Question

The dimensions of resistivity in terms of M, L, T, and Q, where Q stands for the dimension of charge is:
$\text{A}. \quad ML^{3}T^{-1}Q^{-2}$
$\text{B}. \quad ML^{3}T^{-2}Q^{-1}$
$\text{C}. \quad ML^{2}T^{-1}Q^{-1}$
$\text{D}. \quad MLT^{-1}Q^{-1}$

Answer 1

Hint: The principle of homogeneity says that only like quantities can be added or subtracted and in all the equations, the dimensions on the right-hand side and on the left-hand side must be the same. In electrodynamics, resistivity is a unique property of material which is the measure of the resisting power of material towards the flow of current.

Formula used:
$R = \dfrac{{\rho}L}A, \ V=iR, \ V= \dfrac {W_{done}}Q$
Where R is the resistance of a wire of length ‘L’ and cross-section area ‘A’ and the resistivity of the material is $\rho$. W is the work done on charge Q.

Complete step by step answer:
Ohm’s law states, V=iR.
Hence $R=\dfrac{V}{i}$
And $V=\dfrac {W_{done}}Q$
Hence dimensions of $V =\dfrac{ML^2T^{-2}}Q={ML^2T^{-2}}Q^{-1}$ [$W=\int { F.ds }$]
Now, current is the rate of flow of charge, hence dimensions of $R = \dfrac{{ML^2T^{-2}}Q^{-1}}{QT^{-1}}= {ML^2T^{-1}}Q^{-2}$
Finally, using $R = \dfrac{{\rho}L}A \ or \ \rho = \dfrac{{R}A}L$
Putting the dimensions of R, A and L, we get
$\rho = \dfrac{[{ML^2T^{-1}}Q^{-2}] \times [L^2]}{[L]}$
$\rho = {ML^3T^{-1}Q^{-2}}$
Hence, A. is the correct option.

Additional information:
It is to be noted that many physical quantities have the same physical dimensions by they can be completely different in every aspect like work and torque, both have the same dimensions $[M^1 L^{2}T^{-2}]$ which are also the dimensions of heat given.

Note:
Students are advised to learn basic quantities’ dimensions only (for example displacement, angle, intensity, mass, etc.) and complex quantities’ dimensions (for example torque, pressure, force, etc.) can be easily derived by knowing the formulae and proceeding as shown in the above example. Also in the above example for calculating the dimensions of V we have many more formulae like $V=\int { E.dr }$. But it’s our duty to choose the formula so that it becomes the easiest to get evaluated.