Question

At a given place, a mercury barometer records a pressure of $0.70$ m of Hg. What would be the height of the water column if mercury in the barometer is replaced by water? Take the density of mercury to be $13.6 \times {10^3}\,kg\,{m^{ - 3}}$ .

Answer 1

Barometer is a device which is used to record the pressure. Pressure exerted on any substance is given as $\rho gh$ here, $\rho$ is the density of the substance used to record the pressure, $g$ is the acceleration due to gravity and $h$ is the height up to which the substance rises also known as barometric height.

Complete step by step solution: Barometer is used in the weather forecast to show the short term changes in the water.
The density of mercury is $13.6$ times the density of water.
Let ${\rho _m}$ be the density of mercury,
The density of water is ${\rho _w}$ .
The barometric height of the mercury be ${h_m}$
Let the barometric height of the water column in the barometer be ${h_w}$ .
The acceleration due to gravity is $g$ .
As pressure $P$ is given as:
$P = \rho gh$
Therefore, for mercury the pressure will be given as:
$P = {\rho _m}g{h_m}$ --equation $1$
Now, when the mercury is replaced by water the pressure remains the same but the barometric height changes. Thus, we can have the pressure as:
$P = {\rho _w}g{h_w}$ -- equation $2$
As the pressure remains same, thus from equation $1$ and equation $2$ we have
${\rho _w}g{h_w} = {\rho _m}g{h_m}$
$\Rightarrow {h_w} = \dfrac{{{\rho _m}{h_m}}}{{{\rho _w}}}$
As, ${\rho _m} = 13.6 \times {\rho _w}$ , thus we have:
${h_w} = \dfrac{{\left( {13.6{\rho _w}} \right){h_m}}}{{{\rho _w}}}$
$\Rightarrow {h_w} = \left( {13.6} \right){h_m}$
We are given that ${h_m} = 0.70m$ , substituting this value, we get:
${h_m} = 13.6 \times 0.70$
$\Rightarrow {h_m} = 9.52$

The height of the water column is $9.52m$ .

Note: The atmospheric pressure remains the same irrespective of the substance filled in the barometer as a result the substance rises to a height which is inversely proportional to its density. The density of mercury is $13.6$ times the density of water. As a result, the rise in barometric height of water is also $13.6$ times the rise in mercury.