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^3s^ -1 , is : [Take density of water =1000 kg m^ -3 ]
A. 400
From the equation of continuity, the rate of flow is constant: A_P v_P = A_Q v_Q Given A_P = 40 cm ^2 and A_Q = 20 cm ^2, we have: 40 v_P = 20 v_Q v_Q = 2 v_P Applying Bernoulli's equation for a horizontal pipe: P_P + 1 2 v_P^2 = P_Q + 1 2 v_Q^2 P_P - P_Q = 1 2 (v_Q^2 - v_P^2) Substitute the given values (P_P - P_Q = 15 N m ^ -2 , = 1000 kg m ^ -3 ) and v_Q = 2 v_P: 15 = 1 2 1000 ((2 v_P)^2 - v_P^2) 15 = 500 (4 v_P^2 - v_P^2) 15 = 500 3 v_P^2 15 = 1500 v_P^2 v_P^2 = 1 100 v_P = 0.1 m s ^ -1 = 10 cm s ^ -1 The rate of flow of water is: Q = A_P v_P = 40 cm ^2 10 cm s ^ -1 = 400 cm ^3 s ^ -1 Answer: 400
Related: Physics — Mechanical Properties of Fluids · All PYQ Banks