A conductor has a cross-section of `15 mm^(2)` and resistivity of `7.6 xx 10^(-8)Omega` m at `0^@C`. If the temperature coefficient of resistance of the material of the conductor is `5 xx 10^(-3).^(@)C^(-1)`, calculate its resistance for 2 km length of conductor when its temperature is `60^@C`.

Explanation

Here, `A= 15 mm^(2) = 15xx10^(-6) m^(2)`,
`rho=7.6xx10^(-8) Omega m,alpha =5xx10^(-3).^(@)C^(-1)`,
`l_(0)=2 km = 2000m, t_(1)=0^@C,t_(2)=60^@C`,
`R_(0)=rhol/A=((7.6xx10^(-8))xx2000)/(15xx10^(-6)) = 10.12Omega`
`R_(60)=R_(0)[1+alpha (t_(2)-t_(1))]`
`=10.12 [1+5xx10^(-3) xx60]`
`=10.12 xx1.3 =13.16 Omega`