The resistance of a wire is given by
[tex]R= \frac{\rho L}{A} [/tex]
where [tex]\rho [/tex] is the resistivity of the material, L the length of the wire and A its cross-sectional area.
In the problem, [tex]\rho[/tex] and L remain the same, while A changes because the radius changes. The area is given by:
[tex]A=\pi r^2[/tex]
This means that if we double the radius (2r), the area becomes
[tex]A_{new}= \pi (2r)^2 = 4 \pi r^2 =4A[/tex]
And therefore, the new value of the resistance is
[tex]R_{new} = \frac{\rho L}{4 A}= \frac{1}{4}R [/tex]
So, when the radius is doubled, the resistance becomes [tex] \frac{1}{4} [/tex] of its original value.
