The dimension of Planck's constant is equal to that of:

The physical quantity 'Energy Density' has the same dimensional formula as

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).

The force of interaction between two atoms is given by $F = \alpha \beta \exp \left( - \frac{x^2}{\alpha kt} \right)$,where $x$ is the distance,$k$ is the Boltzmann constant,$T$ is temperature,and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is:

Dimensions of $\varepsilon_0$ are

Match the physical quantities in Column-$I$ with their corresponding dimensional formulas in Column-$II$.

Column-$I$	Column-$II$
$(1)$ Torque	$(a)$ $M^1L^1T^{-1}$
$(2)$ Angular momentum	$(b)$ $M^1L^2T^{-1}$
$(3)$ Linear momentum	$(c)$ $M^1L^2T^{-2}$