Answer:
The wire now has less (the half resistance) than before.
Explanation:
The resistance in a wire is calculated as:
[tex]R=\alpha \frac{l}{s}[/tex]
Were:
R is resistance
[tex]\alpha[/tex] is the resistance coefficient
l is the length of the material
s is the area of the transversal wire, in the case of wire will be circular area ([tex]s=\pi r^{2}[/tex]).
So if the lenght and radius are doubled, the equation goes as follows:
[tex]R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }[/tex]
So finally because the circular area is a square function, the resulting equation is half of the one before.