Calculate the electrical conductivity of the material of a conductor of length 3m, area of cross section 0.2 `mm^2` having a resistance of 2 ohm.

Explanation

Conductivity = `sigma = ?`
`l = 3 m`
`A = 0.02 mm^2 = 0.02 xx10^(-6) m^2`
`A = 2 xx10^(-8) m^2`
`R = 2Omega`
`R = rho (l)/4`
`rho =(RA)/l`
`:. sigma = 1/rho = 1/(RA//l) = l/(RA)`
`sigma =3/(2xx2xx10^(-8))=0.7xx10^8`
`= 7.5 xx10^(7) Omega^(-1) m^(-1)`