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(D) Let $A_P = 40 	ext{ cm}^2 = 40 	imes 10^{-4} 	ext{ m}^2$ and $A_Q = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2$.Let $v_P$ and $v_Q$ be the

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$400$

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(D) Let $A_P = 40 	ext{ cm}^2 = 40 	imes 10^{-4} 	ext{ m}^2$ and $A_Q = 20 	ext{ cm}^2 = 20 	imes 10^{-4} 	ext{ m}^2$.Let $v_P$ and $v_Q$ be the velocities at $P$ and $Q$. By the equation of continuity, $A_P v_P = A_Q v_Q = V$ (where $V$ is the rate of flow).So, $v_P = V / A_P$ and $v_Q = V / A_Q$.Using Bernoulli's equation for a horizontal pipe: $P_P + \frac{1}{2} ho v_P^2 = P_Q + \frac{1}{2} ho v_Q^2$.$P_P - P_Q = \frac{1}{2} ho (v_Q^2 - v_P^2) = 15 	ext{ Nm}^{-2}$.$15 = \frac{1}{2} 	imes 1000 	imes (V^2 / A_Q^2 - V^2 / A_P^2) = 500 	imes V^2 	imes (1 / (20 	imes 10^{-4})^2 - 1 / (40 	imes 10^{-4})^2)$.$15 = 500 	imes V^2 	imes (1 / (4 	imes 10^{-6}) - 1 / (16 	imes 10^{-6})) = 500 	imes V^2 	imes (4 - 1) / (16 	imes 10^{-6}) = 500 	imes V^2 	imes 3 / (16 	imes 10^{-6})$.$V^2 = (15 	imes 16 	imes 10^{-6}) / (1500) = 16 	imes 10^{-8} 	ext{ m}^6	ext{s}^{-2}$.$V = 4 	imes 10^{-4} 	ext{ m}^3	ext{s}^{-1} = 4 	imes 10^{-4} 	imes 10^6 	ext{ cm}^3	ext{s}^{-1} = 400 	ext{ cm}^3	ext{s}^{-1}$.

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