A 2.0-m-long piece of wire has a diameter of 2.0 mm and a resistance of 0.018 $\Omega$. The resistivity of the material is
A) $2.8 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m } .$
B) $1.3 \times 10 ^ { - 7 } \Omega \cdot \mathrm { m } .$
C) $1.4 \times 10 ^ { - 8 } \Omega \mathrm { m } .$
D) $1.3 \times 10 ^ { 4 } \Omega \cdot \mathrm { m }$

Question

A 2.0-m-long piece of wire has a diameter of 2.0 mm and a resistance of 0.018  $\Omega$. The resistivity of the material is
A) $2.8 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m } .$
B) $1.3 \times 10 ^ { - 7 } \Omega \cdot \mathrm { m } .$
C) $1.4 \times 10 ^ { - 8 } \Omega \mathrm { m } .$
D) $1.3 \times 10 ^ { 4 } \Omega \cdot \mathrm { m }$

Quizplus · Accepted Answer

The resistivity $\rho$ of a material is given by $\rho = \frac{RA}{L}$, where $R$ is the resistance, $A$ is the cross-sectional area, and $L$ is the length. The area $A$ of the wire can be calculated using the formula for the area of a circle, $A = \pi r^2$, where $r$ is the radius of the wire. Given the diameter is 2.0 mm, the radius is 1.0 mm or $1.0 \times 10^{-3}$ m. Thus, $A = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, \text{m}^2$. Substituting the given values into the resistivity formula gives $\rho = \frac{0.018 \, \Omega \cdot 3.14 \times 10^{-6} \, \text{m}^2}{2.0 \, \text{m}} = 2.8 \times 10^{-8} \, \Omega \cdot \text{m}$.

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