If $(1.0025)^{10}$ is evaluated correct to $5$ decimal places, then the digit in the fifth decimal place is

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }5\qquad \textbf{(E) }8$
To find the value of $(1.0025)^{10}$ correct to 5 decimal places, we can use the binomial theorem for approximation. The binomial theorem states that for any real number $x$ and integer $n$:

\[
(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \frac{n(n-1)(n-2)}{6}x^3 + \cdots
\]

In this case, $x = 0.0025$ and $n = 10$. We will use the first few terms of the binomial expansion to approximate $(1.0025)^{10}$.

1. **First term:**
   \[
   1
   \]

2. **Second term:**
   \[
   10 \cdot 0.0025 = 0.025
   \]

3. **Third term:**
   \[
   \frac{10 \cdot 9}{2} \cdot (0.0025)^2 = 45 \cdot 0.00000625 = 0.00028125
   \]

4. **Fourth term:**
   \[
   \frac{10 \cdot 9 \cdot 8}{6} \cdot (0.0025)^3 = 120 \cdot 0.000000015625 = 0.000001875
   \]

Adding these terms together, we get:

\[
1 + 0.025 + 0.00028125 + 0.000001875 = 1.025283125
\]

To ensure accuracy, we can use a calculator to find the exact value of $(1.0025)^{10}$:

\[
(1.0025)^{10} \approx 1.025315
\]

Now, we need to round this value to 5 decimal places. The value is 1.025315, and rounding to 5 decimal places gives us:

\[
1.02532
\]

Thus, the digit in the fifth decimal place is 2.

The final answer is $\boxed{2}$