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Chapter 15: Problem 15

The surface of contact of two fluids of different densities that are at rest and do not mix is horizontal. Prove this general result ( \(a\) ) from the fact that the potential energy of a system must be a minimum in stable equilibrium, \((b)\) from the fact that at any two points in a horizontal plane in either fluid the pressures are equal.

Short Answer

Expert verified
The surface of contact of two fluids of different densities at rest is horizontal because this orientation minimizes the system's potential energy and ensures that the pressure at any two points in a horizontal plane in either fluid is equal.

Step by step solution

01

Analyzing component (a) - Potential Energy

Consider a small fluid element whose potential energy is given by \(_PE = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height of the fluid element. The potential energy of the entire fluid system is the integral of the potential energy of all such fluid elements. In equilibrium, the potential energy of the system must be at a minimum. Thus, the height \(h\) of each element should be as small as possible, implying a flat, horizontal surface between two fluids.
02

Analyzing component (b) - Pressure Balance

Consider two points in the same horizontal plane in either fluid. Since both points are at the same height, according to Pascal's law, the pressure at these points should be equal. Thus, the pressure throughout a horizontal plane is the same. If the interface between the fluids was not horizontal, this would lead to pressure inconsistencies, contradicting Pascal's law. Therefore, the interface between two fluids at rest must be horizontal.
03

Conclusion

From the principles of Minimum Potential Energy and Pascal's law, it was demonstrated that the surface of contact of two immiscible fluids that are at rest must be horizontal. This is because the system's potential energy is minimized and the pressure in a horizontal plane in each fluid must be equal.

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