Answer:
The resistivity of the wire is:
[tex]\rho=1.60\,\,10^{-8}\,\,\Omega\,m[/tex]
Explanation:
Recall the formula that connects resistance R with the material's resistivity [tex]\rho[/tex] :
[tex]R=\frac{\rho\,\,L}{A}[/tex]
where A stands for the cross-sectional area of the wire, and L for the wire's length.
In our case, the cross-sectional area of a 0.33 mm wire is the area of a circle of 0.165 mm radius (0.000165 m) which we can calculate as:
[tex]A=\pi\,\,R^2=\pi\,\,0.000165^2\,\,m^2=8.55\,\,10^{-8} \,\,m^2[/tex]
Since we don't have the actual resistance, but the information on the current on the wire when applying a potential difference, we use Ohm's Law to get the resistance R of the wire:
[tex]V=I\,\,R\\1.5 = 8 \,\, R\\R = \frac{1.5}{8} \, \Omega\\R= 0.1875\,\,\Omega[/tex]
Now, using the resistance formula shown at the beginning, we solve for the resistivity [tex]\rho[/tex] :
[tex]R=\frac{\rho\,\,L}{A}\\0.1875\,\,\Omega=\frac{\rho\,\,(1\,\,m)}{(8.55\,\,10^{-8}\,m^2)}\\\rho=0.1875\,*\,8.55\,\,10^{-8}} \,\,\Omega\,m\\\rho=1.60\,\,10^{-8}\,\,\Omega\,m[/tex]
The resistivity rho of the piece of the wire will be "1.60 × 10⁻⁸ Ωm".
According to the question,
Voltage, V = 1.5 V
Current in a wire, I = 8.0 A
Diameter of a wire= 0.33 mm
Length or a wire = 1.0 m
We know,
The area of circle,
A = πR²
By substituting the values,
= π × 0.000165²
= 8.55 × 10⁻⁸ m²
By using Ohm's Law,
→ V = IR
By substituting the values,
1.5 = 8 R
R = [tex]\frac{1.5}{8}[/tex]
= 0.1875 Ω
We know that,
→ R = [tex]\frac{\rho L}{A}[/tex]
By substituting the values,
0.1875 = [tex]\frac{\rho \times 1 \ m}{8.55\times 10^{-8}}[/tex]
hence,
Material's resistivity will be:
ρ = 0.1875 × 8.55 × 10⁻⁸
= 1.60 × 10⁻⁸ Ωm
Thus the above answer is right.
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