is $B$.