9. (5th "Hope Cup" Invitational Competition Question) A hemispherical bowl is filled with water. If the bowl's opening is tilted by $30^{\circ}$, the volume of water that overflows is ( ) of the original volume.
A. $\frac{3}{8}$
B. $\frac{11}{12}$
C. $\frac{5}{16}$
D. $\frac{11}{16}$
9. D. Reason: As shown in the figure, from $\angle O A B=\angle O B A=30^{\circ}$, we know $O C=$ $\frac{1}{2} O A=\frac{1}{2} R$.

The volume of the remaining water is the volume of the spherical cap, the height of the spherical cap $h=\frac{1}{2} R, V_{\text {cap }}=$ $\frac{1}{3} \pi\left(\frac{R}{2}\right)^{2}(3 R-h)=\frac{5}{24} R^{3}$, the volume of the hemisphere $V_{*}=\frac{2}{3} \pi R^{3}$, the volume of the overflowed water $V=\frac{11}{24} \pi R^{3}$, the required ratio is $\frac{11}{16}$.