First slide

Fluid dynamics

Question

A cylindrical vessel contains a liquid of density $ρ$ up to a height $h$ . The liquid is closed by a piston of mass $m$ and area of cross-section $A$ . There is a small hole at the bottom of the vessel. The speed $v$ with which the liquid comes out of the hole is:

Moderate

A

$\sqrt{2 g h}$
B

$\sqrt{2 (g h + \frac{m g}{ρ A})}$
C

$\sqrt{2 (g h + \frac{m g}{A})}$
D

$\sqrt{2 g h + \frac{m g}{A}}$

Solution

Applying Bernoulli’s theorem at point 1 and 2,
Difference in pressure energy between 1 and 2 = difference in kinetic energy between 1 and 2
or $ρ h g + \frac{m g}{A} = \frac{1}{2} ρ v^{2}$
or $v = \sqrt{2 g h + \frac{2 m g}{ρ A}}$
$= \sqrt{2 (g h + \frac{m g}{ρ A})}$

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