A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $ 100$ cans, how many rows does it contain?

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 8\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 11$
1. Let's denote the number of rows by \( n \). The number of cans in the \( k \)-th row is given by \( 2k - 1 \). Therefore, the total number of cans in the display is the sum of the first \( n \) odd positive integers.
2. The sum of the first \( n \) odd positive integers is given by the formula:
   \[
   1 + 3 + 5 + \cdots + (2n - 1) = n^2
   \]
   This is a well-known result that can be derived by induction or by recognizing that the sum of the first \( n \) odd numbers forms a perfect square.
3. Given that the total number of cans is 100, we set up the equation:
   \[
   n^2 = 100
   \]
4. Solving for \( n \), we take the square root of both sides:
   \[
   n = \sqrt{100} = 10
   \]
5. Therefore, the number of rows in the display is \( n = 10 \).

The final answer is \( \boxed{10} \).