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An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow?
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Understanding Fluid Dynamics in the Tube
The scenario involves an ideal liquid of density \(\rho\) flowing through a horizontal tube with a central limb and two side limbs, each having equal cross-sectional area \(A\). To determine the force exerted by the liquid on the tube, we can apply Bernoulli's principle and the continuity equation.
Continuity Equation
- Since the flow is streamline and the cross-sectional areas are equal, the velocity of the liquid in each section must remain constant.
- Therefore, the velocity \(V\) in the central limb is the same in the side limbs.
Application of Bernoulli’s Principle
- According to Bernoulli’s equation, the pressure difference in the tube can be derived from the kinetic energy per unit volume of the fluid:
\[
P + \frac{1}{2} \rho V^2 = \text{constant}
\]
- In the side limbs (where the velocity is lower), the pressure will be higher due to the conservation of energy.
Calculating the Force
1. **Pressure in the Central Limb**: Let \(P_c\) be the pressure in the central limb.
2. **Pressure in the Side Limbs**: Let \(P_s\) be the pressure in the side limbs. Using Bernoulli's principle, we can express \(P_s\) as:
\[
P_s = P_c + \frac{1}{2} \rho V^2
\]
3. **Force Calculation**: The force exerted by the liquid on the tube wall can be calculated using the pressure difference across the side limbs.
\[
F = (P_s - P_c) \cdot A = \left(\frac{1}{2} \rho V^2\right) \cdot A
\]
Conclusion
- The liquid exerts a force \(F\) on the tube, which can be calculated using the difference in pressures derived from Bernoulli's principle. This force is a result of the velocity of the liquid and its density, influencing the pressure exerted on the walls of the tube.
The scenario involves an ideal liquid of density \(\rho\) flowing through a horizontal tube with a central limb and two side limbs, each having equal cross-sectional area \(A\). To determine the force exerted by the liquid on the tube, we can apply Bernoulli's principle and the continuity equation.
Continuity Equation
- Since the flow is streamline and the cross-sectional areas are equal, the velocity of the liquid in each section must remain constant.
- Therefore, the velocity \(V\) in the central limb is the same in the side limbs.
Application of Bernoulli’s Principle
- According to Bernoulli’s equation, the pressure difference in the tube can be derived from the kinetic energy per unit volume of the fluid:
\[
P + \frac{1}{2} \rho V^2 = \text{constant}
\]
- In the side limbs (where the velocity is lower), the pressure will be higher due to the conservation of energy.
Calculating the Force
1. **Pressure in the Central Limb**: Let \(P_c\) be the pressure in the central limb.
2. **Pressure in the Side Limbs**: Let \(P_s\) be the pressure in the side limbs. Using Bernoulli's principle, we can express \(P_s\) as:
\[
P_s = P_c + \frac{1}{2} \rho V^2
\]
3. **Force Calculation**: The force exerted by the liquid on the tube wall can be calculated using the pressure difference across the side limbs.
\[
F = (P_s - P_c) \cdot A = \left(\frac{1}{2} \rho V^2\right) \cdot A
\]
Conclusion
- The liquid exerts a force \(F\) on the tube, which can be calculated using the difference in pressures derived from Bernoulli's principle. This force is a result of the velocity of the liquid and its density, influencing the pressure exerted on the walls of the tube.
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An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow?
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An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow?.
An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal liquid of density rho is pushed with velocity V through the central limb of the rube shown in figure . what force does the liquid exert of the tube ? The cross-sectional areas of the three Limbs are equal to A each. Assume streamline flow?.
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