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Question: \(1\) pascal \(\left( {Pa} \right)\) is equal to (A) \(1 \times {10^5}bar\) (B) \(1 \times {10^3...

11 pascal (Pa)\left( {Pa} \right) is equal to
(A) 1×105bar1 \times {10^5}bar
(B) 1×103bar1 \times {10^3}bar
(C) 1×10−3bar1 \times {10^{ - 3}}bar
(D) 1×10−5bar1 \times {10^{ - 5}}bar

Explanation

Solution

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 76cm76cm 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=ρghP = \rho gh, here PP is the pressure exerted due to a column of height hh of a liquid of density ρ\rho , and gg 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 76cm76cm of mercury. We know that the pressure due to a liquid column is given by
P=ρghP = \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 76cm76cm, so we have
h=76cmh = 76cm
We know that 1cm=0.01m1cm = 0.01m. Therefore we get
h=0.76mh = 0.76m.................................(2)
Also, the density of mercury is equal to 13.5g/cc13.5g/cc. So we have
ρ=13.5g/cc\rho = 13.5g/cc
We know that 1g=10−3kg1g = {10^{ - 3}}kg, and 1cc=10−6m31cc = {10^{ - 6}}{m^3}. So we get
ρ=13.5×10−310−6\rho = 13.5 \times \dfrac{{{{10}^{ - 3}}}}{{{{10}^{ - 6}}}}
⇒ρ=13500kg/m3\Rightarrow \rho = 13500kg/{m^3}......................(3)
Also, we know that the value of the acceleration due to gravity is equal to 10m/s210m/{s^2}. So we have
g=10m/s2g = 10m/{s^2}...................................(4)
Putting (2), (3) and (4) in (1) we get
P=13500×9.8×0.76P = 13500 \times 9.8 \times 0.76
On solving we get
P=100548PaP = 100548Pa
Since this is equal to one bar, so we have
1bar=100548Pa1bar = 100548Pa
⇒1Pa=1100548bar≈1×10−5bar\Rightarrow 1Pa = \dfrac{1}{{100548}}bar \approx 1 \times {10^{ - 5}}bar
Thus, one Pascal is equal to 1×10−5bar1 \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.