Given: Pressure difference $P_P - P_Q = 15 \text{ Nm}^{-2}$, Area $A_P = 40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2$, Area $A_Q = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$, Density $\rho = 1000 \text{ kg m}^{-3}$. Let $v_P$ and $v_Q$ be velocities at $P$ and $Q$.
By continuity equation, $A_P v_P = A_Q v_Q$. Thus, $40 v_P = 20 v_Q$, which implies $v_Q = 2 v_P$.
For a horizontal pipe, $P_P + \frac{1}{2} \rho v_P^2 = P_Q + \frac{1}{2} \rho v_Q^2$. Rearranging gives: $$P_P - P_Q = \frac{1}{2} \rho (v_Q^2 - v_P^2)$$ Substituting $v_Q = 2 v_P$: $$15 = \frac{1}{2} \times 1000 \times ((2v_P)^2 - v_P^2) = 500 \times 3 v_P^2 = 1500 v_P^2$$
Solving for $v_P$: $$v_P^2 = \frac{15}{1500} = 0.01 \implies v_P = 0.1 \text{ ms}^{-1}$$ Rate of flow $Q = A_P v_P = 40 \times 10^{-4} \text{ m}^2 \times 0.1 \text{ ms}^{-1} = 4 \times 10^{-4} \text{ m}^3\text{s}^{-1}$. Converting to $\text{cm}^3\text{s}^{-1}$: $$4 \times 10^{-4} \times (100)^3 = 400 \text{ cm}^3\text{s}^{-1}$$
Final Answer: 400
AI generated content. Review strictly for academic accuracy.