To solve this problem, we need to calculate the resistance ($R$) of an aluminum wire based on the given parameters using the formula:

$R = \rho \cdot \frac{L}{A}$

Where:

• $R$ is the resistance,
• $\rho$ is the resistivity of aluminum $= 2.65 \times 10^{-8} \, \Omega \cdot \text{m}$,
• $L$ is the length of the wire $= 40 \, \text{m}$,
• $A$ is the cross-sectional area of the wire.

The cross-sectional area $A$ of a cylindrical wire is given by:

$A = \pi \left(\frac{d}{2}\right)^2$

Where:

• $d$ is the diameter of the wire $= 4.2 \, \text{mm} = 4.2 \times 10^{-3} \, \text{m}$.

Now, let's calculate $A$ and then substitute all values into the formula for resistance. I'll handle the computation next.The calculated resistance of the aluminum wire is approximately $0.0765 \, \Omega$, which can be written as $7.65 \times 10^{-2} \, \Omega$.

Thus, the closest answer is:

A. $7.6 \times 10^{-2} \, \Omega$.

Would you like further details or have any questions?

Here are some related questions to explore:

1. How does the resistivity of a material affect the resistance of a wire?
2. What would happen to the resistance if the wire's length were doubled?
3. How does the diameter of the wire influence its resistance?
4. What is the difference between resistivity and resistance?
5. How does temperature affect the resistivity of aluminum?

Tip: Resistance in a conductor increases as the length increases and decreases with a larger cross-sectional area.