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Byju's Answer

Standard XII

Physics

Accelerating Containers

In the figure...

Question

In the figure, an ideal liquid flows through the tube, which is of uniform cross section. The liquid has velocities $v_{A}$ and $v_{B}$ , and pressure $P_{A}$ and $P_{B}$ at the points A and B respectively. Then

v_{A} = v_{B}

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v_{A} > v_{B}

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P_{A} = P_{B}

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P_{B} > P_{A}

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Solution

The correct options are
A $v_{A} = v_{B}$
D $P_{B} > P_{A}$
As per the continuity equation, $A v = c o n s t a n t$ . As the cross-sectional area remains same we get $v_{A} = v_{B}$ .
And as point B is at the lower level, by Bernoulli's theorem $P_{B} + \frac{1}{2} ρ v^{2} = P_{A} + h + \frac{1}{2} ρ v^{2}$
hence $P_{B} > P_{A}$ .

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