Answer:
Length = 2.92 m
Diameter = 0.11 mm
Explanation:
We have [tex]m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}[/tex] , where:
[tex]l[/tex] is the length
[tex]m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3[/tex]
We divide the first equation by the second equation to get:
[tex]\frac{m}{R} = \frac{d A^2}{\rho}[/tex]
[tex]A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2[/tex]
Using this Area, we find the diameter of the wire:
[tex]D = \sqrt{\frac{4A}{\pi}}[/tex]
[tex]D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}[/tex]
[tex]D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm[/tex]
To find the length, we multiply the two equations stated initially:
[tex]mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}[/tex]
[tex]l^2 = 8.534\\l = 2.92 \ m[/tex]
