Answer:
E) be two times larger.
Explanation:
As we know that the relation between the resistance and the resistivity of the wire is given as:
[tex]R=\rho.\frac{l}{a}[/tex]
where:
[tex]\rho=[/tex] resistivity of the wire
[tex]l=[/tex] length of wire
[tex]a=[/tex] area of wire
[tex]R=[/tex] resistance
Now, when the length of the wire is four times the initial length then for the resistance to remain constant:
[tex]R=\rho.\frac{4l}{a'}[/tex]
where:
[tex]a'=[/tex] area of the new wire
[tex]\rho.\frac{l}{a} =\rho.\frac{4l}{a'}[/tex]
[tex]a'=4a[/tex]
we know that area of the cross section of wire is given as:
[tex]a=\pi.r^2[/tex]
[tex]\pi.r'^2=4\times \pi.r^2[/tex]
[tex]r'=2r[/tex]
Hence the radius must be twice of the initial radius for the resistance to be constant when length is taken four times.