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A horizontally placed hollow tube has a cross-sectional area A at the beginning of the tube that gradually tapers off to A/2 at the end. An incompressible, ideal fluid of density ρ enters the tube with a velocity V at the beginning of the tube. The difference in the pressure across the tube is
- a)2ρV2
- b)ρV2
- c)ρV2/2
- d)3/2 ρV2
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A horizontally placed hollow tube has a cross-sectional area A at the ...
According to equation of continuity velocity at the other end V = 
2V
Applying ρ Bernoulli equation
2VApplying ρ Bernoulli equation
Most Upvoted Answer
A horizontally placed hollow tube has a cross-sectional area A at the ...
Ρ flows through the tube with a constant velocity v. We want to find the rate of flow of the fluid through the tube.
The rate of flow of a fluid is given by the equation:
Q = Av
where Q is the volume flow rate, A is the cross-sectional area, and v is the velocity of the fluid.
To solve this problem, we need to find the cross-sectional area at the end of the tube.
Let's assume that the length of the tube is L.
At the beginning of the tube, the cross-sectional area is A.
At the end of the tube, the cross-sectional area is A/2.
Since the tube is horizontal and the fluid is incompressible, the volume of fluid entering the tube must be equal to the volume of fluid leaving the tube.
The volume of fluid entering the tube is given by the equation:
V1 = A * L
The volume of fluid leaving the tube is given by the equation:
V2 = (A/2) * L
Since the volume of fluid entering the tube is equal to the volume of fluid leaving the tube, we can set up the following equation:
V1 = V2
A * L = (A/2) * L
Simplifying this equation, we get:
A = A/2
Multiplying both sides by 2, we get:
2A = A
Subtracting A from both sides, we get:
A = 0
This means that the cross-sectional area at the beginning of the tube is zero, which is not physically possible. Therefore, our assumption that the cross-sectional area gradually tapers off to A/2 at the end of the tube is incorrect.
In order to find the rate of flow of the fluid through the tube, we need to know the correct relationship between the cross-sectional areas at the beginning and end of the tube. Without this information, we cannot solve the problem.
The rate of flow of a fluid is given by the equation:
Q = Av
where Q is the volume flow rate, A is the cross-sectional area, and v is the velocity of the fluid.
To solve this problem, we need to find the cross-sectional area at the end of the tube.
Let's assume that the length of the tube is L.
At the beginning of the tube, the cross-sectional area is A.
At the end of the tube, the cross-sectional area is A/2.
Since the tube is horizontal and the fluid is incompressible, the volume of fluid entering the tube must be equal to the volume of fluid leaving the tube.
The volume of fluid entering the tube is given by the equation:
V1 = A * L
The volume of fluid leaving the tube is given by the equation:
V2 = (A/2) * L
Since the volume of fluid entering the tube is equal to the volume of fluid leaving the tube, we can set up the following equation:
V1 = V2
A * L = (A/2) * L
Simplifying this equation, we get:
A = A/2
Multiplying both sides by 2, we get:
2A = A
Subtracting A from both sides, we get:
A = 0
This means that the cross-sectional area at the beginning of the tube is zero, which is not physically possible. Therefore, our assumption that the cross-sectional area gradually tapers off to A/2 at the end of the tube is incorrect.
In order to find the rate of flow of the fluid through the tube, we need to know the correct relationship between the cross-sectional areas at the beginning and end of the tube. Without this information, we cannot solve the problem.
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A horizontally placed hollow tube has a cross-sectional area A at the beginning of the tube that gradually tapers off to A/2 at the end. An incompressible, ideal fluid of density ρ enters the tube with a velocity V at the beginning of the tube. The difference in the pressure across the tube isa)2ρV2b)ρV2c)ρV2/2d)3/2 ρV2Correct answer is option 'D'. Can you explain this answer?
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A horizontally placed hollow tube has a cross-sectional area A at the beginning of the tube that gradually tapers off to A/2 at the end. An incompressible, ideal fluid of density ρ enters the tube with a velocity V at the beginning of the tube. The difference in the pressure across the tube isa)2ρV2b)ρV2c)ρV2/2d)3/2 ρV2Correct answer is option 'D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A horizontally placed hollow tube has a cross-sectional area A at the beginning of the tube that gradually tapers off to A/2 at the end. An incompressible, ideal fluid of density ρ enters the tube with a velocity V at the beginning of the tube. The difference in the pressure across the tube isa)2ρV2b)ρV2c)ρV2/2d)3/2 ρV2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontally placed hollow tube has a cross-sectional area A at the beginning of the tube that gradually tapers off to A/2 at the end. An incompressible, ideal fluid of density ρ enters the tube with a velocity V at the beginning of the tube. The difference in the pressure across the tube isa)2ρV2b)ρV2c)ρV2/2d)3/2 ρV2Correct answer is option 'D'. Can you explain this answer?.
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