Answer:
The diameter is [tex]d = 6.5 *10^{-4} \ m[/tex]
Explanation:
From the question we are told that
The length of the cylinder is [tex]l = 120 \ m[/tex]
The resistance is [tex]\ 6.0\ \Omega[/tex]
The resistivity of the metal is [tex]\rho = 1.68 *10^{-8} \ \Omega \cdot m[/tex]
Generally the resistance of the cylindrical wire is mathematically represented as
[tex]R = \rho \frac{l}{A }[/tex]
The cross-sectional area of the cylindrical wire is
[tex]A = \frac{\pi d^2}{4}[/tex]
Where d is the diameter, so
[tex]R = \rho \frac{l}{\frac{\pi d^2}{4 } }[/tex]
=> [tex]d = \sqrt{ \rho* \frac{4 * l }{\pi * R } }[/tex]
[tex]d = \sqrt{ 1.68 *10 ^{-8}* \frac{4 * 120 }{3.142 * 6 } }[/tex]
[tex]d = 6.5 *10^{-4} \ m[/tex]
