Question:
A fluid is flowing through a horizontal pipe of varying cross-section, with speed \(v\ \mathrm{ms^{-1}}\) at a point where the pressure is \(P\) pascal. At another point where pressure is \(\dfrac{P}{2}\) Pascal its speed is \(V\ \mathrm{ms^{-1}}\). If the density of the fluid is \(\rho\ \mathrm{kg\,m^{-3}}\) and the flow is streamline, then \(V\) is equal to :
A fluid is flowing through a horizontal pipe of varying cross-section, with speed \(v\ \mathrm{ms^{-1}}\) at a point where the pressure is \(P\) pascal. At another point where pressure is \(\dfrac{P}{2}\) Pascal its speed is \(V\ \mathrm{ms^{-1}}\). If the density of the fluid is \(\rho\ \mathrm{kg\,m^{-3}}\) and the flow is streamline, then \(V\) is equal to :
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Apply Bernoulli's equation between the two points of the horizontal pipe (no height change): \(P_1+\tfrac12\rho v_1^2=P_2+\tfrac12\rho v_2^2\).
Updated On: Nov 5, 2026
- \(\sqrt{\dfrac{P}{2\rho}+v^2}\)
- \(\sqrt{\dfrac{P}{\rho}+v^2}\)
- \(\sqrt{\dfrac{2P}{\rho}+v^2}\)
- \(\sqrt{\dfrac{P}{\rho}+v}\)
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The Correct Option is B
Solution and Explanation
Horizontal pipe, so no change in height. Bernoulli: \(P+\dfrac12\rho v^2=\dfrac{P}{2}+\dfrac12\rho V^2\).
\(\dfrac12\rho V^2=\dfrac{P}{2}+\dfrac12\rho v^2\Rightarrow V^2=\dfrac{P}{\rho}+v^2\).
\(V=\sqrt{\dfrac{P}{\rho}+v^2}\).
Answer: (2) \(\sqrt{\dfrac{P}{\rho}+v^2}\)
\(\dfrac12\rho V^2=\dfrac{P}{2}+\dfrac12\rho v^2\Rightarrow V^2=\dfrac{P}{\rho}+v^2\).
\(V=\sqrt{\dfrac{P}{\rho}+v^2}\).
Answer: (2) \(\sqrt{\dfrac{P}{\rho}+v^2}\)
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