
Water is flowing through a horizontal pipe of the non-uniform cross-section. At the extreme narrow portion of the pipe, the water will have
A. Maximum speed and least pressure
B. Maximum pressure and least speed
C. Both pressure and speed maximum
D. Both pressure and speed least
Answer
162.3k+ views
Hint: Before going to solve this question let us understand Bernoulli's principle. Bernoulli's principle states that when an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.
Complete step by step solution:
At the extremely narrow portion of the pipe, the speed of flow of water and the pressure will be higher than in other parts of the pipe. The reason behind this continuity equation is the inflow rate is equal to the outflow rate thus the product of AV will remain constant throughout the pipe.
If we consider the water as an ideal fluid then, for the flow of water to be possible, the product of area and volume should be constant, that is,
\[AV = \text{constant}\]
here, if the area increases the velocity decreases.
We know that, from Bernoulli’s equation, we have,
\[P + \rho gh + \dfrac{1}{2}\rho {v^2} = \text{constant}\]
This equation shows that, when the velocity increases, the pressure decreases. Therefore, At the extremely narrow portion of the pipe, the water will have maximum pressure and least speed.
Hence, option B is the correct answer.
Note: In this problem it is important to remember the equation of continuity which tells us about the flow rate of fluid (water) in a pipe and Bernoulli’s equation.
Complete step by step solution:
At the extremely narrow portion of the pipe, the speed of flow of water and the pressure will be higher than in other parts of the pipe. The reason behind this continuity equation is the inflow rate is equal to the outflow rate thus the product of AV will remain constant throughout the pipe.
If we consider the water as an ideal fluid then, for the flow of water to be possible, the product of area and volume should be constant, that is,
\[AV = \text{constant}\]
here, if the area increases the velocity decreases.
We know that, from Bernoulli’s equation, we have,
\[P + \rho gh + \dfrac{1}{2}\rho {v^2} = \text{constant}\]
This equation shows that, when the velocity increases, the pressure decreases. Therefore, At the extremely narrow portion of the pipe, the water will have maximum pressure and least speed.
Hence, option B is the correct answer.
Note: In this problem it is important to remember the equation of continuity which tells us about the flow rate of fluid (water) in a pipe and Bernoulli’s equation.
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