A 1.70 m cylindrical rod of diameter 0.500 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 ∘C ) the ammeter reads 18.7 A , while at 92.0 ∘C it reads 17.2 A . You can ignore any thermal expansion of the rod.

part 1) Find the resistivity and for the material of the rod at 20 ∘C .

part 2) Find the temperature coefficient of resistivity at 20 ∘C for the material of the rod.

Given Information:

Length of rod = L = 1.70 m

diameter of rod = d = 0.500 cm = 0.005 m

Voltage = V = 14 volts

Current at 20° C = I₀ = 18.7 A

Current at 92° C = I = 17.2 A

Room temperature = T₀ = 20° C

Temperature = T = 92° C

Required Information:

Resistivity at 20° C = ρ = ?

Temperature coefficient = α = ?

Resistivity = ρ = 8.63x10⁻⁶ Ω.m

Temperature coefficient = α = 0.00121 per °C

Explanation:

Part 1) Find the resistivity and for the material of the rod at 20° C

We know that resistivity is given by

ρ = R₀A/L

Where R₀ is the resistance of the rod at 20° C, A is the area and L is the length of the cylindrical rod.

Area is given by

A = πr²

A = π(d/2)²

A = π(0.005/2)²

A = 0.0000196 m²

Resistance can be found using Ohm's law

R₀ = V/I₀

R₀ = 14/18.7

R₀ = 0.7486 Ω

ρ = R₀A/L

ρ = 0.7486*0.0000196/1.70

ρ = 8.63x10⁻⁶ Ω.m

Part 2) Find the temperature coefficient of resistivity at 20° C for the material of the rod

The temperature coefficient of resistivity can be found using

α = R/R₀ - 1/(T - T₀)

Where R is the resistance of the rod at 20° C

R = V/I = 14/17.2 = 0.8139 Ω

α = R/R₀ - 1/(T - T₀)

α = (0.8139/0.7486) - 1/(92° - 20°)

α = 0.08722/72°

α = 0.00121 per °C

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray