Question

You are working at a company that manufactures electrical 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 1.60 g of gold into a wire of length L = 2.20 km and uniform diameter. Your supervisor gives you the task of determining the resistance of such a wire (in MΩ) at 20.0°C. (The density of gold is 19.3 ✕ 103 kg/m3.)

Answer 1

Answer:$R=1.424 M\Omega$

Explanation:

Given

mass of gold $m=1.6 gm$

Length of wire $L=2.2 km$

Resistivity of gold $\rho =2.44\times 10^{-8}$

density of gold $=19.3\times 10^3 kg/m^3$

and $mass=volume\times density$

$1.6\times 10^{-3}=volume\times 19.3\times 10^3$

$volume=8.29\times 10^{-8} m^3$

And $Resistance R=\frac{\rho L}{A}$

also be written as

$R=\frac{\rho L^2}{V}$

where $L=length$

$V=volume$

$\rho =resistivity\ of\ gold$

$R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}$

$R=1.42\times 10^{6} \Omega$

$R=1.424 M\Omega$