Question
Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at $A$ and $B$ are $6{\text{ }cm}^{2}$ and $3{\text{ }cm}^{2}$ respectively.
The rate of flow will be $\_\_\_\_$ ${cm}^{3}/s$. (take $g = 10\text{ }m/s^{2}$ )
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Solution
From continuity equation
$$A_{A}V_{A} = A_{B}V_{B} \Rightarrow 6V_{A} = 3V_{B} \Rightarrow V_{B} = 2V_{A} $$Applying Bernoulli’s equation between $A\&\text{ }B$,
$$\begin{matrix} & P_{A} + \frac{1}{2}\rho{\text{ }V}_{A}^{2} = P_{B} + \frac{1}{2}\rho{\text{ }V}_{B}^{2} \\ \Rightarrow \ & \rho\text{ }g \times 0.05 = \frac{1}{2}\rho\left\lbrack {\text{ }V}_{B}^{2} - V_{A}^{2} \right\rbrack = \frac{1}{2}\rho\left( 3{\text{ }V}_{A}^{2} \right) \\ \Rightarrow \ & V_{A} = \sqrt{\frac{2\text{ }g \times 0.05}{3}}\text{ }m/s = \frac{1}{\sqrt{3}}\text{ }m/s = \frac{100}{\sqrt{3}}\text{ }cm/s \\ \Rightarrow \ & \text{~}\text{Volume flow rate}\text{~} = A_{A}{\text{ }V}_{A} = \frac{6 \times 100}{\sqrt{3}}{\text{ }cm}^{3}/sec \\ & \ = 200\sqrt{3}{\text{ }cm}^{3}/sec \end{matrix}$$
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