If the length and width of a rectangle are each increased by $ 10 \%$, then the perimeter of the rectangle is increased by

\[ \textbf{(A)}\ 1 \% \qquad
\textbf{(B)}\ 10 \% \qquad
\textbf{(C)}\ 20 \% \qquad
\textbf{(D)}\ 21 \% \qquad
\textbf{(E)}\ 40 \%
\]
1. Let the initial length and width of the rectangle be \( l \) and \( w \) respectively.
2. The initial perimeter \( P_1 \) of the rectangle is given by:
   \[
   P_1 = 2(l + w)
   \]
3. If the length and width are each increased by \( 10\% \), the new length and width become \( 1.1l \) and \( 1.1w \) respectively.
4. The new perimeter \( P_2 \) of the rectangle is given by:
   \[
   P_2 = 2(1.1l + 1.1w) = 2 \cdot 1.1 (l + w) = 2.2(l + w)
   \]
5. To find the percentage increase in the perimeter, we calculate the ratio of the new perimeter to the initial perimeter:
   \[
   \frac{P_2}{P_1} = \frac{2.2(l + w)}{2(l + w)} = \frac{2.2}{2} = 1.1
   \]
6. The ratio \( 1.1 \) indicates that the new perimeter is \( 1.1 \) times the initial perimeter, which corresponds to a \( 10\% \) increase.

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