Answer:
three times the original diameter
Explanation:
From the wire's resistance formula, we can calculate the relation between the diameter of the wire and its length:
[tex]R=\rho\frac{l}{\pi \frac{d^2}{4}}\\d=\sqrt{\rho \frac{4 l}{\pi R}}\\[/tex]
Here, d is the wire's diameter, [tex]\rho[/tex] is the electrical resistivity of the material and R is the resistance of the wire. We have [tex]l'=9l[/tex]
[tex]d'=\sqrt{\rho \frac{4 l'}{\pi R}}\\d'=\sqrt{\rho \frac{4 (9l)}{\pi R}}\\d'=3\sqrt{\rho \frac{4 l}{\pi R}}\\d'=3d[/tex]