Question

The dimensions of specific resistance are:
A. $\left[ {M{L^{ - 2}}{A^{ - 1}}} \right]$
B. $\left[ {M{L^3}{T^{ - 3}}{A^{ - 2}}} \right]$
C. $\left[ {M{L^3}{T^{ - 2}}{A^{ - 1}}} \right]$
D. $\left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]$

Answer 1

Use the formula for resistance of a wire. This formula gives the relation between the resistance of the conductor, resistivity of specific resistance of material of the wire, length of the wire and cross-sectional area of wire. Substitute the value of resistance using Ohm’s law expression. Substitute dimensions of all the terms in the expression for specific resistance to determine its dimensions.

Formula used:
The resistance $R$ of a wire is
$R = \rho \dfrac{L}{A}$ …… (1)
Here, $\rho$ is the resistivity of the material of the wire, $L$ is length of the wire and $A$ is the cross-sectional area of the wire.
The expression for Ohm’s law is given by
$V = IR$ …… (2)
Here, $V$ is the potential difference across the ends of the conductor, $I$ is the current through the conductor and $R$ is the resistance of the conductor.

Complete step by step answer:
The resistivity of material of a wire is also called specific resistance of the wire.
Rearrange equation (2) for the resistance of the conductor.
$R = \dfrac{V}{I}$
Rearrange equation (1) for the specific resistance of the material of a wire.
$\rho = \dfrac{{RA}}{L}$

Substitute $\dfrac{V}{I}$ for $R$ in the above equation.
$\Rightarrow \rho = \dfrac{{VA}}{{IL}}$ …… (3)
The dimensions of the specific resistance of the wire can be determined from the dimensions of all the physical quantities given in the above equation.

The dimensional formula for potential difference is $\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$. The dimensional formula for cross-sectional area is $\left[ {{L^2}} \right]$. The dimensional formula for current is $\left[ A \right]$. The dimensional formula for length is $\left[ L \right]$.

Substitute $\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$ for $V$, $\left[ {{L^2}} \right]$ for $A$, $\left[ A \right]$ for $I$ and $\left[ L \right]$ for $L$ in equation (3).
$\Rightarrow \rho = \dfrac{{\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]\left[ {{L^2}} \right]}}{{\left[ A \right]\left[ L \right]}}$
$\Rightarrow \rho = \left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]\left[ {{L^2}} \right]\left[ {{A^{ - 1}}} \right]\left[ {{L^{ - 1}}} \right]$
$\therefore \rho = \left[ {M{L^3}{T^{ - 3}}{A^{ - 2}}} \right]$
Therefore, the dimensions of specific resistance are $\left[ {M{L^3}{T^{ - 3}}{A^{ - 2}}} \right]$.

Hence, the correct option is B.

Note: The students should be careful while determining the dimensions of the specific resistance of a material. The dimensions of all the terms in the formula should be taken correctly and the resultant dimensions of the specific resistance should be determined correctly by changing the signs of all the terms in the denominator.