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skyluke89
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
[tex]\rho = \frac{AR}{L} [/tex]
where
[tex]\rho[/tex] is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
[tex]A=\pi r^2[/tex]
where r is the radius of the wire. Substituting in the previous equation ,we find
[tex]\rho = \frac{\pi r^2 R}{L} [/tex]

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
[tex]\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho[/tex]
Therefore, the new resistivity must be 4 times the original one.
david8644

Answer:

4 times the original.

Explanation:

As you know the formula for resitivity of wires is:

[tex]Resisitivity=\frac{Resistance*Area}{length}[/tex]

So the length and the resistance is the same remain the same because we are talking about the same material, the only change woul dbe in the radius, and the radius affects the area:

[tex]Area=r^2\pi[/tex]

That would be the original, since the new one has doubel the radiusthis would be its formula:

[tex]Area=\pi (2r)^2\\Area=4r^2\pi[/tex]

As you can see there will be an area that is 4 times greater than the previous one, this would make the whole resisitivity to becom 4 times greater than the previous one.

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