The resistance of the second wire is 16 R.
where R is the resistance of the first wire.
R = ρ[tex] \frac{l}{A} [/tex]
where l = length of the wire
A = area of the wire
A = [tex] \pi r^{2} [/tex] where, r = [tex] \frac{diameter of wire}{2} [/tex]
Thus, on finding the ratio of resistance of the two wires, we get,
[tex] \frac{R1}{R2} = \frac{l1A2}{l2A1} [/tex]
here, R1 = R
l1 = 8m
l2 = 2m
A1=π[tex] 0.25^{2} [/tex]
A1=π[tex] 0.50^{2} [/tex]
we get. R2 = 16R
