Question

The height of the mercury column in a barometer in a Calcutta laboratory was recorded to be 75 mm. Calculate this pressure in SI and CGS units using the following data: specific gravity of mercury = $ 13.6 $ , density of water $ = {10^3}kg{m^{ - 3}} $ , $ g = 9.8m/{s^2} $ at Calcutta, Pressure $ P = h\rho g $ in usual symbols.

Answer 1

Hint
To find out the pressure, we have to first calculate the density of mercury by using the specific gravity of mercury and density of water. Then, by using the formula, $ P = h\rho g $ , we can calculate the pressure.
Formulae used: $ P = h\rho g $ where $ P $ is the pressure, $ h $ is the height of the mercury column, $ \rho $ is the density of mercury, and $ g $ is the acceleration due to gravity.

Complete step by step answer
The specific gravity or relative density of a material is the ratio of its density to the density of a reference material. The specific gravity of the liquids is measured by taking water as the reference sample. As given in the problem, the specific gravity of mercury $ s.g = 13.6 $ and the density of water $ \rho_{water} = {10^3}kg/{m^3} $ in SI units. So, the density of mercury in SI units is given by,
 $ \rho Hg = s.g \times \rho_{water} $
Then by substituting the known values,
 $ \rho Hg = 13.6 \times {10^3}kg/{m^3} $
 $ \Rightarrow \rho Hg = 13600kg/{m^3}\left( {SI} \right) $
Now, for calculating the pressure, we can substitute the values in the equation. It is given that the height of the mercury column $ h = 75mm = 75 \times {10^{ - 3}}m $ .
Thus, substituting the values in the formula and solving we get,
 $ P = h\rho Hgg = 75 \times {10^{ - 3}} \times 13600 \times 9.8 $
 $ \therefore P = 9996kg{m^{ - 1}}{s^{ - 2}} = 9996Pa $
This is the Pressure in Calcutta laboratory calculated in SI units.
When we redo the calculation in CGS units, we know that the density of water is $ 1g/c{m^3} $ in CGS units. Then, the density of mercury is
 $ \rho Hg = 13.6 \times 1g/c{m^3} = 13.6g/c{m^3}\left( {CGS} \right) $
Similarly, the height of the mercury column in CGS units, $ h = 75mm = 7.5cm $ , and the acceleration due to gravity, $ g = 9.8m/{s^2} = 980cm/{s^2} $ . Hence, the pressure in CGS can be calculated as:
 $ P = h\rho Hgg = 7.5cm \times 13.6g/c{m^{ - 3}} \times 980cm/{s^2} $
 $ \therefore P = 99960gc{m^{ - 1}}{s^{ - 2}} = 99960barye $
Therefore, we obtain the Pressure in the Calcutta laboratory as 9996 Pa and 99960 Ba in SI and CGS units, respectively.

Note
One important point to take care while solving this problem is to maintain all the quantities in the same system of units always. For example, if we are solving in the CGS units, then we may convert the height of the mercury column to cm (centimetre), but we should also remember that the acceleration due to gravity should be $ 980cm{s^{ - 2}} $ .