* 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. 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
Analyze the axial section of the frustum passing through point $\mathrm{O}_{2}$, as shown in figure $\mathrm{a}$, and then make a section through $\mathrm{O}_{2}$ perpendicular to the axis of the frustum, intersecting the axis at point $O$, as shown in figure b. From figure $\mathrm{a}$, it is easy to find that $O O_{2}=4$. The original problem is equivalent to: On the circumference of a circle with center $O$ and radius 1, except for $\mathrm{O}_{2}$, how many more points can be taken such that the circles with these points and point $\mathrm{O}_{2}$ as centers and 3 as the radius have at most one common point. From figure b, $\sin 45^{\circ}<\sin \theta=\frac{3}{4}<\sin 60^{\circ}$, so $45^{\circ}<\theta<60^{\circ}$. Therefore, the maximum number of additional points that can be taken is $\left[\frac{360^{\circ}}{20}\right]-1=3-1=2$, satisfying the above requirements. Choose B.

Figure $\mathbf{a}$
Figure $\mathbf{b}$