As shown in the figure, the stacking situation of the first and second layers of the small balls is such that $\triangle A B E$ is an equilateral triangle with a side length of 1. Construct $E K \perp$ the plane of circle $O_{1}$, then $K$ is the midpoint of $\overparen{A B}$. Thus, $\angle A O_{1} K=45^{\circ} \Rightarrow A K$ $=2 \cdot \frac{\sqrt{2}}{2} \sin 22.5^{\circ}=\sqrt{1-\frac{\sqrt{2}}{2}} \Rightarrow E K=\sqrt{\frac{\sqrt{2}}{2}}$. Since $7 \div \sqrt{\frac{\sqrt{2}}{2}}=\sqrt{49 \sqrt{2}} \approx 8.32$, the small balls can be stacked up to 9 layers, so the maximum number of balls that can be accommodated is 36.
10. In a cylindrical bucket with a diameter of $\sqrt{2}+1$ and a height of 8, the maximum number of balls with a diameter of 1 that can be placed is $\qquad$.