1. For a wire, $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

2. When the wire is stretched, its volume remains constant: $AL = \text{constant}$

3. The new length is given as: $L' = 5L$

4. Since volume is constant, $A' L' = AL$ $A' \cdot 5L = AL$ $A' = \frac{A}{5}$

5. New resistance: $R' = \rho \frac{L'}{A'}$ Substitute $L' = 5L$ and $A' = A/5$: $R' = \rho \frac{5L}{A/5} = 25\rho \frac{L}{A} = 25R$

6. Given original resistance $R = 5\,\Omega$, $R' = 25 \times 5 = 125\,\Omega$

7. Therefore, the correct option is: $\boxed{D\;:\;125\,\Omega}$

Comparison with stored correct answer:

• Stored correct answer: $D$
• Derived answer: $D$
• They match.