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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)430 x 10³ Pa
- b)4.3 x 10³ Pa
- c)43 x 10³ Pa
- d)0.43 x 10³ Pa
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...
To understand the pressure difference between two portions of a horizontal tube with different velocities of water flow, we can use Bernoulli's equation. Bernoulli's equation states that the total mechanical energy per unit volume of a fluid remains constant along a streamline. It can be expressed as:
P + 1/2 ρv² + 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 above a reference point
In this case, we can consider two points along the horizontal tube: one where the velocity is 2 m/s and the pressure is 2 m of Hg (mercury) and another where the velocity is 4 m/s and we need to find the pressure.
Let's consider the first point as point A and the second point as point B.
At point A:
P₁ = 2 m of Hg
v₁ = 2 m/s
At point B:
P₂ = ?
v₂ = 4 m/s
Now, let's use Bernoulli's equation to find the pressure at point B.
P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂
Since the tube is horizontal, the height of the fluid above a reference point (h₁ and h₂) will be the same. Also, we can assume that the density of water (ρ) remains constant.
Therefore, the equation simplifies to:
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
Substituting the given values:
2 + 1/2 * ρ * (2)² = P₂ + 1/2 * ρ * (4)²
Simplifying further:
2 + 2ρ = P₂ + 8ρ
Rearranging the equation:
P₂ = 2 + 2ρ - 8ρ
Since ρ is a positive value (density of water), the term 2ρ-8ρ will result in a negative value. Therefore, the pressure at point B will be less than the pressure at point A.
Hence, the correct answer is option C: 43 x 10 Pa.
P + 1/2 ρv² + 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 above a reference point
In this case, we can consider two points along the horizontal tube: one where the velocity is 2 m/s and the pressure is 2 m of Hg (mercury) and another where the velocity is 4 m/s and we need to find the pressure.
Let's consider the first point as point A and the second point as point B.
At point A:
P₁ = 2 m of Hg
v₁ = 2 m/s
At point B:
P₂ = ?
v₂ = 4 m/s
Now, let's use Bernoulli's equation to find the pressure at point B.
P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂
Since the tube is horizontal, the height of the fluid above a reference point (h₁ and h₂) will be the same. Also, we can assume that the density of water (ρ) remains constant.
Therefore, the equation simplifies to:
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
Substituting the given values:
2 + 1/2 * ρ * (2)² = P₂ + 1/2 * ρ * (4)²
Simplifying further:
2 + 2ρ = P₂ + 8ρ
Rearranging the equation:
P₂ = 2 + 2ρ - 8ρ
Since ρ is a positive value (density of water), the term 2ρ-8ρ will result in a negative value. Therefore, the pressure at point B will be less than the pressure at point A.
Hence, the correct answer is option C: 43 x 10 Pa.
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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)430 x 10³ Pab)4.3x 10³ Pac)43x 10³ Pad)0.43x 10³ PaCorrect answer is option 'C'. Can you explain this answer?
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