A steady flow of an ideal fluid occurs through a horizontal pipe of varying cross-section. The kinetic energy per unit volume of the fluid at section A is 4 times the kinetic energy per unit volume at section B. The ratio of the diameter of the pipe at section A to the diameter at section B ( d_A : d_B ) is:

Step-by-step solution

Kinetic energy per unit volume is given by K = 1 2 v^2 . Given that K_A = 4 K_B , we have: 1 2 v_A^2 = 4 ( 1 2 v_B^2 ) v_A^2 = 4 v_B^2 v_A = 2 v_B According to the equation of continuity, A_A v_A = A_B v_B . Since the cross-sectional area A = d^2 4 , we can write: d_A^2 v_A = d_B^2 v_B ( d_A d_B )^2 = v_B v_A = 1 2 Taking the square root on both sides, we get: d_A d_B = 1 2 Thus, the ratio d_A : d_B is 1 : 2 . Answer: 1 : 2