Fluid specific gravity

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1

a solid blok of SG 0.9 floats such that 75% of its volume is undrewater and 25 pct of it s volume is in fluid X layered above the water. What is the specific gravity of X

2

This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. PURPOSE OF THE EXERCISE:
    This exercise tests your ability to apply Archimedes’ principle in a multi-fluid setting, where the buoyant force is distributed between two fluids with different densities.

  2. ENGINEERING CONTEXT:
    The problem illustrates how to design and predict the floatation behavior of a body positioned at a fluid interface, a technique relevant to designing devices like oil skimmers or multilayer float systems.

  3. PHYSICAL CONTEXT:
    The dominant physics is buoyancy due to the hydrostatic pressure distribution; the block experiences an upward force equal to the weight of fluid displaced in both the water and the overlying fluid X.

  4. ASSUMPTIONS:
    Assume the block is in steady hydrostatic equilibrium with a uniform density for each fluid and that the fluids are immiscible; these assumptions are standard for buoyant equilibrium problems such that the displaced volumes directly yield the corresponding buoyancy forces.

  5. DRAWINGS:
    A helpful drawing would show a vertical cross-section of the block at the interface, with 75% of its volume immersed in water (density = 1000\,\mathrm{kg/m^3}) and 25% immersed in fluid X above water.

  6. SOLUTION TECHNIQUE:
    Let V denote the total volume of the block and g the acceleration due to gravity. Because the block’s specific gravity is 0.9, its density is

    \rho_\mathrm{block} = 0.9\,\rho_\mathrm{water}\,.

    The buoyant forces are generated by volumes 0.75\,V in water and 0.25\,V in fluid X. Equating the block’s weight with the total buoyancy gives:

    \rho_\mathrm{block}\,Vg = (0.75\,V\,\rho_\mathrm{water}g) + (0.25\,V\,\rho_Xg)\,.

    Canceling common factors (V, g) leads to:

    0.9\,\rho_\mathrm{water} = 0.75\,\rho_\mathrm{water} + 0.25\,\rho_X\,.

    Rearranging,

    0.25\,\rho_X = 0.9\,\rho_\mathrm{water} - 0.75\,\rho_\mathrm{water} = 0.15\,\rho_\mathrm{water}\,,

    so that

    \rho_X = \frac{0.15}{0.25}\,\rho_\mathrm{water} = 0.6\,\rho_\mathrm{water}\,.

    Therefore, the specific gravity of fluid X is

    \mathrm{SG}_X = \frac{\rho_X}{\rho_\mathrm{water}} = 0.6\,.

  7. REFLECTION:
    The analysis demonstrates that by superposing the buoyancy contributions from two different fluids and setting them equal to the object’s weight, one can determine a fluid’s specific gravity given the immersed volume fractions. This approach underpins design methods in applications requiring control of floatation at fluid interfaces.

In formulating this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 4.3: Archimedes’ Law Part 1 (HIGHLY RELEVANT)
  2. Chapter 4.3: Archimedes’ Law Part 2 (HIGHLY RELEVANT)
  3. Chapter 4.1: Watercraft Part 1 (HIGHLY RELEVANT)
  4. Chapter 4.1: Watercraft Part 2 (HIGHLY RELEVANT)
  5. Chapter 4.1: Watercraft Part 3 (RELEVANT)

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