Question

Water flows through a horizontal tube as shown in the figure. If the difference of heights of water column in the vertical tubes is $15\text{cm}$, and the areas of cross-section at $\text{A}$ and $\text{B}$ are $4{\text{cm}}^2$ and $2{\text{cm}}^2$ respectively, find the rate of flow of water across any section. $(\text{Take}g=10{\text{m/s}}^2)$

• A. $4\times {10}^{-4}{\text{m}}^3\text{/s}$
• B. $3\times {10}^{-4}{\text{m}}^3\text{/s}$
• C. $2\times {10}^{-4}{\text{m}}^3\text{/s}$
• D. None of these

Answer 1

The correct option is A $4\times {10}^{-4}{\text{m}}^3\text{/s}$

Using continuity equation

$V_A{a}_A=V_B\times a_B$

$\frac{V_B}{V_A}=\frac{a_A}{a_B}......\left(1\right)$

Using Bernoulli’s equation between $\text{A}$ and $\text{B}$

$\Rightarrow \frac{1}{2}\rho V_A^2+\rho g{z}_A+P_A=\frac{1}{2}\rho V_B^2+\rho g{z}_B+P_B$

As both $\text{A}$ and $\text{B}$ are at same level

$\therefore z_A=z_B$

$\Rightarrow \frac{1}{2}\rho V_A^2+P_A=\frac{1}{2}\rho V_B^2+P_B$

$\Rightarrow P_A-P_B=\frac{1}{2}\times \rho (V_B^2-V_A^2)$

$\Rightarrow \rho g\Delta h=\frac{1}{2}\rho \{{\left(\frac{V_B}{V_A}\right)}^2-1\}{V}_A^2$

From equation $\left(1\right)$,

$\Rightarrow \rho g\Delta h=\frac{1}{2}\rho \{{\left(\frac{a_A}{a_B}\right)}^2-1\}{V}_A^2$

$\Rightarrow 2\times 10\times (15\times {10}^{-2})=\{{\left(\frac{4\times {10}^{-4}}{2\times {10}^{-4}}\right)}^2-1\}{V}_A^2$

$\therefore V_A=1\text{m/s}$

$\text{Rate of flow}$ $=V_A{a}_A=1\times (4\times {10}^{-4})$

$=4\times {10}^{-4}{\text{m}}^3\text{/s}$

Hence, option $\left(\text{A}\right)$ is correct.

Shortcut: $Q=\frac{a_A{a}_B}{\sqrt{a_A^2-a_B^2}}\sqrt{2gh}$