6. A frustum of height 8 contains a sphere $O_{1}$ with a radius of 2, the center $O_{1}$ is on the axis of the frustum, and the sphere $O_{1}$ is tangent to the top base and the side of the frustum. Inside the frustum, another sphere $O_{2}$ with a radius of 3 can be placed such that the sphere $O_{2}$ is tangent to the sphere $O_{1}$, the bottom base, and the side of the frustum at only one point. Apart from the sphere $O_{2}$, the maximum number of additional spheres with a radius of 3 that can be placed inside the frustum is
(A) 1 ;
(B) 2 ;
(C) 3 ;
(D) 4 .
6. (B)
Draw the axial section of the frustum through $\mathrm{O}_{2}$, as shown in Figure a. Then, draw a section through $\mathrm{O}_{2}$ perpendicular to the axis of the frustum, intersecting the axis at circle $O$. From Figure a, it is easy to find that $O O_{2}=4$.
This problem is equivalent to: On a circle with center $O$ and radius 4, how many more points can be placed, excluding $O_{2}$, such that circles with these points and $O_{2}$ as centers and 3 as the radius have at most one common point. As shown in Figure b.
$\sin 45^{\circ}<\sin \theta=\frac{3}{4}<\sin 60^{\circ}, \therefore 45^{\circ}<\theta<60^{\circ}$, so the maximum number of additional points that can be placed is $\left[\frac{360^{\circ}}{2 \theta}\right]-1=3-1=2$. This satisfies the above requirements.
Thus, the frustum can contain at most 2 more spheres with a radius of 3. Therefore, the answer is (B).