Answer
(a) $2.4W/m$
(b) $2.0W/m$
Work Step by Step
(a) We know that
$\frac{P}{L}=I^2(\rho/A)$
We plug in the known values to obtain:
$\frac{P}{L}=(35A)^2(1.68\times 10^{-8}\Omega m/8.45\times 10^{-6}m^2)$
$\frac{P}{L}=2.4W/m$
(b) We know that
$\frac{P}{L}=I^2(\frac{\rho}{A})$
We plug in the known values to obtain:
$\frac{P}{L}=(25A)^2(1.68\times 10^{-8}\Omega m/5.27\times 10^{-6}m^2)$
$\frac{P}{L}=2.0W/m$