If the resistance of a conductor is `5 Omega` at `50^(@)C` and `7Omega` at `100^(@)C` then the mean temperature coefficient of resistance of the material is

Answer

`3.3 xx 10^(-3)^(@)C^(-4)`,`5.14 Omega`

Explanation

`alpha = (R_(t_2)- R_(t_1))/(R_(t_1)(t_(2) - t_(2))) = (7 - 6)/(6(100 - 50)) = (1)/(300)`
`= 3.3 xx 10^(-3).^(@)C^(-4)`
if `r_(0)` = resistance of the conductor at `0^(@)C` then `R_(50) = R_(0) ( 1+ alpha xx 50)`
or `R_(0) = (R_(50))/(1 + alpha xx 50) = (6)/(1+(1)/(300) xx 50) = (36)/(7)`
`= 5.14 Omega`