What is the sum of the digits of the decimal form of the product $ 2^{1999}\cdot 5^{2001}$?

$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$
1. First, we simplify the given expression \(2^{1999} \cdot 5^{2001}\). Notice that we can factor out \(5^2\) from \(5^{2001}\):
   \[
   2^{1999} \cdot 5^{2001} = 2^{1999} \cdot 5^{1999} \cdot 5^2
   \]

2. Next, we recognize that \(2^{1999} \cdot 5^{1999} = (2 \cdot 5)^{1999} = 10^{1999}\). Therefore, the expression simplifies to:
   \[
   10^{1999} \cdot 25
   \]

3. Now, we need to understand the decimal form of \(10^{1999} \cdot 25\). The term \(10^{1999}\) is a 1 followed by 1999 zeros:
   \[
   10^{1999} = 1\underbrace{000\ldots000}_{1999 \text{ zeros}}
   \]

4. Multiplying this by 25 shifts the decimal point 1999 places to the right and appends 25 at the end:
   \[
   10^{1999} \cdot 25 = 25\underbrace{000\ldots000}_{1999 \text{ zeros}}
   \]

5. The sum of the digits of this number is simply the sum of the digits of 25, since all the zeros do not contribute to the sum:
   \[
   2 + 5 = 7
   \]

The final answer is \(\boxed{7}\).