10. As shown in Figure 3, fill in the 10 spaces in the annulus with the numbers $1,2, \cdots, 10$. Add up the absolute values of the differences between all adjacent cells (cells that share a common edge). If this sum is to be maximized, then the maximum value is $\qquad$
10.50.

Let the two numbers in adjacent cells be $a$ and $b$ ($a > b$). Then $|a-b| = a-b$, and there are 10 differences. To maximize the sum of these 10 differences, the 10 minuends $a$ should be as large as possible, and the 10 subtrahends $b$ should be as small as possible. Since each number is adjacent to two other numbers, we can set $10, 9, 8, 7, 6$ as the minuends twice, and $5, 4, 3, 2, 1$ as the subtrahends twice, which satisfies the condition. Therefore, the maximum value we seek is
$$
\begin{array}{l}
2(10+9+8+7+6)- \\
2(5+4+3+2+1) \\
=50 .
\end{array}
$$

Figure 7 is one way to fill in the numbers that meets the condition.