The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at A is 1 . 5 cm 2 , and B is 25 mm 2 , if the speed of liquid at B is 60 cm s - 1 then ( P A – P B ) is (Given P A and P B are liquid pressures at A and B points. Density ρ = 1000 kg m - 3 A and B are on the axis of tube)

Step-by-step solution

Using Bernoulli's theorem, P A + 1 2 ρ V A 2 = P B + 1 2 ρ V B 2       . . . ( i ) Using equation of continuity, A A V A = A B V B So, 1 . 5 × 10 - 4 × V A = 25 × 10 - 6 × V B V A = 25 × 10 - 6 1 . 5 × 10 - 4 × 0 . 6 = 1 10   m   s - 1 Substituting in equation (i) ⇒ P A - P B = 1000 2 0 . 6 2 - 0 . 1 2 = 1000 2 × 0 . 7 × 0 . 5 = 175   P a