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Question: The resistance of a wire of length \[100\,{\text{cm}}\] and \[7 \times {10^{ - 3}}\,{\text{cm}}\] ra...

The resistance of a wire of length 100 cm100\,{\text{cm}} and 7×10−3 cm7 \times {10^{ - 3}}\,{\text{cm}} radius is 6 ohms. Its specific resistance is
A. 924×10−8 Ω⋅cm924 \times {10^{ - 8}}\,\Omega \cdot {\text{cm}}
B. 92.4×10−8 Ω⋅cm92.4 \times {10^{ - 8}}\,\Omega \cdot {\text{cm}}
C. 900×10−8 Ω⋅cm900 \times {10^{ - 8}}\,\Omega \cdot {\text{cm}}
D. 224×10−8 Ω⋅cm224 \times {10^{ - 8}}\,\Omega \cdot {\text{cm}}

Explanation

Solution

Use the formula for area of the circle to determine the cross- sectional area of the wire using the radius of the wire. Then use the formula for the resistance of the wire. This formula 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.

Formula used:
The resistance RR of a wire is given by
R=ρLAR = \rho \dfrac{L}{A} …… (1)
Here, ρ\rho is the resistivity of the material of the wire, LL is the length of the wire and AA is the cross-sectional area of the wire.
The cross- sectional area AA of a wire is given by
A=πR2A = \pi {R^2} …… (2)
Here, RR is the radius of the wire.

Complete step by step answer:
We have given that the length of the wire is 100 cm100\,{\text{cm}} and the radius of the wire is 7×10−3 cm7 \times {10^{ - 3}}\,{\text{cm}}.
L=100 cmL = 100\,{\text{cm}}
R=7×10−3 cmR = 7 \times {10^{ - 3}}\,{\text{cm}}
Let us determine the cross-sectional area AA of the wire.
Substitute 3.143.14 for π\pi and 7×10−3 cm7 \times {10^{ - 3}}\,{\text{cm}} for RR in equation (2).
A=(3.14)(7×10−3 cm)2A = \left( {3.14} \right){\left( {7 \times {{10}^{ - 3}}\,{\text{cm}}} \right)^2}
⇒A=153.86×10−6 cm2\Rightarrow A = 153.86 \times {10^{ - 6}}\,{\text{c}}{{\text{m}}^2}
Hence, the cross-sectional area of the wire is 153.86×10−6 cm2153.86 \times {10^{ - 6}}\,{\text{c}}{{\text{m}}^2}.
The resistivity of the material of the wire is also known as the specific resistance of the wire.

We can determine the specific resistance of the wire using equation (1). Rearrange the equation (1) for the specific resistance ρ\rho of the wire.
ρ=RAL\rho = \dfrac{{RA}}{L}
Substitute 6 Ω6\,\Omega for RR, 153.86×10−6 cm2153.86 \times {10^{ - 6}}\,{\text{c}}{{\text{m}}^2} for AA and 100 cm100\,{\text{cm}} for LL in the above equation.
ρ=(6 Ω)(153.86×10−6 cm2)100 cm\rho = \dfrac{{\left( {6\,\Omega } \right)\left( {153.86 \times {{10}^{ - 6}}\,{\text{c}}{{\text{m}}^2}} \right)}}{{100\,{\text{cm}}}}
⇒ρ=923.16×10−8 Ω⋅cm2\Rightarrow \rho = 923.16 \times {10^{ - 8}}\,\Omega \cdot {\text{c}}{{\text{m}}^2}
∴ρ≈924×10−8 Ω⋅cm2\therefore \rho \approx 924 \times {10^{ - 8}}\,\Omega \cdot {\text{c}}{{\text{m}}^2}
Therefore, the specific resistance of the material of the wire is 924×10−8 Ω⋅cm2924 \times {10^{ - 8}}\,\Omega \cdot {\text{c}}{{\text{m}}^2}.

Hence, the correct option is A.

Note: The students may convert the units of the length and radius of the wire from the CGS system of units (cm) to the SI system of units (m). If the unit conversion is done then also we will arrive at the final correct answer. But the unit of the final answer should also be converted in the SI system of units. Hence, in order to avoid this lengthy procedure, unit conversion is not done as the final answer given in the options is also in the CGS system of units.