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A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,?
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A block of ice, of constant thickness, floats on with 1cm of ice and s...
Analysis of Ice Block Floating in Sea Water
Ice block in sea water is a classic example of buoyancy and Archimedes' principle. Let's analyze the given scenario step by step.
Given Information
- Density of sea water = 1030 kg/m³
- Density of ice = 920 kg/m³
- Total thickness of ice and sea water above the water level = 1 cm
Calculating the Volume of Ice Block
Since the ice block is floating in equilibrium, the weight of the displaced water is equal to the weight of the ice block. Let's assume the thickness of the ice block above the water level is h cm.
- Volume of the ice block = Volume of water displaced
- Volume of ice block = Volume of sea water displaced
- Density of ice * Volume of ice = Density of sea water * Volume of sea water
Solving for the Thickness of Ice Block
Given that the total thickness of ice and sea water above the water level is 1 cm, we can express the thickness of ice block in terms of h.
- Thickness of ice block + Thickness of sea water = 1 cm
- Density of ice * (1-h) = Density of sea water * h
Substitute the Density Values
Now, substitute the density values of ice and sea water into the equation to solve for the thickness of the ice block.
- 920 * (1-h) = 1030 * h
- 920 - 920h = 1030h
- 920 = 1950h
- h = 0.472 cm
Conclusion
Therefore, the thickness of the ice block above the water level is approximately 0.472 cm. This analysis demonstrates the principles of buoyancy and density in the context of an ice block floating in sea water.
Ice block in sea water is a classic example of buoyancy and Archimedes' principle. Let's analyze the given scenario step by step.
Given Information
- Density of sea water = 1030 kg/m³
- Density of ice = 920 kg/m³
- Total thickness of ice and sea water above the water level = 1 cm
Calculating the Volume of Ice Block
Since the ice block is floating in equilibrium, the weight of the displaced water is equal to the weight of the ice block. Let's assume the thickness of the ice block above the water level is h cm.
- Volume of the ice block = Volume of water displaced
- Volume of ice block = Volume of sea water displaced
- Density of ice * Volume of ice = Density of sea water * Volume of sea water
Solving for the Thickness of Ice Block
Given that the total thickness of ice and sea water above the water level is 1 cm, we can express the thickness of ice block in terms of h.
- Thickness of ice block + Thickness of sea water = 1 cm
- Density of ice * (1-h) = Density of sea water * h
Substitute the Density Values
Now, substitute the density values of ice and sea water into the equation to solve for the thickness of the ice block.
- 920 * (1-h) = 1030 * h
- 920 - 920h = 1030h
- 920 = 1950h
- h = 0.472 cm
Conclusion
Therefore, the thickness of the ice block above the water level is approximately 0.472 cm. This analysis demonstrates the principles of buoyancy and density in the context of an ice block floating in sea water.
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A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,?
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A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,?.
A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of ice, of constant thickness, floats on with 1cm of ice and sea water, appearing above the water level. If the densities sea water are the total thickness of the 920kgm³, 1030kgm³ respectively, ice block is,?.
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