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Bernoulli's Principle

An ideal flui...

Question

An ideal fluid is flowing through the given tubes which is placed on a horizontal surface. If the liquid has velocities $v_{A}$ and $v_{B}$ , and pressures $P_{A}$ and $P_{B}$ at points $A$ and $B$ respectively, then the correct relation is:
(The figure shown below is as if the system is seen from the top)

v_{A} > v_{B}, P_{A} < P_{B}

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v_{A} < v_{B}, P_{A} > P_{B}

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v_{A} = v_{B}, P_{A} = P_{B}

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v_{A} > v_{B}, P_{A} = P_{B}

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Solution

The correct option is C $v_{A} = v_{B}, P_{A} = P_{B}$
Applying continuity equation:
$v_{A} A_{A} = v_{B} A_{B}$
$A_{A} = A_{B}$ , so $v_{B} = v_{A}$
Applying Bernoulli's equation:
$P_{A} + ρ g h_{A} + \frac{1}{2} ρ v_{A}^{2} = P_{B} + ρ g h_{B} + \frac{1}{2} ρ v_{B}^{2}$
Since $v_{A} = v_{B}$ , and pipes are placed on a horizontal surface $(h_{A} = h_{B})$
$\Rightarrow P_{A} = P_{B}$

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Bernoulli's Principle

Standard XII Physics

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An ideal fluid is flowing through the given tubes which is placed on a horizontal surface. If the liquid has velocities vA and vB, and pressures PA and PB at points A and B respectively, then the correct relation is: (The figure shown below is as if the system is seen from the top)

The correct option is C vA=vB,PA=PBApplying continuity equation: vAAA=vBAB AA=AB, so vB=vA Applying Bernoulli's equation: PA+ρghA+12ρv2A=PB+ρghB+12ρv2B Since vA=vB, and pipes are placed on a horizontal surface (hA=hB) ⇒PA=PB