Question

A wire of length \[1\,{\text{m}}\] and of radius \[0.1\,{\text{mm}}\] has a resistance of \[100\,\Omega \]. The resistivity of the material is
A.\[0.0214\,\Omega \cdot {\text{m}}\]
B.\[3.14 \times {10^{ - 8}}\,\Omega \cdot {\text{m}}\]
C. \[3.14 \times {10^{ - 6}}\,\Omega \cdot {\text{m}}\]
D.\[0.00214\,\Omega \cdot {\text{m}}\]

Answer 1

Hint: Use the equation for the resistance of the wire. This equation gives the relation between the resistance of the wire, resistivity of the material of the wire, length of the wire and cross-sectional area of the wire.

Formulae used:
The expression for the resistance of the wire is
\[R = \rho \dfrac{L}{A}\] …… (1)
Here, \[R\] is the resistance of the wire, \[L\] is the length of the wire, \[A\] is the cross-sectional area of the wire and \[\rho \] is the resistivity of the material of the wire.
The equation for the area of the circle is
\[A = \pi {r^2}\] …… (2)
Here, \[A\] is the area of the circle and \[r\] is the radius of the circle.

Complete step by step answer:
The wire has length, radius and resistance \[1\,{\text{m}}\], \[0.1\,{\text{mm}}\] and \[100\,\Omega \] respectively.
Convert the unit of the radius \[r\] of the wire in the SI system of units.
\[r = \left( {0.1\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{m}}}}{{1\,{\text{mm}}}}} \right)\]
\[ \Rightarrow r = {10^{ - 3}}\,{\text{m}}\]
Hence, the radius of the wire is \[{10^{ - 3}}\,{\text{m}}\].
Calculate the cross sectional area of the wire.
Substitute \[3.14\] for \[\pi \] and \[{10^{ - 3}}\,{\text{m}}\] for \[r\] in equation (2).
\[A = \left( {3.14} \right){\left( {0.1\,{\text{mm}}} \right)^2}\]
\[ \Rightarrow A = \left( {3.14} \right){\left[ {\left( {0.1\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{m}}}}{{1\,{\text{mm}}}}} \right)} \right]^2}\]
\[ \Rightarrow A = 3.14 \times {10^{ - 8}}\,{{\text{m}}^2}\]
Hence, the cross-sectional area of the wire is \[3.14 \times {10^{ - 8}}\,{{\text{m}}^2}\].
Calculate the resistivity of the material of the wire.
Rearrange equation (1) for the resistivity \[\rho \] of the material of the wire.
\[\rho = \dfrac{{RA}}{L}\]
Substitute \[100\,\Omega \] for \[R\], \[3.14 \times {10^{ - 8}}\,{{\text{m}}^2}\] for \[A\] and \[1\,{\text{m}}\] for \[L\] in the above equation.
\[\rho = \dfrac{{\left( {1000\,\Omega } \right)\left( {3.14 \times {{10}^{ - 8}}\,{{\text{m}}^2}} \right)}}{{1\,{\text{m}}}}\]
\[ \Rightarrow \rho = 3.14 \times {10^{ - 6}}\,\Omega \cdot {\text{m}}\]
Therefore, the resistivity of the material of the wire is \[3.14 \times {10^{ - 6}}\,\Omega \cdot {\text{m}}\].
Hence, the correct option is C.

Note:Always remember that resistivity of a material is not a variable quantity. The resistivity of a specific material remains the same in all conditions.It is the characteristic property of the material used for making a wire.Hence, it has different values for the different material.