First slide

Fluid dynamics

Question

A tank is filled with water to a height H. The tank has two orifices A and B at depths H/4 and H/2 respectively below the free surface of water. Area of orifices A and B are respectively ‘a’ and ‘a/2’. If $Q_{A} and Q_{B}$ represent rate of volume flow of water through A and B, then $Q_{A} / Q_{B}$ is

Easy

A

$2 \sqrt{2}$
B

$\sqrt{2}$
C

$1 / 2 \sqrt{2}$
D

$2$

Solution

$Q = S \sqrt{2 g h}$ where S is the cross-sectional area of orifice and h is the depth of orifice below the free surface of water,
$∴ \frac{Q_{A}}{Q_{B}} = \frac{S_{A} \sqrt{2 g h_{A}}}{S_{B} \sqrt{2 g h_{B}}} = \frac{a \sqrt{2 g \frac{H}{4}}}{(a / 2) \sqrt{2 g \frac{H}{2}}} = \sqrt{2}$

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