Answer
a) $0.634$
b) $1.01\;\rm \Omega$
c) $1.59\;\rm \Omega$
d) $327\;\rm W, \;515\;W$
e) $2.04\;\rm mm$
f) Yes
Work Step by Step
a)
We know that the resistance of a wire is given by $R=\dfrac{\rho l}{A}$
So the ratio of the resistance of a copper wire to that of an aluminum wire is
$$\dfrac{R_{Cu}}{R_{Al}}=\dfrac{\dfrac{\rho_{Cu} l}{A}}{\dfrac{\rho_{Al} l}{A}}=\dfrac{\rho_{Cu}}{\rho_{Al}}$$
since the two wires are having the same length and the same.
Plugging from table 18-1;
$$\dfrac{R_{Cu}}{R_{Al}} =\dfrac{1.68\times10^{-8}}{2.65\times10^{-8}}=\color{red}{\bf 0.634}$$
b) $$R_{Cu}=\dfrac{\rho_{Cu}l_{Cu}}{A_{cu}}=\dfrac{\rho_{Cu}l_{Cu}}{\pi r_{Cu}^2}=\dfrac{\rho_{Cu}l_{Cu}}{\pi \left(\dfrac{D_{Cu}}{2}\right)^2}$$
whereas $r=0.5D$ and $D$ is the diameter.
Plugging the known;
$$R_{Cu}= \dfrac{1.68\times10^{-8}\cdot 125}{\pi \left(\dfrac{1.63\times10^{-3}}{2}\right)^2}=\color{red}{\bf1.01}\;\rm \Omega$$
c)
Using the last formula above;
$$R_{Al} =\dfrac{\rho_{Al}l_{Al}}{\pi \left(\dfrac{D_{Al}}{2}\right)^2}=\dfrac{2.65\times10^{-8}\cdot 125}{\pi \left(\dfrac{1.63\times10^{-3}}{2}\right)^2}=\color{red}{\bf 1.59}\;\rm \Omega$$
d) We know that the power is given by
$$P=I^2R$$
Thus,
$$P_{Cu}=I^2R_{Cu}=18^2\cdot 1.01=\color{red}{\bf 327}\;\rm W$$
$$P_{Al}=I^2R_{Al}=18^2\cdot 1.59=\color{red}{\bf 515}\;\rm W$$
e)
$$R_{Al}=R_{Cu}=\dfrac{\rho_{Al}l_{Al}}{\pi \left(\dfrac{D_{Al}}{2}\right)^2}=1.01$$
and solving for $D_{Al}$;
$$ \left(\dfrac{D_{Al}}{2}\right)^2=\dfrac{\rho_{Al}l_{Al}}{1.01\pi }$$
$$ D_{Al} =\sqrt{\dfrac{4\rho_{Al}l_{Al}}{1.01\pi }}=\sqrt{\dfrac{4\cdot 2.65\times10^{-8}\cdot 125}{1.01\pi }}=2.04\times10^{-3}\;\rm m$$
$$ D_{Al} =\color{red}{\bf 2.04}\;\rm mm$$
f)
The assumption in section (18-4) is that an aluminum wire of the same mass and length as the copper wire would have a smaller resistance.
So, first, we need to find the mass of our copper wire. We need to use the mass density law;
$$\rho_{Cu}=\dfrac{m_{Cu}}{V_{Cu}}$$
Thus,
$$m_{Cu}=\rho_{Cu}V_{Cu}=\rho_{Cu}A_{Cu}l_{Cu}=\pi \rho_{Cu}l_{Cu}\left(\dfrac{D_{Cu}}{2}\right)^2$$
$$m_{Cu}= \pi\cdot 8900\cdot 125\cdot \left(\dfrac{ 1.63\times10^{-3} }{2}\right)^2=\bf2.32 \;\rm kg$$
Now we need to find the cross-sectional area of an aluminum wire that has the same length as the copper wire and the same mass.
$$m_{Al}=m_{Cu}=\rho_{Al}V_{Al}=\rho_{Al}A_{Al}l_{Al} $$
Solving for $A_{Al}$;
$$A_{Al}=\dfrac{m_{Al}}{\rho_{Al} l_{Al} }$$
Plugging the known;
$$A_{Al}=\dfrac{2.32}{2700\cdot 125 }=\bf 6.87\times 10^{-6}\;\rm m^2$$
Thus,
$$R_{Al}=\dfrac{\rho_{Al} l_{Al}}{A_{Al}}=\dfrac{2.65\times10^{-8}\cdot 125}{6.87\times 10^{-6}}=\color{red}{\bf 0.482}\;\rm \Omega$$
It is obvious now that an aluminum wire with the same mass and length as a copper wire will have less resistance. Actually, its resistance is less than half of copper's wire resistance which its resistance, in this case, is 1.01 $\Omega$