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# The magnitude of atmospheric pressure is equal to the pressure exerted by a:(A) $76mm$ tall column of mercury(B) $760mm$ tall column of water(C) $76cm$ tall column of mercury(D) $760cm$ tall column of mercury

Last updated date: 19th Jun 2024
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Hint: Calculate the pressure for each of the given heights of the mercury column and then tally it with the standard pressure of the atmosphere.
Formula used: The formula used for calculating the atmospheric pressure is $p = \rho gh$; where $p$is the pressure, $\rho$is the density of mercury and $h$is the height of the mercury column.
For calculation purposes we have used, $\rho = 13600kg{m^{ - 3}}$ and $1mb = 100N{m^{ - 2}}$.

Complete step by step answer: For solving the given question, we first need to know what is the magnitude of the atmospheric pressure.
The magnitude of atmospheric pressure can be represented in different units. They are listed as follows:
(A) $1013.25 \times {10^3}dynesc{m^{ - 2}}$
(B) $1013.25millibars$
(C) $1013.25hPa$
(D) $101.325kPa$
All the standard devices that are used to measure the atmospheric pressure make use of mercury to make the pressure measurements. This is because expansion and contraction of mercury occurs evenly with changes in temperature and pressure. Another important reason for using mercury is that it is the only metal that is in liquid state at room temperature.
So, it is quite clear from the above discussion that water cannot be used for pressure measurement and hence option$(2)$is not appropriate.
Now, in the other three options we have three given values of the mercury column height. So, to check for the correct option we have to calculate the resultant pressure values for each of the given values. To calculate pressure, we use:
$p = \rho gh$ $\to eqn.1$; where all the terms have the physical meanings as stated above.
When $h = 76mm = 0.076m$, we have:
$p = \left( {13600 \times 9.8 \times 0.076} \right)N{m^{ - 2}} = 10129.28N{m^{ - 2}} = 101.2928mb$
The obtained value is not equal to standard atmospheric pressure, so option $(1)$is incorrect.
When $h = 76cm = 0.76m$, we have:
$p = \left( {13600 \times 9.8 \times 0.76} \right)N{m^{ - 2}} = 101292.8N{m^{ - 2}} = 1012.928mb \approx 1013mb$
The obtained value is almost equal to standard atmospheric pressure, so option $(3)$is correct.
When $h = 760cm = 76m$,we have
$p = \left( {13600 \times 9.8 \times 76} \right)N{m^{ - 2}} = 10129280N{m^{ - 2}} = 10129.28mb$

The obtained value is not equal to standard atmospheric pressure, so option $(4)$is incorrect.

Note:
Here, the formula used, i.e., $p = \rho gh$ comes from the definition of pressure as we define pressure as the ratio of force per unit area.