A metal wire of dimeter 4mm and length 100 m has a resistacne of `0.408 Omega at 10^@ and 0.508Omega at 120^2 C` find the value of
a. temperature cofficinet of resistance
b. its resistivities of `0^@C and 120^@C`

Explanation

Given that `r =4//2 mm = 2mm = 2xx10^(-3)m,l = 100m,`
`t_1 = 10^@C, t_2 = 120^@C, R_(t1) = 0.408 Omega, R_(t2) = 0.508 Omega`
i. Temperature coefficient fo resistance is given by
`alpha = (R_1 - R_2)/(R_(t1)(t_2 - t_1)) = (0.508 - 0.408)/(0.409(120 - 10)) = 2.2xx10^(-3)C^(-1)`
ii. We know that `R_(t1) = R_0(1+alphat_1)`
or `R_0 = (R-(t1))/(1 + alphat_1) = (0.408)/(1+22xx10^(-3)xx10) = (0.408)/(1.-22) = 0.4Omega`
iii. Resistivity at `0^@C` is given by
`rho_0 = (R_0A)/(l) =l (R_0pir^2)/(l)`
`= (0.4xx3.14xx(2xx10^(-3))^(2))/(100) Omegam = 5.02xx10^(-8)Omega m`
Resistivity at `120^(@)C` is given by
`rho_(0) = rh0(1 +alphat) = 5.502xx10^(-8)(1+2.2xx10^(-3xx120)Omega`
`5.02xx10^(-8)x1.264Omega = 6.35xx10^(-08)Omega`