Answer:
The conductivity of Nichrome is [tex]2\times 10^6\ S/m[/tex].
Explanation:
Given:
Potential difference (V) = 2.0 V
Current flowing (I) = 4.0 A
Length of wire (L) = 1.0 m
Area of cross section of wire (A) = 1.0 mm² = 1 × 10⁻⁶ m² [1 mm² = 10⁻⁶ m²]
We know, from Ohm's law, that the ratio of voltage and current is always a constant and equal to the resistance of the resistor. Therefore, the resistance of the nichrome wire is given as:
[tex]R=\frac{V}{A}=\frac{2.0}{4.0}=0.5\ \Omega[/tex]
Now, resistance of the nichrome wire in terms of its resistivity, length and area of cross section is given as:
[tex]R=\rho\frac{L}{A}[/tex]
Where, [tex]\rho\to resistivity\ of\ Nichrome[/tex]
Now, plug in all the values given and solve for [tex]\rho[/tex]. This gives,
[tex]0.5\ \Omega=\rho\frac{1.0\ m}{1\times 10^{-6}\ m^2}\\\\\rho=\frac{0.5\times 1\times 10^{-6}}{1.0}=0.5\times 10^{-6}\ \Omega-m[/tex]
Now, conductivity of a material is the reciprocal of its resistivity. Therefore, the conductivity of Nichrome is given as:
[tex]\sigma=\frac{1}{\rho}=\frac{1}{0.5\times 10^{-6}}=2\times 10^6\ S/m[/tex]
Conductivity is measured in Siemens per meter (S/m)
Therefore, the conductivity of Nichrome is [tex]2\times 10^6\ S/m[/tex].
