Calculate the pressure due to a column of 0.4 m of (a) water, (b) an oil of sp. gr. 0.9, and (c) mercury of sp. gr. 13.6.

Hydrostatic Pressure of a 0.4m Column

Problem Statement

Calculate the pressure due to a column of 0.4 m of (a) water, (b) an oil of sp. gr. 0.9, and (c) mercury of sp. gr. 13.6. Take density of water, $\rho = 1000 \, \text{kg/m}^3$.

Given Data

Height of liquid column, $Z = 0.4 \, \text{m}$
Density of water, $\rho_{\text{water}} = 1000 \, \text{kg/m}^3$
Specific gravity of oil, $S_{\text{oil}} = 0.9$
Specific gravity of mercury, $S_{\text{mercury}} = 13.6$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

The formula for hydrostatic pressure ($p$) due to a liquid column is given by:

$$ p = \rho g Z $$

Where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $Z$ is the height of the fluid column.

(a) Pressure due to Water

Using the density of water directly.

$$ p_{\text{water}} = \rho_{\text{water}} \times g \times Z $$ $$ p_{\text{water}} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.4 \, \text{m} $$ $$ p_{\text{water}} = 3924 \, \text{N/m}^2 $$

(b) Pressure due to Oil

First, we find the density of the oil using its specific gravity.

$$ \rho_{\text{oil}} = S_{\text{oil}} \times \rho_{\text{water}} $$ $$ \rho_{\text{oil}} = 0.9 \times 1000 \, \text{kg/m}^3 $$ $$ \rho_{\text{oil}} = 900 \, \text{kg/m}^3 $$

Now, we calculate the pressure.

$$ p_{\text{oil}} = \rho_{\text{oil}} \times g \times Z $$ $$ p_{\text{oil}} = 900 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.4 \, \text{m} $$ $$ p_{\text{oil}} = 3531.6 \, \text{N/m}^2 $$

(c) Pressure due to Mercury

First, we find the density of mercury using its specific gravity.

$$ \rho_{\text{mercury}} = S_{\text{mercury}} \times \rho_{\text{water}} $$ $$ \rho_{\text{mercury}} = 13.6 \times 1000 \, \text{kg/m}^3 $$ $$ \rho_{\text{mercury}} = 13600 \, \text{kg/m}^3 $$

Now, we calculate the pressure.

$$ p_{\text{mercury}} = \rho_{\text{mercury}} \times g \times Z $$ $$ p_{\text{mercury}} = 13600 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.4 \, \text{m} $$ $$ p_{\text{mercury}} = 53366.4 \, \text{N/m}^2 $$

Final Results:

(a) Pressure of water: $ 3924 \, \text{N/m}^2 $

(b) Pressure of oil: $ 3531.6 \, \text{N/m}^2 $

Explanation of Hydrostatic Pressure

Hydrostatic Pressure is the pressure exerted by a fluid at rest due to the force of gravity. It is the weight of the fluid column above the point of measurement. The pressure at any given depth in a fluid is constant and acts equally in all directions. As shown by the formula $p = \rho g Z$, the pressure is directly proportional to three factors: the density of the fluid ($\rho$), the acceleration due to gravity ($g$), and the vertical height (or depth) of the fluid column ($Z$).

Physical Meaning

This problem clearly demonstrates the effect of fluid density on pressure. Even though the height of the liquid column is the same in all three cases (0.4 m), the resulting pressures are vastly different:

Oil, being less dense than water, exerts the least pressure.
Water exerts a moderate pressure.
Mercury, which is 13.6 times denser than water, exerts by far the highest pressure.

This is why mercury was historically used in barometers and manometers; its high density means that even large atmospheric pressures can be balanced by a relatively short, manageable column of liquid.