Water flows through a horizontal tube as shown in the figure. The difference in height between the w

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 $6 \mathrm{~cm}^2$ and $3 \mathrm{~cm}^2$ respectively. The rate of flow will be $\_\_\_\_$ $\mathrm{cm}^3 / \mathrm{s}$. (take $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Professor PC Thomas Classes · Answer

From the equation of continuity, $A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \frac{A_1}{A_2}v_1 = \frac{6}{3}v_1 = 2v_1$. Using Bernoulli's equation, the pressure difference is $P_1 - P_2 = \frac{1}{2}ho(v_2^2 - v_1^2) = ho g \Delta h$. Thus, $\frac{1}{2}ho(4v_1^2 - v_1^2) = ho g \Delta h \Rightarrow \frac{3}{2}v_1^2 = g\Delta h \Rightarrow v_1 = \sqrt{\frac{2g\Delta h}{3}}$. Substituting values: $v_1 = \sqrt{\frac{2 	imes 1000 	ext{ cm/s}^2 	imes 5 	ext{ cm}}{3}} = \frac{100}{\sqrt{3}} 	ext{ cm/s}$. The rate of flow is $Q = A_1 v_1 = 6 	imes \frac{100}{\sqrt{3}} = 200\sqrt{3} 	ext{ cm}^3/	ext{s}$.