If $A$ denotes the area of free surface of a liquid and $h$ the depth of an orifice of area of cross-section $a,$ below the liquid surface, then find the velocity $v$ of flow through the orifice.

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Solution

Since velocity $= \sqrt{2 g h}$
and by the equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
So, $v = \frac{a}{A} \sqrt{2 g h}$

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If A denotes the area of free surface of a liquid and h the depth of an orifice of area of cross-section a, below the liquid surface, then find the velocity v of flow through the orifice.

Since velocity =√2ghand by the equation of continuity,A1v1=A2v2So, v=aA√2gh

If $A$ denotes the area of free surface of a liquid and $h$ the depth of an orifice of area of cross-section $a,$ below the liquid surface, then find the velocity $v$ of flow through the orifice.

Since velocity $= \sqrt{2 g h}$
and by the equation of continuity,
$A_{1} v_{1} = A_{2} v_{2}$
So, $v = \frac{a}{A} \sqrt{2 g h}$