Respuesta :
Answer:
Resistance of gold wire, [tex]R=1977 \times 10^3 ohm[/tex]
Explanation:
In this question we have given
Density of gold, [tex]d=19.3\times 10^3 \frac{kg}{m^3}[/tex]
resistivity of gold, [tex]r=2.44\times 10^{-8} ohm.m[/tex]
Length of wire, [tex]L= 2.05 km[/tex]
Temperature, [tex]T= 20^oC[/tex]
We know that relation between volume and density is given as
[tex]Density= \frac{mass}{Volume}[/tex]
Therefore, volume occupied by one gram gold is given as,
[tex]V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3[/tex].........(1)
We Know that Volume of gold wire which is cylindrical in shape is given by following formula
[tex]V=\pi \times r^2 \times L[/tex]......(2)
Here,
[tex]A= \pi \times r^2[/tex]...........(3)
here A is the cross sectional area of cylendrical gold wire
From equation 2 and 3
we got
[tex]V=A \times L[/tex]...............(4)
on comparing equation 1 and equation 4, we got,
[tex]A \times L=5.181\times 10^{-8} m^3[/tex]
[tex]A=\frac{5.181\times 10^{-8} m^3}{2050 m}[/tex]
[tex]A=2.53\times 10^{-11}m^2[/tex]
we know that resistance and resistivity are related by following formula,
[tex]Resistance = resistivity\times \frac{L}{A}[/tex]................(5)
Put values of resistivity, A and L in equation 5, we got
[tex]R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}[/tex]
[tex]R=1977 \times 10^3 ohm[/tex]
Therefore resistance of gold wire, [tex]R=1977 \times 10^3 ohm[/tex]
