Wire resistance and cross-sectional area

Steps:

• Resistance formula: $R = \rho L / A$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.
• For same $L$ and $R$, $\rho_{cu} / A_{cu} = \rho_{alloy} / A_{alloy}$.
• Given $\rho_{cu} = \frac{1}{2} \rho_{alloy}$, so $A_{alloy} / A_{cu} = \rho_{alloy} / \rho_{cu} = 2$.
• Area $A = \pi (d/2)^2$, so $(d_{alloy} / d_{cu})^2 = 2$, thus $d_{alloy} / d_{cu} = \sqrt{2}$.

Why A is correct:

• Diameters ratio equals square root of areas ratio, which is $\sqrt{2}$ from inverse resistivity relation in Ohm's law.

Why the others are wrong:

• B: Assumes direct proportionality to resistivity, ignoring area scaling.
• C: Overcomplicates by multiplying ratio by $\sqrt{2}$ unnecessarily.
• D: Squares the resistivity ratio instead of taking square root for diameters.

Final answer: A