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Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is 15 Nm$^{-2}$. The area of cross-section at P and Q are 40 cm$^2$ and 20 cm$^2$, respectively. The rate of flow of water through the pipe, in cm$^3$s$^{-1}$, is: [Take density of water = 1000 kg m$^{-3}$]
Detailed Solution
Equation of continuity: $A_1v_1=A_2v_2 \Rightarrow 40v_1=20v_2 \Rightarrow v_2=2v_1$. Bernoulli's principle: $P_1-P_2=\frac{1}{2}\rho[(2v_1)^2-v_1^2] \Rightarrow \frac{15\times2}{10^3}=3v_1^2 \Rightarrow v_1=0.1$ m/s. Rate of flow $=A_1v_1=40\times10^{-4}\times0.1=4\times10^{-4}$ m$^3$/s $=400$ cm$^3$/s.