If R be the resistance of the wire of resistivity ρ whose length is l and the area of cross-section is A, then its resistance R is given by, \(R=\rho\frac lA\). Therefore, \(R=\rho\frac{l^2}{lA}=\rho\frac{l^2}V\), where Al = V is the volume of the conductor. If the wire is stretched to double its length, then the new length is l' = 2l and new resistance can be written as

\(R'=\rho'\frac{l'^2}{V'}=\rho\frac{\left(2l\right)^2}V\), where V' = V and ρ' = ρ, because resistivity and the volume of wire remains the same.

∴ \(R'=4\rho\frac{l^2}V=4\rho\frac{l^2}{Al}=4R\) \(\left[\because R=\rho\frac{l^2}{Al}\right]\)

Thus, when a wire is stretched to double its length, its resistance becomes 4 times its original resistance and resistivity remains the same.

**Click here for the difference between resistance and resistivity.**