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 ^3 s ⁻¹ , is : [Take density of water =1000 kg m ⁻³ ]

Step-by-step solution

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 fl