We use the Equation of Continuity ($A_1v_1 = A_2v_2$) and Bernoulli's Equation for horizontal flow ($P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$).
Given $A_P = 40 \times 10^{-4} m^2$ and $A_Q = 20 \times 10^{-4} m^2$. Let $Q$ be the flow rate. Then $v_P = \frac{Q}{A_P}$ and $v_Q = \frac{Q}{A_Q}$. Since $A_P = 2A_Q$, we have $v_Q = 2v_P$.
The pressure difference is $P_P - P_Q = \frac{1}{2}\rho(v_Q^2 - v_P^2)$. Substituting $v_Q = 2v_P$: $$15 = \frac{1}{2} \times 1000 \times ((2v_P)^2 - v_P^2)$$ $$15 = 500 \times 3v_P^2$$ $$v_P^2 = \frac{15}{1500} = 0.01 \implies v_P = 0.1 m/s$$
Flow rate $Q = A_P \times v_P = 40 \times 10^{-4} m^2 \times 0.1 m/s = 4 \times 10^{-4} m^3/s$. Converting to $cm^3/s$: $4 \times 10^{-4} \times 10^6 = 400 cm^3/s$.
Final Answer: 400
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