A tube with cross sections at A and B , having areas of 4 mm ^2 and 2 mm ^2 respectively, is kept vertical with B positioned upward. The distance separating A and B is 15 16 cm . Water enters through B at a rate of 1 cm ^3 s ⁻¹ . Determine the pressure difference P_A - P_B . Note that the speed decreases as the water falls down. Take g = 10 m s ⁻² .

Step-by-step solution

Volume flow rate Q = 1 cm ^3 s ⁻¹ = 10⁻⁶ m ^3 s ⁻¹ Area at A , A_A = 4 mm ^2 = 4 10⁻⁶ m ^2 Area at B , A_B = 2 mm ^2 = 2 10⁻⁶ m ^2 Velocity at A , v_A = Q A_A = 10⁻⁶ 4 10⁻⁶ = 0.25 m/s Velocity at B , v_B = Q A_B = 10⁻⁶ 2 10⁻⁶ = 0.5 m/s Applying Bernoulli's theorem at points A and B : P_A + 1 2 v_A^2 + g z_A = P_B + 1 2 v_B^2 + g z_B P_A - P_B = 1 2 (v_B^2 - v_A^2) + g (z_B - z_A) Given z_B - z_A = h = 15 16 cm = 15 16 10⁻² m 1 2 (v_B^2 - v_A^2) = 1 2 1000 (0.5^2 - 0.25^2) = 500 (0.25 - 0.0625) = 500 0.1875 = 93.75