$1$ pascal $\left( {Pa} \right)$ is equal to
(A) $1 \times {10^5}bar$
(B) $1 \times {10^3}bar$
(C) $1 \times {10^{ - 3}}bar$
(D) $1 \times {10^{ - 5}}bar$

Hint: One bar of pressure is approximately equal to the pressure of the atmosphere of the earth at the sea level. The pressure of the earth’s atmosphere is measured using mercury, and its value is equal to the pressure of $76cm$ of the mercury column. Use the formula of the pressure due to a liquid column, and substitute the values of the quantities in their SI units to get the pressure in pascal.
Formula used: The formula used to solve this question is given by
$P = \rho gh$, here $P$ is the pressure exerted due to a column of height $h$ of a liquid of density $\rho $, and $g$ is the acceleration due to gravity.

Complete step-by-step solution:
We know that the atmospheric pressure of earth is measured in units of bar. At sea level, it is equal to $76cm$ of mercury. We know that the pressure due to a liquid column is given by
$P = \rho gh$......................(1)
Since Pascal is an SI unit, we have to substitute all the three values present on the RHS of the above expression, in their respective SI units.
Since height of the mercury column is equal to $76cm$, so we have
$h = 76cm$
We know that $1cm = 0.01m$. Therefore we get
$h = 0.76m$.................................(2)
Also, the density of mercury is equal to $13.5g/cc$. So we have
$\rho = 13.5g/cc$
We know that $1g = {10^{ - 3}}kg$, and $1cc = {10^{ - 6}}{m^3}$. So we get
$\rho = 13.5 \times \dfrac{{{{10}^{ - 3}}}}{{{{10}^{ - 6}}}}$
$ \Rightarrow \rho = 13500kg/{m^3}$......................(3)
Also, we know that the value of the acceleration due to gravity is equal to $10m/{s^2}$. So we have
$g = 10m/{s^2}$...................................(4)
Putting (2), (3) and (4) in (1) we get
$P = 13500 \times 9.8 \times 0.76$
On solving we get
$P = 100548Pa$
Since this is equal to one bar, so we have
$1bar = 100548Pa$
$ \Rightarrow 1Pa = \dfrac{1}{{100548}}bar \approx 1 \times {10^{ - 5}}bar$
 Thus, one Pascal is equal to $1 \times {10^{ - 5}}bar$.

Hence, the correct answer is option D.

Note: The atmospheric pressure is also expressed in atm along with the bar. But these both are not the standard units of the pressure. The SI unit of pressure is Pascal only. But still we use the unit atm to express the atmospheric pressure.