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Step-by-Step Solution
Step 1: Recall the definitions and relationships
Electrical conductivity, denoted typically by $ \sigma $, is defined as the reciprocal of electrical resistivity $ \rho $:
$ \sigma = \dfrac{1}{\rho} \quad \text{and} \quad \rho = \dfrac{R A}{L}, $
where:
$ R $ is the resistance of a material of length $ L $ and cross-sectional area $ A $.
$ \rho $ is the resistivity.
$ \sigma $ is the conductivity.
Step 2: Express resistance in terms of potential difference and current
We know that $ R = \dfrac{V}{I} $. Substituting this into $ \rho = \dfrac{R A}{L} $, we obtain:
$ \rho = \dfrac{V}{I} \cdot \dfrac{A}{L}. $
Hence,
$ \sigma = \dfrac{1}{\rho} = \dfrac{1}{\frac{V}{I} \cdot \frac{A}{L}} = \dfrac{L I}{V A}. $
Step 3: Determine the dimensional formula of the involved quantities
(a) Dimensions of $ L $: $ [L] $.
(b) Dimensions of $ I $: $ [I] $ (electric current).
(c) Dimensions of $ A $ (area): $ [L^2] $.
(d) Dimensions of $ V $ (potential difference): To find this, recall that $V = \dfrac{\text{Work/energy}}{\text{charge}}$. Work/energy has dimensions $ [M L^2 T^{-2}] $, and charge can be expressed as current $ \times $ time, $ [I][T] $. Therefore,
$ V = \dfrac{[M L^2 T^{-2}]}{[I][T]} = [M L^2 T^{-3} I^{-1}]. $
Step 4: Substitute dimensional forms into conductivity
$ \sigma = \dfrac{[L][I]}{ \bigl[\,M L^2 T^{-3} I^{-1}\bigr] [L^2]}.
$
Combine the terms carefully:
In the numerator: $ [L][I] $.
In the denominator: $ [M L^2 T^{-3} I^{-1}][L^2] = [M] [L^4] [T^{-3}] [I^{-1}].
So,
$ \sigma = \dfrac{[L][I]}{[M L^4 T^{-3} I^{-1}]} = [M^{-1} L^{-3} T^{3} I^{2}].
$
Step 5: Conclude the dimensional formula
Therefore, the dimensions of electrical conductivity $ \sigma $ are
$ [M^{-1} L^{-3} T^{3} I^{2}]. $
Answer
The correct dimension is $ M^{-1} L^{-3} T^{3} I^{2} $.
