Water flows in streamline motion through a horizontal pipe of circular cross-section as shown. The pressure difference of water between P and Q is 15 Nm⁻². The areas of cross-section at P and Q are 40 cm² and 20 cm² respectively. The rate of flow of water through the pipe (in cm³s⁻¹) is: [density of water =1000 kg m⁻³]

Solution & Explanation

Principle: streamline flow obeys the continuity equation and Bernoulli's theorem. Continuity: A_P·v_P = A_Q·v_Q. So (40)·v_P = (20)·v_Q → v_Q = 2·v_P. The pipe is horizontal, so Bernoulli gives P_P − P_Q = (1/2)·ρ·(v_Q² − v_P²). Substitute v_Q = 2v_P: P_P − P_Q = (1/2)·ρ·(4v_P² − v_P²) = (1/2)·ρ·3v_P². Put numbers: 15 = (1/2)·(1000)·3·v_P² = 1500·v_P². So v_P² = 15/1500 = 0.01 → v_P = 0.1 m/s. Rate of flow Q = A_P·v_P = (40×10⁻⁴ m²)·(0.1 m/s) = 4×10⁻⁴ m³/s. Convert: 4×10⁻⁴ m³/s = 400 cm³/s. This matches option D (400). Trap: use A_P (the wider area) consistently with v_P; mixing P and Q values gives a wrong rate.