Question

What is the barometric height of a liquid of density $3.4\,gc{m^{ - 3}}$ at a place where the mercury barometer is$70\,cm$?
A. $70\,cm$
B. $140\,cm$
C. $280\,cm$
D. $340\,cm$

Answer 1

Hint: Here first we will calculate the pressure for a mercury barometer with the given height where the density of the liquid is the same as that of the mercury. Now the pressure will remain the same in the barometer but the change will change then from this we can calculate the height of the barometer in the given density.

Complete step by step answer:
As per the problem we know the height of the mercury barometer is equal to $70\,cm$. Now we have to find the barometric height of a liquid of density $3.4\,gc{m^{ - 3}}$.We know for mercury barometer, height of the barometer is give as,
$h = 70cm\,of\,Hg$
In meter the height will be,
$h = 0.70m\,$

Now the density of the mercury liquid, $\rho = 13.6\,gc{m^{ - 3}}$
Now representing density in kilogram per meter cube,
$1g = {10^{ - 3}}kg$
$\Rightarrow 1c{m^3} = {\left( {{{10}^{ - 2}}} \right)^3}{m^3}$
Now,
$\rho = 13.6 \times {10^{ - 3}}kg \times {\left( {{{10}^{ - 2}}} \right)^{ - 6}}{m^{ - 3}}$
Hence the density in its SI unit will be,
$\rho = 13.6 \times {10^3}kg{m^{ - 3}}$
And the acceleration due to gravity, $g = 9.8m{s^{ - 2}}$
We know pressure as,
$P = \rho gh$
Where, Pressure = $P$, Density = $\rho $, Acceleration due to gravity = $g$ and Height = $h$.

Now putting the given known values we will get,
$P = 13.6 \times {10^3}kg{m^{ - 3}} \times 9.8m{s^{ - 2}} \times 0.70m$
$\Rightarrow P = 93296Pa$
The pressure will remain the same in the barometer. If we change the density of liquid then the height will change and we need to calculate the change height. Now the pressure will be,
$P = \rho 'gh'$
Where, $\rho ' = 3.4\,gc{m^{ - 3}} = 3.4 \times {10^3}\,kg{m^{ - 3}}$, $P = 93296\,Pa$ and $g = 9.8\,m{s^{ - 2}}$.
Now putting the given known values we will get,
$93296Pa = 3.4 \times {10^3}kg{m^{ - 3}} \times 9.8m{s^{ - 2}} \times h'$
Rearranging the equation we will get,
$\dfrac{{93296Pa}}{{3.4 \times {{10}^3}kg{m^{ - 3}} \times 9.8m{s^{ - 2}}}} = h'$
$ \therefore h' = 2.8\,m = 280\,cm$

Therefore the correct option is C.

Note: The SI unit of pressure is Pascal that is denoted as Pa. A barometer is an instrument which is used to measure the atmospheric pressure which is also called barometric pressure. Remember that atmospheric pressure changes with distance above and below the sea level. It is also used to measure the altitude and two main types of barometers are mercury and aneroid.