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The ratio of height of a mercury column in a barometer at a place to the height of the liquid column at the same place is \[1:4\]. Find the density of the liquid.

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Last updated date: 19th Jun 2024
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Answer
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Hint Since pressure is the physical quantity relating both height and density use the pressure formula to combine the data of both liquid and mercury. After obtaining the simplified formula substitute the given data and find the density of the liquid.

Complete step-by-step solution
Let, \[{h_1}\] and \[{h_2}\] be the height of the mercury column and liquid column respectively.
 \[{\rho _1}\] and\[{\rho _2}\] be the respective density of mercury and the liquid in the columns.
From the question we have,
 \[{h_1}:{h_2} = {\text{ }}1:4\] which means \[{h_1} = {\text{ }}H\] and \[{h_2} = {\text{ }}4H\]

According to Pascal's law, it states that pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid present in that container as well as the walls of the container. In other words, pressure remains the same throughout the fluid if the height remains constant.

We know that,
 $P = \rho gh$
Here,
\[\rho \]is the density of the fluid
g is the acceleration due to gravity
h is the height of the fluid
So, using the above formula,
Pressure of mercury is
 ${P_1} = {\rho _1}g{h_1}$
And pressure of liquid is
 ${P_2} = {\rho _2}g{h_2}$
Since pressure is same at any point on the same horizontal level,
 $
  {P_1} = {P_2} \\
  \therefore {\rho _1}g{h_1} = {\rho _2}g{h_2} \\
 $
We know that density of mercury ${\rho _1} = 13.6gc{m^{ - 3}}$
On substituting the known data,
 $
  13.6gH = {\rho _2}g4H \\
  {\rho _2} = \dfrac{{13.6}}{4} = 3.4gc{m^{ - 3}} \\
 $

The density of the liquid in the column of the barometer is \[3.4gc{m^{ - 3}}\].

Note The density of common liquids should be known. The acceleration due to gravity gets cancelled as it remains the same for both pressures. The ratio should be taken into common terms.