Determining the fractional volume above the surface of a liquid is important in understanding buoyancy and flotation. The fractional volume (\(V_f\)) can be calculated using the formula:
\[ V_f = \dfrac{\rho_l - \rho_o}{\rho_l} \cdot \dfrac{m}{\rho_o} \]
Where:
- \(V_f\) is the fractional volume above the surface
- \(\rho_l\) is the density of the liquid (in kilograms per cubic meter, kg/m³)
- \(\rho_o\) is the density of the object (in kilograms per cubic meter, kg/m³)
- \(m\) is the mass of the object (in kilograms, kg)
Example 1: Fractional Volume of a Wooden Block in Water
Problem: A wooden block with a mass of 2 kg and a density of 600 kg/m³ is floating in water with a density of 1000 kg/m³. What is the fractional volume of the block above the surface of the water?
Calculation:
Given:
- \(\rho_l = 1000 \, \text{kg/m}^3\)
- \(\rho_o = 600 \, \text{kg/m}^3\)
- \(m = 2 \, \text{kg}\)
Using the formula:
\[ V_f = \dfrac{\rho_l - \rho_o}{\rho_l} \cdot \dfrac{m}{\rho_o} = \dfrac{1000 - 600}{1000} \cdot \dfrac{2}{600} = \dfrac{400}{1000} \cdot \dfrac{2}{600} = 0.4 \cdot \dfrac{2}{600} = 0.4 \cdot 0.0033 = 0.00132 \]
Answer: The fractional volume of the wooden block above the surface of the water is 0.00132, or 0.132%.
Example 2: Fractional Volume of an Iceberg in Seawater
Problem: An iceberg with a mass of 500,000 kg and a density of 920 kg/m³ is floating in seawater with a density of 1025 kg/m³. What is the fractional volume of the iceberg above the surface of the seawater?
Calculation:
Given:
- \(\rho_l = 1025 \, \text{kg/m}^3\)
- \(\rho_o = 920 \, \text{kg/m}^3\)
- \(m = 500000 \, \text{kg}\)
Using the formula:
\[ V_f = \dfrac{\rho_l - \rho_o}{\rho_l} \cdot \dfrac{m}{\rho_o} = \dfrac{1025 - 920}{1025} \cdot \dfrac{500000}{920} = \dfrac{105}{1025} \cdot \dfrac{500000}{920} = 0.1024 \cdot 543.48 = 55.7 \]
Answer: The fractional volume of the iceberg above the surface of the seawater is 55.7, or 5570%.
Example 3: Fractional Volume of a Plastic Bottle in Oil
Problem: A plastic bottle with a mass of 0.05 kg and a density of 950 kg/m³ is floating in oil with a density of 800 kg/m³. What is the fractional volume of the plastic bottle above the surface of the oil?
Calculation:
Given:
- \(\rho_l = 800 \, \text{kg/m}^3\)
- \(\rho_o = 950 \, \text{kg/m}^3\)
- \(m = 0.05 \, \text{kg}\)
Using the formula:
\[ V_f = \dfrac{\rho_l - \rho_o}{\rho_l} \cdot \dfrac{m}{\rho_o} = \dfrac{800 - 950}{800} \cdot \dfrac{0.05}{950} = \dfrac{-150}{800} \cdot \dfrac{0.05}{950} = -0.1875 \cdot 0.0000526 = -0.0000098 \]
Answer: The fractional volume of the plastic bottle above the surface of the oil is -0.0000098, or -0.00098%.
Example 4: Fractional Volume of a Metal Cube in Mercury
Problem: A metal cube with a mass of 10 kg and a density of 13600 kg/m³ is floating in mercury with a density of 13560 kg/m³. What is the fractional volume of the metal cube above the surface of the mercury?
Calculation:
Given:
- \(\rho_l = 13560 \, \text{kg/m}^3\)
- \(\rho_o = 13600 \, \text{kg/m}^3\)
- \(m = 10 \, \text{kg}\)
Using the formula:
\[ V_f = \dfrac{\rho_l - \rho_o}{\rho_l} \cdot \dfrac{m}{\rho_o} = \dfrac{13560 - 13600}{13560} \cdot \dfrac{10}{13600} = \dfrac{-40}{13560} \cdot \dfrac{10}{13600} = -0.00295 \cdot 0.000735 = -0.00000217 \]
Answer: The fractional volume of the metal cube above the surface of the mercury is -0.00000217, or -0.000217%.
Example 5: Fractional Volume of a Styrofoam Block in Water
Problem: A Styrofoam block with a mass of 0.2 kg and a density of 100 kg/m³ is floating in water with a density of 1000 kg/m³. What is the fractional volume of the Styrofoam block above the surface of the water?
Calculation:
Given:
- \(\rho_l = 1000 \, \text{kg/m}^3\)
- \(\rho_o = 100 \, \text{kg/m}^3\)
- \(m = 0.2 \, \text{kg}\)
Using the formula:
\[ V_f = \dfrac{\rho_l - \rho_o}{\rho_l} \cdot \dfrac{m}{\rho_o} = \dfrac{1000 - 100}{1000} \cdot \dfrac{0.2}{100} = \dfrac{900}{1000} \cdot \dfrac{0.2}{100} = 0.9 \cdot 0.002 = 0.0018 \]
Answer: The fractional volume of the Styrofoam block above the surface of the water is 0.0018, or 0.18%.
