Two copper wires have the same length. The cross-sectional area of one wire is four times that of the smaller wire. If the smaller wire has a resistance of 40 Ω, what is the resistance of the larger wire?

Resistance is inversely proportional to cross-sectional area: $R = \rho \frac{L}{A}$. For two copper wires of identical length $L$, $R_1 A_1 = R_2 A_2$. If the smaller wire has area $A_1 = \frac{1}{4} A_2$ and resistance $R_1 = 40\,\Omega$, the resistance of the larger wire with 4 times the area is: $R_2 = R_1 \times \frac{A_1}{A_2} = 40 \times \frac{1}{4} = 10\,\Omega$.

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