To solve this problem it is necessary to apply the concepts related to pressure as a unit that measures the force applied in a specific area as well as pressure as a measurement of the density of the liquid to which it is subjected, its depth and the respective gravity.
The two definitions of pressure can be enclosed under the following equations
[tex]P = \frac{F}{A}[/tex]
Where
F= Force
A = Area
[tex]P = \rho gh[/tex]
Where,
[tex]\rho =[/tex] Density
g = Gravity
h = Height
Our values are given as,
[tex]d = 0.8m \rightarrow r = 0.4m[/tex]
[tex]A = \pi r^2 = \pi * 0.4^2 = 0.503m^2[/tex]
If we make a comparison between the lid and the tube, the diameter of the tube becomes negligible.
Matching the two previous expressions we have to
[tex]\frac{F}{A} = \rho g h[/tex]
Re-arrange to find h
[tex]h = \frac{F}{A\rho g}[/tex]
[tex]h = \frac{390}{(0.503)(1000)(9.8)}[/tex]
[tex]h = 0.079m[/tex]
[tex]h = 7.9cm[/tex]
Therefore the height of water in the tube is 7.9cm
