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Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15 are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height of
- a)50 cm
- b)25 cm
- c)10 cm
- d)Zero
Correct answer is option 'B'. Can you explain this answer?
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Two cubes of size 1.0 m sides, one of relative density 0.60 and anothe...
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Two cubes of size 1.0 m sides, one of relative density 0.60 and anothe...
Given:
- Size of cubes = 1.0 m
- Relative density of one cube = 0.60
- Relative density of another cube = 1.15
To find:
- Height of the lighter cube above water surface in equilibrium
Solution:
1. Understanding relative density:
- Relative density is the ratio of the density of a substance to the density of a reference substance at a specific temperature and pressure.
- The reference substance is usually water for solids and liquids, and air for gases.
- Relative density has no units as it is a ratio of two densities.
2. Finding the densities of the cubes:
- Density of water = 1000 kg/m³ (reference substance)
- Density of the first cube = 0.60 x 1000 = 600 kg/m³
- Density of the second cube = 1.15 x 1000 = 1150 kg/m³
3. Understanding buoyancy:
- When an object is placed in a fluid, it experiences an upward force called buoyant force.
- The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.
4. Applying Archimedes' principle:
- When the cubes are placed in water, they experience buoyant force.
- The buoyant force acting on a cube is equal to the weight of water displaced by the cube.
- The weight of water displaced by a cube is equal to the volume of the cube submerged in water multiplied by the density of water.
- The volume of the cube submerged in water is equal to the volume of the cube minus the volume of the cube above water surface.
5. Equating forces:
- In equilibrium, the weight of the cube is balanced by the buoyant force acting on the cube.
- The weight of the cube is equal to the volume of the cube multiplied by the density of the cube multiplied by the acceleration due to gravity (g).
6. Solving for the height:
- Equating the weight of the cube to the buoyant force acting on the cube, we get:
- Volume of the cube submerged in water x Density of water x g = Volume of the cube x Density of the cube x g
- Simplifying and rearranging, we get:
- Height of the cube above water surface = (Density of the cube / Density of water - 1) x Side length of the cube
- For the lighter cube, we have:
- Height = (0.60 / 1000 - 1) x 1.0 = -0.40 m
- The negative value indicates that the cube will sink completely in water.
- For the denser cube, we have:
- Height = (1.15 / 1000 - 1) x 1.0 = 0.15 m = 15 cm
- Therefore, the height of the lighter cube above water surface in equilibrium is 15 cm.
Answer:
- Height of the lighter cube above water surface in equilibrium = 15 cm.
- Size of cubes = 1.0 m
- Relative density of one cube = 0.60
- Relative density of another cube = 1.15
To find:
- Height of the lighter cube above water surface in equilibrium
Solution:
1. Understanding relative density:
- Relative density is the ratio of the density of a substance to the density of a reference substance at a specific temperature and pressure.
- The reference substance is usually water for solids and liquids, and air for gases.
- Relative density has no units as it is a ratio of two densities.
2. Finding the densities of the cubes:
- Density of water = 1000 kg/m³ (reference substance)
- Density of the first cube = 0.60 x 1000 = 600 kg/m³
- Density of the second cube = 1.15 x 1000 = 1150 kg/m³
3. Understanding buoyancy:
- When an object is placed in a fluid, it experiences an upward force called buoyant force.
- The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.
4. Applying Archimedes' principle:
- When the cubes are placed in water, they experience buoyant force.
- The buoyant force acting on a cube is equal to the weight of water displaced by the cube.
- The weight of water displaced by a cube is equal to the volume of the cube submerged in water multiplied by the density of water.
- The volume of the cube submerged in water is equal to the volume of the cube minus the volume of the cube above water surface.
5. Equating forces:
- In equilibrium, the weight of the cube is balanced by the buoyant force acting on the cube.
- The weight of the cube is equal to the volume of the cube multiplied by the density of the cube multiplied by the acceleration due to gravity (g).
6. Solving for the height:
- Equating the weight of the cube to the buoyant force acting on the cube, we get:
- Volume of the cube submerged in water x Density of water x g = Volume of the cube x Density of the cube x g
- Simplifying and rearranging, we get:
- Height of the cube above water surface = (Density of the cube / Density of water - 1) x Side length of the cube
- For the lighter cube, we have:
- Height = (0.60 / 1000 - 1) x 1.0 = -0.40 m
- The negative value indicates that the cube will sink completely in water.
- For the denser cube, we have:
- Height = (1.15 / 1000 - 1) x 1.0 = 0.15 m = 15 cm
- Therefore, the height of the lighter cube above water surface in equilibrium is 15 cm.
Answer:
- Height of the lighter cube above water surface in equilibrium = 15 cm.
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Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15 are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height ofa)50 cmb)25 cmc)10 cmd)ZeroCorrect answer is option 'B'. Can you explain this answer?
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Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15 are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height ofa)50 cmb)25 cmc)10 cmd)ZeroCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15 are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height ofa)50 cmb)25 cmc)10 cmd)ZeroCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density = 1.15 are connected by weightless wire and placed in a large tank of water. Under equilibrium the lighter cube will project above the water surface to a height ofa)50 cmb)25 cmc)10 cmd)ZeroCorrect answer is option 'B'. Can you explain this answer?.
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