Answer
a. Resistance of wire is given as $R=\rho\times L\times A$ For equal sized wires of length L and cross section A, but different values of resistivity, $\frac{R_{Al}}{R_{Cu}}=\frac{\rho_{Al}}{\rho_{Cu}}$
b. $R_{Al}=2.38\Omega$
Work Step by Step
b. Use equation found in part a, solve for resistance of copper wire:
$R_{Al}=R_{Cu}\times\frac{\rho_{Al}}{\rho_{Cu}}$
$R_{Al}=1.5\Omega\times\frac{2.73E-8}{1.72E-8}=2.38\Omega$
