Problem Statement
Find the density of a metallic body which floats at the interface of mercury of sp. gr. 13.6 and water such that 40% of its volume is sub-merged in mercury and 60% in water.
Given Data
- Volume in mercury, \( V_{Hg} = 0.40 \times V_{body} \)
- Volume in water, \( V_{water} = 0.60 \times V_{body} \)
- Specific gravity of mercury, \( S.G._{Hg} = 13.6 \)
- Density of mercury, \( \rho_{Hg} = 13600 \, \text{kg/m}^3 \)
- Density of water, \( \rho_{water} = 1000 \, \text{kg/m}^3 \)
Diagram of the Floating Body
Solution
For the body to be in equilibrium, its total weight must be balanced by the total buoyant force from both the water and the mercury.
1. Buoyant Force from Water (\(F_{B, water}\))
This is the weight of the water displaced by 60% of the body's volume.
2. Buoyant Force from Mercury (\(F_{B, Hg}\))
This is the weight of the mercury displaced by 40% of the body's volume.
3. Equilibrium Equation
The weight of the body is \( W_{body} = \rho_{body} \times g \times V_{body} \). Now, we set the weight equal to the sum of the buoyant forces.
We can cancel out \( g \times V_{body} \) from all terms.
The density of the metallic body is \( 6040 \, \text{kg/m}^3 \).
Explanation of Concepts
Archimedes' Principle at an Interface: When an object floats at the interface between two immiscible fluids (like water and mercury), it experiences a separate buoyant force from each fluid. The total buoyant force is the sum of the force from the upper fluid and the force from the lower fluid.
Equilibrium Condition: For the object to float in a stable position, the total upward buoyant force must be exactly equal to the object's total downward weight. By expressing the buoyant forces in terms of the displaced volumes and the object's weight in terms of its unknown density, we can create an equation to solve for that density.
