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Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2 m/s is 2 m of Hg. What will be the pressure in the portion where velocity is 4 m/s?
- a)2.84×105Pa
- b)2.78×105Pa
- c)2.66×105 Pa
- d)2.60×105Pa
Correct answer is option 'C'. Can you explain this answer?
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Water is flowing through a horizontal tube. The pressure of the liquid...
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Water is flowing through a horizontal tube. The pressure of the liquid...
P2=P1+1/2d(v1²-v2²)
=2mHg+1/2×10³ ×(2²-4²)
=-6000+2Hg
2.66×1 0⁵
=2mHg+1/2×10³ ×(2²-4²)
=-6000+2Hg
2.66×1 0⁵
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Water is flowing through a horizontal tube. The pressure of the liquid...
The Bernoulli's equation states that the total pressure in a fluid system is constant. It can be written as:
P + (1/2)ρv^2 + ρgh = constant
Where:
P is the pressure of the fluid
ρ is the density of the fluid
v is the velocity of the fluid
g is the acceleration due to gravity
h is the height of the fluid column
In the given problem, the pressure in the portion where velocity is 2 m/s is 2 m of Hg. We can assume that the height of the fluid column (h) is the same in both portions. Therefore, the pressure in the portion where velocity is 4 m/s can be calculated using the Bernoulli's equation:
P1 + (1/2)ρv1^2 + ρgh = P2 + (1/2)ρv2^2 + ρgh
Since P1 = 2 m of Hg and v1 = 2 m/s, and we want to find P2 when v2 = 4 m/s, we can substitute these values into the equation:
2 + (1/2)ρ(2^2) + ρgh = P2 + (1/2)ρ(4^2) + ρgh
Simplifying the equation, we get:
2 + 2ρ + ρgh = P2 + 8ρ + ρgh
Subtracting ρgh from both sides, we get:
2 + 2ρ = P2 + 8ρ
Subtracting 2ρ from both sides, we get:
2 = P2 + 6ρ
Since ρ is the density of the fluid, it does not change within the system. Therefore, we can treat it as a constant. Rearranging the equation, we get:
P2 = 2 - 6ρ
Substituting the value of ρ, we get:
P2 = 2 - 6(mercury density)
The density of mercury is approximately 13,600 kg/m^3. Substituting this value, we get:
P2 = 2 - 6(13,600)
P2 = 2 - 81,600
P2 ≈ -81,598
Therefore, the pressure in the portion where velocity is 4 m/s is approximately -81,598 m of Hg.
P + (1/2)ρv^2 + ρgh = constant
Where:
P is the pressure of the fluid
ρ is the density of the fluid
v is the velocity of the fluid
g is the acceleration due to gravity
h is the height of the fluid column
In the given problem, the pressure in the portion where velocity is 2 m/s is 2 m of Hg. We can assume that the height of the fluid column (h) is the same in both portions. Therefore, the pressure in the portion where velocity is 4 m/s can be calculated using the Bernoulli's equation:
P1 + (1/2)ρv1^2 + ρgh = P2 + (1/2)ρv2^2 + ρgh
Since P1 = 2 m of Hg and v1 = 2 m/s, and we want to find P2 when v2 = 4 m/s, we can substitute these values into the equation:
2 + (1/2)ρ(2^2) + ρgh = P2 + (1/2)ρ(4^2) + ρgh
Simplifying the equation, we get:
2 + 2ρ + ρgh = P2 + 8ρ + ρgh
Subtracting ρgh from both sides, we get:
2 + 2ρ = P2 + 8ρ
Subtracting 2ρ from both sides, we get:
2 = P2 + 6ρ
Since ρ is the density of the fluid, it does not change within the system. Therefore, we can treat it as a constant. Rearranging the equation, we get:
P2 = 2 - 6ρ
Substituting the value of ρ, we get:
P2 = 2 - 6(mercury density)
The density of mercury is approximately 13,600 kg/m^3. Substituting this value, we get:
P2 = 2 - 6(13,600)
P2 = 2 - 81,600
P2 ≈ -81,598
Therefore, the pressure in the portion where velocity is 4 m/s is approximately -81,598 m of Hg.
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Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2 m/s is 2 m of Hg. What will be the pressure in the portion where velocity is 4 m/s?a)2.84×105Pab)2.78×105Pac)2.66×105Pad)2.60×105PaCorrect answer is option 'C'. Can you explain this answer?
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Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2 m/s is 2 m of Hg. What will be the pressure in the portion where velocity is 4 m/s?a)2.84×105Pab)2.78×105Pac)2.66×105Pad)2.60×105PaCorrect answer is option 'C'. 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 Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2 m/s is 2 m of Hg. What will be the pressure in the portion where velocity is 4 m/s?a)2.84×105Pab)2.78×105Pac)2.66×105Pad)2.60×105PaCorrect answer is option 'C'. 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 Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2 m/s is 2 m of Hg. What will be the pressure in the portion where velocity is 4 m/s?a)2.84×105Pab)2.78×105Pac)2.66×105Pad)2.60×105PaCorrect answer is option 'C'. Can you explain this answer?.
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