To calculate the pressure exerted by the liquid at a point 5 cm above the height of the liquid in the bottle, we'll first need to find the height of the liquid column and then apply the hydrostatic pressure formula.
-
Determine the height of the liquid column:
Since the bottle is half filled and the height of the bottle is 28 cm, the height of the liquid (h) is: \[ h = \frac{28 , \text{cm}}{2} = 14 , \text{cm} \] -
Convert the height of the liquid to meters: \[ h = 14 , \text{cm} = 0.14 , \text{m} \]
-
Density of the liquid:
The density \( \rho \) of the liquid is given as 0.9 g/cm³. To convert it to kg/m³: \[ \rho = 0.9 , \text{g/cm}^3 = 0.9 \times 1000 , \text{kg/m}^3 = 900 , \text{kg/m}^3 \] -
Calculate the pressure at the base of the liquid column using the hydrostatic pressure formula: \[ P = \rho g h \] where \( g \) (acceleration due to gravity) is approximately \( 9.81 , \text{m/s}^2 \).
Substituting the values, we get: \[ P = 900 , \text{kg/m}^3 \times 9.81 , \text{m/s}^2 \times 0.14 , \text{m} \]
Let's calculate that: \[ P = 900 \times 9.81 \times 0.14 = 1,239.54 , \text{Pa} \]
-
Calculate the pressure at the point 5 cm above the liquid: Now, we need to find the pressure at a point 5 cm above the liquid level. Since we are above the liquid column, the pressure at this point will be the atmospheric pressure minus the pressure exerted by the height of the liquid column.
The total height from the base of the liquid column to the point of interest is: \[ h' = 14 , \text{cm} + 5 , \text{cm} = 19 , \text{cm} = 0.19 , \text{m} \]
So, we can find the pressure at the height of 19 cm: \[ P' = \rho g h' \] \[ P' = 900 \times 9.81 \times 0.19 = 1,675.11 , \text{Pa} \]
-
Find the pressure relative to atmospheric pressure:
Pressure at a point above the liquid will be less than the hydrostatic pressure calculated, simply because it is not within the liquid. In practice, the pressure of the air will still be present in addition to the hydrostatic pressure of the liquid directly below perturbed to atmospheric pressure:
Thus, unless specified to take atmospheric pressure into account:
- The pressure at the point in question is: \[ P_{above} = P_{atm} - P_{liquid} \]
- The pressure due to 14 cm of liquid below 5cm above is effectively just = 1,675.11 Pa from the atmosphere point directly rather than from any column height above liquid level.
Hence the calculated pressure at the point 5 cm above the liquid will remain \( 900 , \text{kg/m}^3 \) multiplied by \( 9.81 \) hence effectively being less than full atmospheric potential pressure existence.
This leads to the approximation derived as simply the effective hydrostatic contribution unique from liquid to general atmospheric potential pressure being 1675.11 Pa remaining as an equivalent representation of pressure contribution among liquid levels, continually measured at the final resultant desired above the liquid column accurately constituted.