Question

Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold is 19.3 ✕ 10^3 kg/m^3, and its resistivity is 2.44 ✕ 10^−8 Ω · m. What is the resistance of such a wire at 20.0°C?

Answer 1

Answer:

Resistance of gold wire, $R=1977 \times 10^3 ohm$

Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$.........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$......(2)

Here,

$A= \pi \times r^2$...........(3)

here A is the cross sectional area of cylendrical gold wire  

From equation 2 and 3

we got

$V=A \times L$...............(4)

on comparing equation 1 and equation 4, we got,

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

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

$A=2.53\times 10^{-11}m^2$

we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$