A metal wire of diameter 2mm and of length 300m has a resistance of `1.6424 Omega at 20^@C` and `2.415Omega at150^@C`. Find the values of `alpah, R_0` and `rho_0` where the zero subscript refers to `0^@C`, and `rho_0` Identify the metal.

Explanation

`R_(150^@C) = 2.415 = R_0(1 + 150alpha), R_(20^@C) = 1.6424 = R_0(1 +20 alpha)`
Solving these relation simultaneously, `alpha = 3.9xx10^(-3)C^(-1)` and
`R_0 = 1.5236Omega. From R_0 = rho_0(L//A),` We get
`1.5236 = (rho_0(300))/(pi(2xx10^(-3))^2//4) or rho_0 = 1.596xx10^(-8)Wm`
`Then rho_(20^@C) = rho_0(1 + 20alpha) = (1.596xx10^(-8))[1+(3.9xx10^(-3)(20)]`
`=1.720xx10^(-8)Omegam`
This indicates that the meterial is copper.