Question

Water flows through a horizontal tube as shown in figure. If the difference in heights of the water columns in the vertical tubes is $4\text{cm}$, and the cross-sectional area $A_1$ and $A_2$ are $6{\text{cm}}^2$ and $3{\text{cm}}^2$ respectively, find the rate of flow of water across any section.

• A. $306.72{\text{cm}}^3/\text{s}$
• B. $204.2{\text{cm}}^3/\text{s}$
• C. $102.48{\text{cm}}^3/\text{s}$
• D. $372.6{\text{cm}}^3/\text{s}$

Answer 1

The correct option is A $306.72{\text{cm}}^3/\text{s}$

Rate of flow of water is the discharge $Q$ given as,
$Q=Av$
From continuity equation,
$A_1{v}_1=A_2{v}_2$
$\Rightarrow Q=A_1{v}_1=A_2{v}_2=\text{constant}$
For the given venturimeter set-up, the velocity $v_1$ is
$v_1=\sqrt{\frac{2gh}{{\left(\frac{A_1}{A_2}\right)}^2-1}}$
Here, $h=4\text{cm}$ is the difference in heights of the water columns.
Flow rate through the tube is,
$Q=A_1{v}_1$
$=A_1\times \sqrt{\frac{2gh}{{\left(\frac{A_1}{A_2}\right)}^2-1}}$
Here, $g=9.8\times 100=980{\text{cm/s}}^2,$ and $\frac{A_1}{A_2}=\frac{6}{3}=2$
$\Rightarrow Q=6\times \sqrt{\frac{2\times 980\times 4}{2^2-1}}$
$\therefore Q=306.72{\text{cm}}^3/\text{s}$