Water flows through a horizontal tube as shown in the figure. The difference in height between the water colunms in vertical tubes is 5 cm and the area of cross-sections at A and B are $$6cm^{2}$$ and $$3cm^{2}$$ respectively. The rate of flow will be ____ $$cm^{3/s}$$. $$(take g=10m/s^{2})$$

Using the continuity equation ($$A_1 v_1 = A_2 v_2$$),

$$6 \cdot v_1 = 3 \cdot v_2 \implies v_2 = 2v_1$$

$$P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$$ (Bernoulli's equation for horizontal flow)

$$\rho gh = \frac{1}{2} \rho ( (2v_1)^2 - v_1^2 )$$

$$gh = \frac{1}{2} (3v_1^2)$$

$$1000 \times 5 = \frac{3}{2} v_1^2$$

$$v_1 = \sqrt{\frac{10000}{3}} = \frac{100}{\sqrt{3}} \text{ cm/s}$$

The rate of flow ($$Q$$) is given by $$A_1 \times v_1$$, 

$$Q = 6 \times \frac{100}{\sqrt{3}} = \frac{600}{\sqrt{3}}$$

$$Q = \frac{600\sqrt{3}}{3} = 200\sqrt{3} \text{ cm}^3/\text{s}$$