Match the following two columns:

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(p)$ $M L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(q)$ $M L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(r)$ $M L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(s)$ None of these

Let $[{\varepsilon _0}]$ denote the dimensional formula of the permittivity of free space and $[{\mu _0}]$ denote the dimensional formula of the permeability of free space. If $M = \text{mass}$,$L = \text{length}$,$T = \text{time}$,and $I = \text{electric current}$,then:

The dimensional formula of angular impulse is:

The physical quantity which has the same dimensional formula as that of $\frac{\text{Energy}}{\text{Mass} \times \text{Length}}$ is:

Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ Viscous force	$(i)$ $[M^1 L^1 T^{-2}]$
$(b)$ Coefficient of viscosity	$(ii)$ $[M^1 L^{-1} T^{-1}]$
	$(iii)$ $[M^1 L^{-1} T^{-2}]$

There is another useful system of units,besides the $SI/MKS$. $A$ system,called the $CGS$ (centimeter-gram-second) system. In this system,Coulomb's law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm$ $(= 10^{-2} \ m)$,$F$ in dynes $(= 10^{-5} \ N)$ and the charges in electrostatic units $(esu)$,where $1 \ esu$ of charge $= \frac{1}{[3]} \times 10^{-9} \ C$. The number $[3]$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times 10^8 \ m/s$. An approximate value of $c$ then is $c = 3 \times 10^8 \ m/s$.
$(i)$ Show that the Coulomb law in $CGS$ units yields $1 \ esu$ of charge $= 1 \ (dyne)^{1/2} \ cm$. Obtain the dimensions of units of charge in terms of mass $M$,length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$.
$(ii)$ Write $1 \ esu$ of charge $= xC$,where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi \epsilon_0}} = \frac{{10^{-9}}}{{{x^2}}} \frac{N \ m^2}{C^2}$. With $x = \frac{1}{[3]} \times 10^{-9}$,we have $\frac{1}{{4\pi \epsilon_0}} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$ or $\frac{1}{{4\pi \epsilon_0}} = (2.99792458)^2 \times 10^9 \frac{N \ m^2}{C^2}$ (exactly).