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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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