Question

Find out the pressure due to a column of $0.4m$ of
(A) Water.
(B) An oil of specific gravity $0.9$.
(C) Mercury of specific gravity $13.6$.
Let us assume that the density of water is
$\rho =1000\dfrac{kg}{{{m}^{3}}}$

Answer 1

Hint: The pressure of the column has been found by taking the product of the density of the fluid, acceleration due to gravity and height of the fluid. Substitute the density of each of the materials. Then find out the pressure. The pressure has been expressed in kilopascal. This will help you in answering this question.

Complete step by step answer:
The height of the liquid column is given as,
$h=0.4m$
Let the acceleration due to gravity be,
$g=10m{{s}^{-2}}$
The pressure of the column has been found by taking the product of the density of the fluid, acceleration due to gravity and height of the fluid. This can be written as,
$P=\rho gh$
As we all know the density of water can be written as,
${{\rho }_{w}}=1000\dfrac{kg}{{{m}^{3}}}$
Substituting the values in the equation will give,
$P=1000\times 10\times 0.4=4kPa$
For oil, the density of oil can be written as,
${{\rho }_{o}}=900kg{{m}^{-3}}$
Substituting the values in the equation will give,
$P=900\times 10\times 0.4=3.6kPa$
The density of the mercury has been mentioned as,
${{\rho }_{m}}=13.6\times {{10}^{3}}kg{{m}^{-3}}$
Substituting this in the equation of pressure will give,
$P=13.6\times {{10}^{3}}\times 10\times 0.4=54.4kPa$
Therefore the pressure due to a column of $0.4m$ of water, oil and the mercury are found to be $4kPa,3.6kPa,54.4kPa$.

Note: Pressure can be explained as the physical force which is exerted on a body. The force acted will be normal to the surface of the bodies per unit area. The fundamental formula for pressure is the ratio of force to the area of the surface. The unit of pressure is expressed in Pascal. There are different kinds of pressures: Atmospheric, Differential, absolute and Gauge Pressure.