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Question: A nichrome wire 50cm long and \(1m{{m}^{2}}\) cross-section carries a current of 4A when connected t...

A nichrome wire 50cm long and 1mm21m{{m}^{2}} cross-section carries a current of 4A when connected to a 2V battery. The resistivity of nichrome wire is
A. 1×10−6Ω−m1\times {{10}^{-6}}\Omega -m
B. 4×10−7Ω−m4\times {{10}^{-7}}\Omega -m
C. 3×10−7Ω−m3\times {{10}^{-7}}\Omega -m
D. 2×10−7Ω−m2\times {{10}^{-7}}\Omega -m

Explanation

Solution

We could derive the equation for resistivity from the proportionality relation of Resistance of a conductor with its length and area of cross-section. In that relation, the resistivity acts as the proportionality constant. Substituting that relation for resistance in ohm’s law will help you get the required equation to solve this question.

Formula used:
R=ρlAR=\rho \dfrac{l}{A}
The ohm’s law,
V=IRV=IR

Complete answer:
Resistivity is used as measurement of resistance provided by a given size of a specific material to electrical conduction. Now, let us derive the equation for resistivity which is represented by ρ.
We know that, the resistance R of a conductor is directly proportional to its length (l) and inversely proportional to its cross-sectional area (A), that is,
R∝lAR\propto \dfrac{l}{A}
Therefore,
R=ρlAR=\rho \dfrac{l}{A} …………….. (1)
Where, ρ is a proportionality constant called resistivity. Remember that, ρ depends on the material of the conductor but not on its dimensions.
Now let us recall the Ohm’s law, which is given by,
V=IRV=IR …………….. (2)
Where, V and I are voltage and current respectively.
From (1) and (2) we have,
ρlA=VI\rho \dfrac{l}{A}=\dfrac{V}{I}
ρ=V×AI×l\rho =\dfrac{V\times A}{I\times l} ………….. (3)
We are directly given the values to be substituted in the question,
Voltage V=2V
Area of cross-section A= 1mm21m{{m}^{2}}
A=1mm2=10−6m2A=1m{{m}^{2}}={{10}^{-6}}{{m}^{2}}
Current I=4A
Length l=50cm
l=50cm=50×10−2m=0.5ml=50cm=50\times {{10}^{-2}}m=0.5m
Now we have to simply substitute these values in equation (3),
ρ=2V×10−6m24A×0.5m\rho =\dfrac{2V\times {{10}^{-6}}{{m}^{2}}}{4A\times 0.5m}
ρ=1×10−6Ω−m\rho =1\times {{10}^{-6}}\Omega -m

So, the correct answer is “Option A”.

Note:
While dealing with numerical problems like this, make sure that the given values are in their SI units, if not, you have to convert them to their respective SI units so as to avoid further confusion. You may then convert the solution according to the unit given in the question. However, in this question the options are given in SI units.