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26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how?
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26. A ball of relative density 0.8 fall into water from a height of 2 ...
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26. A ball of relative density 0.8 fall into water from a height of 2 ...
Solution:
Given data:
Relative density of the ball = 0.8
Height of the fall = 2 m
To find:
The depth to which the ball will sink in water.
Assumptions:
1. The ball is a perfect sphere.
2. The ball does not experience any air resistance during its fall.
3. The water is calm and there are no other forces acting on the ball.
The buoyant force experienced by an object submerged in a fluid is given by Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. Mathematically, it can be expressed as:
Buoyant force = Density of fluid x Volume of fluid displaced x g
Here, the fluid is water and the object is the ball. The relative density of the ball is the ratio of its density to the density of water. Since the relative density is less than 1, the ball will float in water.
When the ball is completely submerged in water, the buoyant force acting on the ball will be equal to its weight. The weight of the ball is given by:
Weight of the ball = Mass of the ball x g
Since the ball is falling freely, it will accelerate due to gravity, and its weight can be expressed as:
Weight of the ball = Mass of the ball x g
Using the equation of relative density, we can relate the density of the ball to the density of water:
Density of the ball = Relative density x Density of water
Since the ball is completely submerged, its volume is equal to the volume of water displaced. The volume of the ball can be calculated using its radius:
Volume of the ball = (4/3) x π x (radius)^3
Now, we can equate the weight of the ball to the buoyant force:
Mass of the ball x g = Density of water x Volume of the ball x g
Simplifying the equation:
Mass of the ball = Density of water x Volume of the ball
Substituting the values:
Mass of the ball = 1000 kg/m^3 x (4/3) x π x (radius)^3
Since the ball is falling freely, the gravitational potential energy is converted into kinetic energy. At the point of impact, all the kinetic energy is converted into potential energy. Therefore, the gravitational potential energy can be calculated as:
Gravitational potential energy = Mass of the ball x g x height of fall
Setting the gravitational potential energy equal to the buoyant force:
Mass of the ball x g x height of fall = Density of water x Volume of the ball x g
Simplifying the equation:
Height of fall = Density of water x Volume of the ball / Mass of the ball
Substituting the values:
Height of fall = 1000 kg/m^3 x (4/3) x π x (radius)^3 / (1000 kg/m^3 x (4/3) x π x (radius)^3 x 0.8)
Simplifying further:
Height of fall = 1 / 0.8
Height of fall = 1.25 m
Therefore, the ball will sink to a depth of 1.25 m in water.
Answer: The depth to which the ball will sink is 1.25 m.
Given data:
Relative density of the ball = 0.8
Height of the fall = 2 m
To find:
The depth to which the ball will sink in water.
Assumptions:
1. The ball is a perfect sphere.
2. The ball does not experience any air resistance during its fall.
3. The water is calm and there are no other forces acting on the ball.
The buoyant force experienced by an object submerged in a fluid is given by Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. Mathematically, it can be expressed as:
Buoyant force = Density of fluid x Volume of fluid displaced x g
Here, the fluid is water and the object is the ball. The relative density of the ball is the ratio of its density to the density of water. Since the relative density is less than 1, the ball will float in water.
When the ball is completely submerged in water, the buoyant force acting on the ball will be equal to its weight. The weight of the ball is given by:
Weight of the ball = Mass of the ball x g
Since the ball is falling freely, it will accelerate due to gravity, and its weight can be expressed as:
Weight of the ball = Mass of the ball x g
Using the equation of relative density, we can relate the density of the ball to the density of water:
Density of the ball = Relative density x Density of water
Since the ball is completely submerged, its volume is equal to the volume of water displaced. The volume of the ball can be calculated using its radius:
Volume of the ball = (4/3) x π x (radius)^3
Now, we can equate the weight of the ball to the buoyant force:
Mass of the ball x g = Density of water x Volume of the ball x g
Simplifying the equation:
Mass of the ball = Density of water x Volume of the ball
Substituting the values:
Mass of the ball = 1000 kg/m^3 x (4/3) x π x (radius)^3
Since the ball is falling freely, the gravitational potential energy is converted into kinetic energy. At the point of impact, all the kinetic energy is converted into potential energy. Therefore, the gravitational potential energy can be calculated as:
Gravitational potential energy = Mass of the ball x g x height of fall
Setting the gravitational potential energy equal to the buoyant force:
Mass of the ball x g x height of fall = Density of water x Volume of the ball x g
Simplifying the equation:
Height of fall = Density of water x Volume of the ball / Mass of the ball
Substituting the values:
Height of fall = 1000 kg/m^3 x (4/3) x π x (radius)^3 / (1000 kg/m^3 x (4/3) x π x (radius)^3 x 0.8)
Simplifying further:
Height of fall = 1 / 0.8
Height of fall = 1.25 m
Therefore, the ball will sink to a depth of 1.25 m in water.
Answer: The depth to which the ball will sink is 1.25 m.
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26. A ball of relative density 0.8 fall into water from a height of 2 ...
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26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how?
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26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how? 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 26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how?.
26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how? 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 26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 26. A ball of relative density 0.8 fall into water from a height of 2 The depth to which the ball will sink is a) 8 m)(a) 4 m (c) 6 m?and how?.
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