Question

Figure shows a stream of water emerging from the opening of a tap. As the water falls through a height $h=PQ$, the cross-sectional area of the stream decreases from $A$ to $a$. Obtain the expression for the rate of flow of water through the opening of the tap.

• A. $a^2{\left[\frac{2gh}{A^2-a^2}\right]}^{1/2}$
• B. $A^2{\left[\frac{2gh}{A^2-a^2}\right]}^{1/2}$
• C. $aA{\left[\frac{2gh}{A^2-a^2}\right]}^{1/2}$
• D. $A^2{\left[\frac{gh}{A^2-a^2}\right]}^{1/2}$

Answer 1

The correct option is C $aA{\left[\frac{2gh}{A^2-a^2}\right]}^{1/2}$

Let $V$ be the velocity of water at $P$ and $v$ at $Q$.
Applying the equation of continuity for $P$ and $Q$,
$\Rightarrow AV=av...\left(i\right)$
Applying Bernoulli's equation for points $P$ and $Q$,
$P_P+\rho g{h}_P+\frac{1}{2}\rho V^2=P_Q+\frac{1}{2}\rho g{h}_Q+\frac{1}{2}\rho v^2$
Since the flow stream through the pipe is open to atmosphere,
$P_P=P_Q=P_0$ (atmospheric pressure)
$\Rightarrow P_0+\rho g(h_P-h_Q)+\frac{1}{2}\rho V^2=P_0+\frac{1}{2}\rho v^2$
$\Rightarrow V^2=v^2-2gh...\left(ii\right)$
$[h=PQ=h_P-h_Q]$
From Eq.$\left(i\right)$, $v=\frac{AV}{a}$ and substituting this in Eq.$\left(ii\right)$, we get
$V={\left[\frac{2gh{a}^2}{A^2-a^2}\right]}^{1/2}$
Hence, rate of flow,
$Q=AV$
$=aA{\left[\frac{2gh}{A^2-a^2}\right]}^{1/2}$