11. (2003 National High School Competition Question) Place eight balls, each with a radius of 1, in two layers inside a cylinder, such that each ball is tangent to its adjacent balls, and is also tangent to one base and the side of the cylinder, then the height of the cylinder is equal to
11. Fill in $\sqrt[8]{8}+2$. Reason: As shown in the figure, from the given information, the centers of the four balls in the upper and lower layers $A^{\prime} 、 B^{\prime} 、 C^{\prime} 、 D^{\prime}$ and $A 、 B 、 C 、 D$ are the vertices of two squares with side lengths of 2, and the circumcircles $\odot O^{\prime}$ and $\odot O$ of these squares form the top and bottom bases of a cylinder.
At the same time, the projection of $A^{\prime}$ on the lower base must be the midpoint $M$ of $\overparen{A B}$.
In $\triangle A^{\prime} A B$, $A^{\prime} A=A^{\prime} B=A B=2$. Let the midpoint of $A B$ be $N$, then $A^{\prime} N=\sqrt{3}$. Also, $O M=O A=\sqrt{2}, O N=1$, so $M N=\sqrt{2}-1, A^{\prime} M=\sqrt{A^{\prime} N^{2}-M N^{2}}=\sqrt[4]{8}$. Therefore, the height of the original cylinder is $\sqrt[4]{8}+2$.