A1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 14.0 v across its ends, while an ammeter measures the current through it. you observe that at room temperature (20.0 degrees c) the ammeter reads 18.5 a, while at 92.0 degrees c it reads 17.2 a. you can ignore any thermal expansion of the rod.

find the resistivity and for the material of the rod at 20 degrees c.

find the temperature coefficient of resistivity at 20 degrees c for the material of the rod.

Resistance = ρ * (L/A) and Rf = Ri * ([1 + α * (Tf – Ti)] 
ρ = Resistivity L = length in meters A = cross sectional area in m^2 α = temperature coefficient of resistivity 
L = 1.50 m Area = π * r^2 r = d/2 = 0.25 cm = 2.5 * 10^-3 m Area = π * (2.5 * 10^-3)^2 

The cylindrical rod is similar to a resistor. Since the current is decreasing, the resistance must be increasing. This means the resistance is increasing as the temperature increases. Resistance = Voltage ÷ Current At 20˚, R = 15 ÷ 18.5 At 92˚, R = 15 ÷ 17.2 

Now you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures. Resistance = ρ * (L/A) ρ = Resistance * (A/L) 
At 20˚, ρ = (15 ÷ 18.5) * [π * (2.5 * 10^-3)^2] ÷ 1.5 = At 92˚, ρ = (15 ÷ 17.2) * [π * (2.5 * 10^-3)^2] ÷ 1.5 = 
Now you know the resistivity at the two temperatures. Let’s determine the temperature coefficient of resistivity for the material of the rod. 
Rf = Ri * ([1 + α * (Tf – Ti)] Rf = 15 ÷ 17.2, Ri = 15 ÷ 18.5, Tf = 92˚, Ti = 20˚ 
15 ÷ 17.2 = 15 ÷ 18.5 * [1 + α * (92 – 20)] Multiply both sides by (18.5 ÷ 15) (18.5 ÷ 15) * (15 ÷ 17.2) = 1 + α * 72 Subtract 1 from both sides (18.5 ÷ 15) * (15 ÷ 17.2) – 1 = α * 72 Divide both sides by 72 α = 1.05 * 10^-3 

I’m not sure about this question