Answer:
[tex]\rho=4.02\times 10^{-8}\ \Omega-m[/tex]
Explanation:
Let us assume that the radius of the wire, r = 0.8 mm = 0.0008 m
EMF of the battery, V = 12 V
Slope of I versus 1/d, m = 600 A-m
The resistance of any material is given by :
[tex]R=\rho \dfrac{d}{A}[/tex]
d is the length of wire
Since, [tex]I=\dfrac{V}{R}[/tex]
[tex]I=\dfrac{VA}{\rho d}[/tex]
[tex]I=\dfrac{VA}{\rho}.(\dfrac{1}{d})[/tex]
[tex]y=slope\times x[/tex]
[tex]\dfrac{VA}{\rho}=600[/tex]
[tex]\rho=\dfrac{VA}{600}[/tex]
[tex]\rho=\dfrac{12\times \pi \times (0.0008)^2}{600}[/tex]
[tex]\rho=4.02\times 10^{-8}\ \Omega-m[/tex]
So, the resistivity of the material of which the wire is made is [tex]4.02\times 10^{-8}\ \Omega-m[/tex].