6. Given that line $l$ intersects sphere $O$ at exactly one point $P$, and two half-planes $\alpha$ and $\beta$ starting from line $l$ intersect sphere $O$ to form two circular sections with radii $1$ and $\sqrt{3}$, respectively. The dihedral angle $\alpha-l-\beta$ has a plane angle of $\frac{5 \pi}{6}$. Then the radius of sphere $O$ is ( ).
(A) $\sqrt{7}$
(B) $2 \sqrt{7}$
(C) $\sqrt{10}$
(D) $2 \sqrt{10}$
6. B.

From the problem and the symmetry of the sphere, we know that the cross-sectional diagram is as shown in Figure 4.

Let $O_{1} P$ and $O_{2} P$ be the radii of the two cross-sections, with $O_{1} P=1, O_{2} P=\sqrt{3}$, $O_{1} O=d_{1}, O_{2} O=d_{2}$, and the radius of the sphere be $R$.
Then, from the problem and the cosine rule, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
d_{1}^{2}+1=d_{2}^{2}+3=R^{2}, \\
d_{1}^{2}+d_{2}^{2}-2 d_{1} d_{2} \cos \frac{\pi}{6}=1+3-2 \sqrt{3} \cos \frac{5 \pi}{6}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
d_{1}^{2}=27, \\
d_{2}^{2}=25 .
\end{array}\right.
\end{array}
$$

Therefore, the radius of the sphere $R=2 \sqrt{7}$.