Question

A wire with a resistance of 5Ω is drawn out through a die so that its new length is four times its original length. What is the resistance of the longer wire when the resistivity and density of the material are unchanged?

Answer 1

$$R=\rho \frac{l}{A}$$

The density of the wire remains the same, hence the volume of the wire would be unchanged

$$V=lA=const$$

The new resistance of the wire

$$R'=\rho'\frac{l'}{A'}=\rho\frac{4l}{A/4}=16R\\ =16*5\:\Omega=80\:\Omega$$