Question

You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched into incredibly long, thin wires. The company has developed a new technique that will stretch 1.00 g of gold into a wire of length L 5 2.40 km and uniform diameter. Your supervisor gives you the task of determining the resis- tance of such a wire at 20.08C.(the density of gold is 19.3 ✕ 103 kg/m3.)

Answer 1

Explanation:

We know that the relation between volume and density is as follows.

      Volume = $\frac{\text{mass}}{\text{density}}$

So,       V = $\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}$

               = $5.181 \times 10^{-8} m^{3}$

Now, we will calculate the area as follows.

      Area = $\frac{\text{volume}}{\text{length}}$

               = $\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}$

               = $2.15 \times 10^{-11} m^{2}$

Formula to calculate the resistance is as follows.

         R = $\rho \frac{l}{A}$

             = $\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}$

             = $2.71 \times 10^{6} ohm$

Thus, we can conclude that the resistance of given wire is $2.71 \times 10^{6} ohm$.