Question

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 the velocity $v$ of flow through the orifice is given by:

• A. $v=\sqrt{\left(2gh\right)}$
• B. $v=\sqrt{\left(2gh\right)}\sqrt{\left(\frac{A^2}{A^2-a^2}\right)}$
• C. $v=\sqrt{\left(2gh\right)}\sqrt{\left(\frac{A}{A-a}\right)}$
• D. $v=\sqrt{\left(2gh\right)}\sqrt{\left(\frac{A^2-a^2}{A^2}\right)}$

Answer 1

The correct option is B $v=\sqrt{\left(2gh\right)}\sqrt{\left(\frac{A^2}{A^2-a^2}\right)}$

$v_1=$ velocity of surface of liquid

$v_2=$ velocity of liquid far om orifice

Acc. to continuity theorem.

$A{v}_1=a{v}_2\Rightarrow \frac{A{v}_1}{a}=v_2$

$\Rightarrow \frac{a{v}_2}{A}=v_1$

Acc. to Bernoulli's Theorem

$P_0+\rho gh+\frac{1}{2}\rho v_1^2=P_0+\rho g\left(0\right)+\frac{1}{2}\rho v_L^2$

$\rho gh+\frac{1}{2}\rho [\frac{a^2{v}_1^2}{A^2}-v_1^2]=0$

$\frac{v_1^2}{2}\left[\frac{A^2-a^2}{A^2}\right]=gh$

$v_1=\sqrt{2gh}\sqrt{\frac{A^2}{A^2-a^2}}$

Answer: $\sqrt{2gh}\sqrt{\frac{A^2}{A^2-a^2}}$