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

Hint: To solve this particular question, we should know the formula for the specific resistance of the material and how to write the dimensional formula for the specific resistance. All physical quantities can be expressed as a combination of seven fundamental or base quantities. The seven fundamental quantities are represented as Mass$(M)$, length$(L)$, Time$(T)$, electric current$(A)$, Thermodynamic temperature$(K)$, Luminous intensity $(cd)$and Amount of substance $(mol)$

Formula used:
$R = \rho \dfrac{l}{A}$(Where $R$stands for the resistance of the conductor, $l$stands for the length of the conductor, $A$stands for the area of cross section of the conductor and $\rho $stands for the specific resistance of the material)
$R = \dfrac{V}{I}$(Where $V$stands for the potential difference and $I$stands for the current through the conductor)

Complete step by step answer:
The resistance of a conductor is directly proportional to the length of a conductor and it is inversely proportional to the area of cross section of the conductor.
This can be written as,
$R \propto \dfrac{l}{A}$
The proportionality sign can be eliminated by introducing a constant,
$R = \rho \dfrac{l}{A}$
Where this constant $\rho $is called the electrical resistivity or the specific resistance of the conductor.
From this equation, the specific resistance can be written as,
$\rho = \dfrac{{RA}}{l}$
According to Ohm’s law, Resistance can be written as the ratio of the potential difference across the conductor to the current flowing through the conductor.
This resistance can be written as,
$R = \dfrac{V}{I}$
Substituting this value of resistance in the expression for specific resistance, we get
$\rho = \dfrac{V}{I}\dfrac{A}{l}$
The dimension of potential difference can be written as,
$V = \left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
The dimension of current can be written as,
$I = \left[ A \right]$
The dimension for area can be written as
$A = \left[ {{L^2}} \right]$
The dimension of length can be written as,
$l = \left[ L \right]$
Substituting all this dimensional formulae in the expression of specific resistance, we get
$\rho = \dfrac{{VA}}{{Il}} = \dfrac{{\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]\left[ {{L^2}} \right]}}{{\left[ A \right]\left[ L \right]}} = \left[ {M{L^3}{T^{ - 3}}{A^{ - 2}}} \right]$

So, the correct answer is “Option B”.

Note:
To answer this question, we must have an idea about the dimensional formulae of basic quantities. One can even derive the dimensional formula for quantities like area, volume etc. The dimensional formula for potential difference can be derived as follows:
The potential difference can be written as, $V = \dfrac{W}{Q}$
Where $W$stands for the work done and $Q$stands for the charge.
Work done is the product of force and displacement.
Force is the product of mass and acceleration
Hence the potential difference can be written as,
 $V = \dfrac{{xma}}{Q}$
Substituting the dimensional formula for the displacement$x$, mass$m$, acceleration $a$and charge$Q$,
We get
$V = \left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$