Awire with a resistance of 5.3 ω is drawn out through a die so that its new length is 4 times its original length. find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

The resistance of a wire is defined as
R = (ρL)/A
where
ρ = resistivity
L = length
A =  cross-sectional area

Before the wire is drawn, let 
L₁ = the length
A₁ =  the cross-sectional area.

The volume of the wire and its resistivity remain unchanged after it is drawn.
After it is drawn, the new length is
L₂ = 4L₁
To preserve the volume, the new cross-sectional area is
A₂ = (L₁A₁)/L₂ = (L₁A₁)/(4L₁) = A₁/4

The initial resistance is given as
R₁ = (ρL₁)/A₁ = 5.3 Ω

The final resistance is


84.8 Ω

Since resistance is based on cross-sectional area, the new wire would be thinner and its resistance will go up. Neglecting any change to the atomic structure of the wire, and a constant volume over the stretching, the cross section would be reduced by 4 times and its resistance would go up 4 times to 21.2 Ohms