Question

A liquid haying area of free surface A has an orifice at a depth h with an area a, below the liquid surface, then the velocity $\nu$ of flow through the orifice is:

• A. $\nu=\sqrt{2gh}$
• B. $\nu=\sqrt{2gh}\sqrt{\frac{A^2}{A^2-a^2}}$
• C. $\nu=\sqrt{2gh}\sqrt{\frac{A}{A-a}}$
• D. $\nu=\sqrt{2gh}\sqrt{\frac{A^2-a^2}{A^2}}$

Answer 1

Answer: (A)

Applying Bernoulli's theorem, $\frac{P}{\rho} + \frac{1}{2} ( \nu')^2 + gh = \frac{P}{\rho} + \frac{1}{2} \nu^2 + 0$ where $\nu'$ is velocity of all surfaces ofliquid and V is velocity of efflux. From equation of continuity $A \nu' = a \nu$ $\Rightarrow $ $\nu' = \frac{a \nu}{A}$
$\therefore $ $\frac{1}{2} \left(\frac{a \nu}{A} \right)^2 + gh = \frac{1}{2} \nu^2 $ $\Rightarrow \, \nu = \sqrt{2gh} \sqrt{\left( \frac{A^2}{A^2 - a^2} \right) } $