(6) Suppose the axial section of an inverted conical container is an equilateral triangle. Water is poured into this container, and a solid sphere with radius $r$ is placed inside. At this moment, the sphere is tangent to the container walls and the water surface. Then, the height of the water level after removing the sphere is ( ).
(A) $r$
(B) $2 r$
(C) $\sqrt[3]{12} r$
(D) $\sqrt[3]{15} r$
6) D Hint: As shown in the figure, the axial section of the cone is $P A B$, with the center of the sphere being $O$, we can get
$$
\begin{array}{c}
P C=O C+P O=r+2 r=3 r, A C=\sqrt{3} r, \\
V_{P A B}=\frac{1}{3} \pi(\sqrt{3} r)^{2} \cdot 3 r=3 \pi r^{3} .
\end{array}
$$

Let the height of the water surface $E F$ after the ball is removed be $h$. Then
$$
V_{\text {PEF }}=3 \pi r^{3}-\frac{4}{3} \pi r^{3}=\frac{5}{3} \pi r^{3} \text {. }
$$

Since $\frac{V_{P E F}}{V_{P A B}}=\frac{h^{3}}{P C^{3}}$, we have $h^{3}=15 r^{3}, h=\sqrt[3]{15} r$.