A1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 13.0 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 17.1 a. you can ignore any thermal expansion of the rod. 1.find the resistivity and for the material of the rod at 20 ^\circ {\rm c}. (rho= ? ω*m)

2.find the temperature coefficient of resistivity at 20 ^\circ {\rm c} for the material of the rod.(α= ? (c)^-1)

ρ(20°C) = 9.7 * 10^-6 Ω m

α = 0.0013 /°C

Explanation:

Step 1: Data given

length of the rod = 1.70 meters

diameter = 0.55 cm

Potential difference = 13.0 V

Temperature = 20°C →18.7 A

At 92°C → 17.1 A

Step 2: Calculate the resistivity

ρ =  (RA)/L

 ⇒ with R = the resistance of the rod

   ⇒ R = V/I0 = 13/18.7 = 0.695 Ω

 ⇒ with L = length in meters  = 1.70 m

 ⇒ with A = cross sectional area in m^2  =  π * r²  

    ⇒ r = d/2 = 0.550/2 = 0.275 cm = 0.00275m

ρ(20°C) = (R*π * r²  )/L

ρ(20°C) = (0.695 * π 0.00275²)/1.7

ρ(20°C) = 9.7 * 10^-6 Ω m

ρ(92°C) = (R*π * r²  )/L

  ⇒ R = 13/17.1 = 0.76

ρ(92°C) =(0.76 * π 0.00275²)/1.7

ρ(92°C) = 1.06 *10^-5 Ω m

2.Find the temperature coefficient of resistivity at 20 ^\circ {\rm C} for the material of the rod.(α= ? (C)^-

RT = R0*[1 + α(T-T0)]

⇒ with RT = the resistance of the rod at tmperature T = 92.0°C

⇒ with R0= the resistance of the rod at temperature T = 20°C

⇒ with α = the temperature coefficient of resisivity

⇒ with T = 92°C

⇒ with t0 = 20°C

RT = V/I = 13.0/ 17.1 =0.76 Ω

α =((RT/R0)-1)/(T-T0)

α = ((0.76/0.695)-1)/(92-20)

α = 0.0013 /°C

the answer is b, have negligible mass

i think it would be 5