Question

Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is $4\text{cm}$, and the cross-sectional area $A_1$ and $A_2$ are $4{\text{cm}}^2$ and $2{\text{cm}}^2$ respectively, find the rate of flow of water across any section.
Take $g=9.8{\text{m/s}}^2,\sqrt{2613.33}=51.12$ for calculation.

• A. $122.43{\text{cm}}^3/\text{s}$
• B. $204.48{\text{cm}}^3/\text{s}$
• C. $298.6{\text{cm}}^3/\text{s}$
• D. $302.2{\text{cm}}^3/\text{s}$

Answer 1

The correct option is B $204.48{\text{cm}}^3/\text{s}$

For a venturimeter, velocity of liquid is given as
$v_1=\sqrt{\frac{2gh}{{\left(\frac{A_1}{A_2}\right)}^2-1}}$
Where, $h=4\text{cm}$ is difference in height of liquid columns in the vertical tubes
$A_1=4{\text{cm}}^2$
$A_2=2{\text{cm}}^2$
$v_1$ is the velocity of the liquid through larger cross-section.

Discharge of the liquid is given as
$Q=A_1{v}_1$
$\Rightarrow Q_1=A_1\times \sqrt{\frac{2gh}{{\left(\frac{A_1}{A_2}\right)}^2-1}}$
$\Rightarrow Q_1=4\times \sqrt{\frac{2\times 980\times 4}{2^2-1}}$
$\because g=9.8\times 100=980{\text{cm/s}}^2$
$\Rightarrow Q_1=4\times \sqrt{2613.33}$
$\therefore Q_1=4\times 51.12=204.48{\text{cm}}^3/\text{s}$
Hence, the rate of flow of water is $204.48{\text{cm}}^3/\text{s}$