The relationship between pressure (p) and depth of fluid (h) in the static condition is:
p ∝ h
The question asks about the relationship between pressure (\(p\)) and the depth (\(h\)) of a fluid when the fluid is in a static condition (not moving). This is a fundamental concept in fluid mechanics, specifically hydrostatics.
In a static fluid, the pressure at a certain depth below the surface is given by the formula:
\(p = p_0 + \rho g h\)
Let's break down what each term means:
In the formula \(p = p_0 + \rho g h\), for a specific fluid at a given location, the atmospheric pressure (\(p_0\)), the fluid density (\(\rho\)), and the acceleration due to gravity (\(g\)) are essentially constant values. Therefore, the pressure \(p\) is linearly dependent on the depth \(h\).
If we consider the gauge pressure, which is the pressure relative to atmospheric pressure (\(p - p_0\)), the formula becomes:
\(p - p_0 = \rho g h\)
This clearly shows that the gauge pressure is directly proportional to the depth \(h\):
\(p - p_0 \propto h\)
Even when considering the total pressure \(p\), since \(p_0\), \(\rho\), and \(g\) are constants, the term \(\rho g h\) is the part of the pressure that increases with depth. This increase is directly proportional to \(h\).
Therefore, the relationship between pressure and depth in a static fluid is a direct proportionality when considering the pressure change from the surface or gauge pressure, and a linear relationship for the total pressure.
Let's look at the given options:
The option \(p \propto h\) best describes the core relationship: as depth increases, pressure increases proportionally (above the atmospheric pressure).
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