Hydrostatic Paradox 

The hydrostatic paradox explains the very interesting concept in fluid mechanics that pressure exerted by a fluid at rest at any depth is determined by only the height of the fluid column above that point and does not depend on the shape or volume of the container. The whole idea clashes with our general notion about how pressure behaves in various containers.

1.0What is the Hydrostatic Paradox?

The meaning of the Hydrostatic Paradox is that pressure at any depth is independent of the size, shape, and aspect of the container, with the liquid's height being at par. This paradoxical-looking result indicates that the shape and volume of the containing vessel do not affect the pressure at a particular depth.

2.0Principle of Hydrostatic Paradox 

The Hydrostatic Paradox is based on the idea that hydrostatic pressure is determined by the height of the liquid column, its density, and gravitational force and does not depend on the shape of the container used. Mathematically, the pressure at any given depth is described by the formula:

$P=\rho gh$

Here:

• 𝑃 = pressure at the point,
• 𝜌 = density of the fluid,
• 𝑔 = acceleration due to gravity and
• ℎ = height of the fluid column above the point.

S.I. unit of pressure is Pascal (Pa). 

Key Takeaway

1. The difference in pressure between two points at different depths in a fluid, you can use the following formula: $\Delta P=\rho g\Delta h$. Here, P is the difference in pressure, and h is the different depths in a fluid. 
2. The formula for pressure can also be written as: $P=P_0+\rho gh$. This formula is used to calculate the pressure of the fluid on an open surface like an ocean or lake. Here, P₀ is the atmospheric pressure.
3. The formula $P=\rho gh$ is used to calculate the pressure in a fluid at rest in a closed column where the atmospheric pressure is assumed to be zero. 

3.0Hydrostatic Law

The law states that in a stationary fluid, Hydrostatic pressure due to liquid on the vertical wall of the container increases with the increase in depth and has a direct relation with its corresponding column height. 

In simpler words the deeper you go into the fluid, the greater pressure the fluid will exert. It happens primarily due to the weight of the fluid increases with the depth. This expresses why pressure becomes greater by sinking deeper into a fluid. The expression for pressure at hydrostatic is:

$P=\rho gh$

This law is an important concept in fluid dynamics to understand why containers of different shapes filled with the same height exert the same hydrostatic pressure at the bottom. 

4.0Gauge Pressure and Atmospheric pressure

As mentioned earlier, Hydrostatic law is the concept used only for stationary fluids. In real-life scenarios, the gauge pressure gives the actual pressure exerted by the fluid on an object, relative to the atmospheric pressure. This is the actual pressure experienced by the object submerged in a fluid. 

The gauge pressure does not include the effect of atmospheric pressure, which is assumed to be constant and acting uniformly on all surfaces. Hence, gauge pressure represents the real, effective pressure exerted by the fluid, which is critical for understanding fluid dynamics in systems like tanks, pipes, and machinery. 

Mathematically the Gauge pressure $(P_{gauge})$ with respect to atmospheric pressure $(P_{atm})$ is represented as: $P_{gauge}=\rho gh-P_{atm}$

5.0Examples of Hydrostatic Paradox 

Problem 1: A U-tube manometer contains mercury (ρ_Hg=13600 kg/m³) and water (ρ_water=1000 kg/m3). The difference in the height of mercury columns in the two arms of the U-tube is 0.1 meters. Calculate the pressure difference between the two arms.

Solution: It is given that: 

ρ_Hg=13600 kg/m³, h = 0.1m, g = 9.8 m/s² 

The pressure difference($\Delta$P) due to the height difference(h) in a fluid of density p is given by: 

$\Delta P=\rho gh$

$\Delta P=13600\times 9.8\times 0.1$

$\Delta P=13328Pa$

Problem 2: A rectangular gate in a dam is 2 meters high and 4 meters wide. The gate is submerged in water such that the top edge of the gate is 3 meters below the water surface. Calculate the hydrostatic force acting on the gate. Use the density of water ρ=1000 kg/m³ and g=9.8 m/s².

Solution: 

Hydrostatic Pressure at the top and bottom of the Gate: 

1. Top of the gate: 

$h_{top}=3m$

$P_{top}=\rho g{h}_{top}=1000\times 9.8\times 3=29400Pa$

2. Bottom of the gate: 

$h_{bottom}=3+2=5m$

$P_{bottom}=\rho g{h}_{bottom}=1000\times 9.8\times 5=49000Pa$

Average Pressure on the gate: 

$P_{avg}=\frac{P_{top}+P_{bottom}}{2}=\frac{29400+49000}{2}=39200Pa$

Hydrostatic force $(F)=P_{avg}\times Area$

Now the area of Gate =

$Height\times width=4\times 2=8m^2$

$F=39200\times 8=3,13,600N$

Problem 3: A solid cube with a side length of 0.5 meters and density ρ_cube=500 kg/m³ is completely submerged in a container with three immiscible fluids: oil (ρ_oil=800 kg/m3, water (ρ_water=1000 kg/m³), and mercury (ρ_Hg=13600 kg/m³). The cube is half submerged in oil, with the remaining part in water and mercury. Determine the hydrostatic pressure acting on the top and bottom faces of the cube and calculate the net buoyant force acting on the cube. Assume g=9.8 m/s².

Solution: According to the question 

The total height of the cube is 0.5m 

The cube is half submerged in oil, so the depth in oil is 0.25m 

The remaining 0.25m is submerged in water and mercury. 

Hydrostatic Pressure on the top face: The top face is at the interface between the oil and air.

Depth of the top face h_top in oil = 0.25 meters.

Hydrostatic pressure at the top face due to oil: P_top = ρ_oilgh_top = 800×9.8×0.25 = 1960 Pa

Hydrostatic Pressure on the Bottom Face: The bottom face is submerged in water and mercury.

Depth of the bottom face h_bottom in oil + water: 0.25 meters (oil) + 0.25 meters (water) = 0.5 meters.

Pressure due to oil: P_oil  = 800×9.8×0.25 = 1960 Pa

Pressure due to water: P_water = 1000×9.8×0.25 = 2450 Pa

Total pressure at the bottom face: P_bottom = P_oil +P_water = 1960+2450 = 4410 Pa

The buoyant force is the weight of the displaced fluid.

The volume of the cube V_cube = 0.5×0.5×0.5=0.125 m³

Weight of displaced oil: W_oil = ρ_oilgV_oil = 800×9.8×(0.5×0.5×0.25) = 490 N

Weight of displaced water: W_water = ρ_watergV_water = 1000×9.8×(0.5×0.5×0.25)= 613.5 N

Total buoyant force:

F_byoyant = W_oil + W_water = 490 + 613.5 = 1102.5N