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A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is
- a)P
- b)3P/4
- c)P/2
- d)P/4
Correct answer is option 'D'. Can you explain this answer?
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A liquid is flowing in a horizontal uniform capillary tube under a con...
In fluid dynamics, the flow rate of a liquid through a capillary tube can be described using Poiseuille's law. According to this law, the flow rate \( Q \) is given by the formula:
Q = (π * r^4 * ΔP) / (8 * η * L)
where:
- \( Q \) = Flow rate
- \( r \) = Radius of the tube
- \( ΔP \) = Pressure difference
- \( η \) = Viscosity of the liquid
- \( L \) = Length of the tube
Now, let's analyze the problem step by step.
1. Initial Conditions
- Let the initial radius be \( r \) and length be \( L \).
- The initial flow rate \( Q_1 = (π * r^4 * P) / (8 * η * L) \).
2. New Conditions
- The radius is doubled (\( 2r \)) and the length is also doubled (\( 2L \)).
- The new flow rate \( Q_2 \) becomes:
Q_2 = (π * (2r)^4 * P') / (8 * η * (2L))
- Simplifying \( Q_2 \):
Q_2 = (π * 16r^4 * P') / (16 * η * L) = (π * r^4 * P') / (η * L)
3. Condition for Doubling the Flow Rate
To find the pressure \( P' \) that doubles the flow rate, set \( Q_2 = 2Q_1 \):
(π * r^4 * P') / (η * L) = 2 * [(π * r^4 * P) / (8 * η * L)]
Cancelling out common terms and simplifying gives:
P' = P/4
Conclusion
Thus, to double the flow rate when both the radius and length of the capillary tube are doubled, the pressure difference required is:
P' = P/4
Therefore, the correct answer is option **D: P/4**.
Q = (π * r^4 * ΔP) / (8 * η * L)
where:
- \( Q \) = Flow rate
- \( r \) = Radius of the tube
- \( ΔP \) = Pressure difference
- \( η \) = Viscosity of the liquid
- \( L \) = Length of the tube
Now, let's analyze the problem step by step.
1. Initial Conditions
- Let the initial radius be \( r \) and length be \( L \).
- The initial flow rate \( Q_1 = (π * r^4 * P) / (8 * η * L) \).
2. New Conditions
- The radius is doubled (\( 2r \)) and the length is also doubled (\( 2L \)).
- The new flow rate \( Q_2 \) becomes:
Q_2 = (π * (2r)^4 * P') / (8 * η * (2L))
- Simplifying \( Q_2 \):
Q_2 = (π * 16r^4 * P') / (16 * η * L) = (π * r^4 * P') / (η * L)
3. Condition for Doubling the Flow Rate
To find the pressure \( P' \) that doubles the flow rate, set \( Q_2 = 2Q_1 \):
(π * r^4 * P') / (η * L) = 2 * [(π * r^4 * P) / (8 * η * L)]
Cancelling out common terms and simplifying gives:
P' = P/4
Conclusion
Thus, to double the flow rate when both the radius and length of the capillary tube are doubled, the pressure difference required is:
P' = P/4
Therefore, the correct answer is option **D: P/4**.
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A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled isa)Pb)3P/4c)P/2d)P/4Correct answer is option 'D'. Can you explain this answer?
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