Question

In the following $I$ refers to the current and the other symbols have the usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity?
$
A.\,M{L^{ - 3}}{T^{ - 3}}{I^2}\\
B.\,{M^{ - 1}}{L^{ - 3}}{T^3}{I^2}\\
C.\,{M^{ - 1}}{L^3}{T^3}I\\
D.\,{M^{ - 1}}{L^{ - 3}}{T^3}I
$

Answer 1

Hint: In this question we have to do the dimensional analysis and we have to find the dimension of the given quantity which is electrical conductivity. First, we will find the relation of the electrical conductivity with such quantities which have simple dimensions and then we will manipulate the dimensions to calculate the dimension of the given quantity.

Complete step by step answer:
As we stated that we have to find the relation of the electrical conductivity, the electrical conductivity is given by the following mathematical representation:
$\sigma = \dfrac{1}{\rho }$, Where $\sigma $ is the electrical conductivity and $\rho $ is the electrical conductivity.
The relation for the electrical conductivity is given below and we will use this relation to put its value in the relation of electrical conductivity:
$\rho = \dfrac{{Ra}}{l}$, putting this value, we get:
$
\sigma = \dfrac{1}{{\dfrac{{Ra}}{l}}}\\
\implies \sigma = \dfrac{l}{{Ra}}
$
Now we have this formula in which we have length area and resistance, we have to use the relation of the resistance to meet this formula more simplified:
The relation of resistance is given by Ohm's law. The Ohm's law is given below and when we put this value in the equation we get:
$
V = IR\\
R = \dfrac{V}{I}
$, Putting this value
$
\sigma = \dfrac{l}{{\left( {\dfrac{V}{I}} \right)a}}\\
\implies \sigma = \dfrac{{I \times l}}{{Va}}
$
Again, to simplify this equation, we will put the relation of potential.
$V = \dfrac{W}{Q}$, putting this value,
$
\sigma = \dfrac{{I \times l}}{{\dfrac{W}{Q}a}}\\
\implies \sigma = \dfrac{{Q \times I \times l}}{{Wa}}
$
Now, we will use the equation of Q, and put it in the above relation:
$
\sigma = \dfrac{{It \times I \times l}}{{Wa}}\\
\implies \sigma = \dfrac{{lt{I^2}}}{{Wa}}
$
Now we will find the dimension of the work done,
$W = F.x$
$
W = [ML{T^{ - 2}}][L]\\
\implies W = [M{L^2}{T^{ - 2}}]
$
And the dimension of the Area: $[{L^2}]$
Putting all these dimensions in the equation of electrical conductivity:
$
\sigma = \dfrac{{[L][T][{I^2}]}}{{[M{L^2}{T^{ - 2}}][{L^2}]}}\\
\implies \sigma = \dfrac{{[L][T][{I^2}]}}{{[M{L^2}{T^{ - 2}}][{L^2}]}}\\
\implies \sigma = [{M^{ - 1}}{L^{ - 3}}{T^3}{I^2}]
$
Now that we have found the dimension of electrical conductivity well matched with the options and say which is the correct option.

Hence the correct option is B.

Note:
These questions require strong knowledge of relation between the quantities if a student knows the dimensional formula of potential, then one can solve book questions faster than the student which has to do all this calculation so that they can use the dimensional formula of a much simpler quantity.