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Gauge pressure is an important concept in fluid mechanics. It represents the pressure in a system compared to the surrounding atmospheric pressure. In other words, it's the pressure exerted by a fluid above atmospheric pressure.
The gauge pressure is particularly useful because it shows the additional pressure applied by just the fluid, excluding any atmospheric effects.In the context of the U-shaped tube, gauge pressure at the water-mercury interface is determined by the water column in the tube.
To understand how this pressure is calculated, consider the common formula:

• \( P = \rho g h \)

Here, \( \rho \) is the water density (1000 kg/m³), \( g \) is the gravitational acceleration (9.81 m/s²), and \( h \) is the height of the water column (0.15 m).
By substituting these values, we find a gauge pressure of 1471.5 Pa at the interface between water and mercury. This straightforward calculation is essential for determining pressure effects in fluids.

The U-shaped tube is a fascinating tool in fluid dynamics, often used to demonstrate principles like pressure balance and fluid statics. In our problem, it holds both mercury and water, allowing us to explore how pressure changes in such systems. When a fluid, like water, is poured into one arm of the tube, it affects the mercury level. The mercury moves downward on the water-filled side and upward on the opposite side. This movement is due to the pressure from the water column, creating a need to calculate the pressure balances. In exercises involving a U-shaped tube, understanding how different fluid densities affect each other is key. Mercury, being denser than water, requires less volume to exert an equal pressure to a larger column of water.
This behavior is leveraged in our scenario to figure out the displacement of mercury on the tube's right side.

Mercury density is crucial in this system's behavior. With a density of 13600 kg/m³, mercury is much denser than water. This high density means mercury can exert significant pressure even in small amounts.
The concept of mercury's density helps explain why the mercury rises on the opposite side when water is added to one arm of the U-shaped tube.To maintain pressure equilibrium within the fluid system, mercury compensates for the added water pressure by moving accordingly. This movement demonstrates the principle of pressure balance across different fluids. Understanding the varying densities is essential to accurately calculate displacement and pressure at the interfaces.Applying the formula for pressure equality:

• \( \rho_{\text{water}} g h_{\text{water}} = \rho_{\text{mercury}} g h_{\text{mercury}} \)

allows us to solve for the displacement of mercury based on the water column's height.

Pressure calculation is the process of determining the pressure exerted by a fluid within a system. In our exercise, we look closely at calculating pressures at various points within a U-shaped tube containing both water and mercury.To arrive at the pressure values for each fluid, we use the formula \( P = \rho g h \), where \( \rho \) represents fluid density, \( g \) is gravitational acceleration, and \( h \) is the height of the fluid column.
In practice, this formula helps us find the gauge pressure at the interface of the two fluids.Moreover, understanding how to balance pressures is immensely useful in fluid statics problems. The pressure exerted by the water directly influences the mercury's position in the other arm. Using the principles of pressure calculation, we ensure the pressures equalize across both ends of the tube.Finally, knowing how to compute such pressures allows us to determine the vertical distance between fluid levels, reaffirming our grasp of fluid statics principles.