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Calculate the area of the cross section of a wire of length 2 m, resistance 46 \[\Omega \] and resistivity$1.84 \times {10^{ - 6}}\Omega m$.

Answer
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Hint: These are basic level problems which can be easily solved by identifying the values to use the formula area of wire having resistance and resistivity i.e. $area = \dfrac{{\left( {resistivity \times length} \right)}}{{resis\tan ce}}$ to find the area of cross-section of given wire.

Complete Step-by-Step solution:
In this question there are 2 terms Resistance and resistivity which can be explained as; resistance is the measurement of the opposition to current flow in a wire it is measured in ohms. And resistivity is a characteristic property of the material of a wire resistivity helps in comparing the materials on the basis of their ability to conduct electric currents.
The given values are
Length of the wire = 2m, resistance = 46$\Omega $ , resistivity =$1.84 \times {10^{ - 6}}\Omega m$
The area of the cross section of a wire of length can be find by the formula $area = \dfrac{{\left( {resistivity \times length} \right)}}{{resis\tan ce}}$
Substituting the given values in the above formula
$ \Rightarrow $$area = \dfrac{{\left( {1.84 \times {{10}^{ - 6}} \times 2} \right)}}{{46}}$
$ \Rightarrow $$area = 8 \times {10^{ - 8}}{m^2}$
Hence the area of cross-section of wire of length 2 cm and resistance 46 \[\Omega \] is $8 \times {10^{ - 8}}{m^2}$.

Note: Do you know how the resistance of a wire depends on the area of cross-section of that wire let’s find out how; The thickness of the wires will influence the sum of the resistance. Larger wires are more cross-sectional in area. Fluid passes at a faster rate into a larger pipe than it flows through a small funnel. This can be due to the reduced resistance present in the larger pipe. Similarly, the longer the wire, the less resistance it would provide to the electric charge stream. If all other factors are the same, the charge can pass across thicker wires with larger cross-sectional areas than through thinner wires at higher rates.