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 \(6 cm^2\) and \(3 cm^2\) respectively.
The rate of flow will be ___________ \(cm ^3 / s\). (take \(g =10 m / s ^2\) )

From continuity equation
\( A_{A}V_{A} = A_{B}V_{B} \Rightarrow 6V_{A} = 3V_{B} \Rightarrow V_{B} = 2V_{A} \)
Applying Bernoullis equation between A & B,
\( P_{A} + \frac{1}{2}\rho V_{A}^{2} = P_{B} + \frac{1}{2}\rho V_{B}^{2} \)
\( \Rightarrow \rho g \times 0.05 = \frac{1}{2}\rho [V_{B}^{2} - V_{A}^{2}] = \frac{1}{2}\rho (3V_{A}^{2}) \)
\( \Rightarrow V_{A} = \sqrt{\frac{2g \times 0.05}{3}} m/s = \frac{1}{\sqrt{3}} m/s = \frac{100}{\sqrt{3}} cm/s \)
Volume flow rate \( = A_{A}V_{A} = \frac{6 \times 100}{\sqrt{3}} cm^{3}/sec = 200\sqrt{3} cm^{3}/sec \)