(a) [tex]1.72\cdot 10^{-5} \Omega m[/tex]
The resistance of the rod is given by:
[tex]R=\rho \frac{L}{A}[/tex] (1)
where
[tex]\rho[/tex] is the material resistivity
L = 1.20 m is the length of the rod
A is the cross-sectional area
The radius of the rod is half the diameter: [tex]r=0.570 cm/2=0.285 cm=2.85\cdot 10^{-3} m[/tex], so the cross-sectional area is
[tex]A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2[/tex]
The resistance at 20°C can be found by using Ohm's law. In fact, we know:
- The voltage at this temperature is V = 15.0 V
- The current at this temperature is I = 18.6 A
So, the resistance is
[tex]R=\frac{V}{I}=\frac{15.0 V}{18.6 A}=0.81 \Omega[/tex]
And now we can re-arrange the eq.(1) to solve for the resistivity:
[tex]\rho=\frac{RA}{L}=\frac{(0.81 \Omega)(2.55\cdot 10^{-5} m^2)}{1.20 m}=1.72\cdot 10^{-5} \Omega m[/tex]
(b) [tex]8.57\cdot 10^{-4} /{\circ}C[/tex]
First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is
I = 17.5 A
So the resistance is
[tex]R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega[/tex]
The equation that gives the change in resistance as a function of the temperature is
[tex]R(T)=R_0 (1+\alpha(T-T_0))[/tex]
where
[tex]R(T)=0.86 \Omega[/tex] is the resistance at the new temperature (92.0°C)
[tex]R_0=0.81 \Omega[/tex] is the resistance at the original temperature (20.0°C)
[tex]\alpha[/tex] is the temperature coefficient of resistivity
[tex]T=92^{\circ}C[/tex]
[tex]T_0 = 20^{\circ}[/tex]
Solving the formula for [tex]\alpha[/tex], we find
[tex]\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C[/tex]
