A steady flow of a liquid of density ρ is shown in figure. At point 1 , the area of crosssection is 2 A and the speed of flow of liquid is 2 m s - 1 . At point 2 , the area of crosssection is A . Between the points 1 and 2 , the pressure difference is 100 N m - 2 and the height difference is 10 cm . The value of ρ is (Acceleration due to gravity = 10 m s - 2 )

Step-by-step solution

According to the equation of continuity, A 1 v 1 = A 2 v 2 ⇒ v 2 = 2 2   m   s - 1 For a steady flow, applying Bernoulli's equation at point 1 and 2 , P 1 + ρ g h 1 + 1 2 ρ v 1 2 = P 2 + ρ g h 2 + 1 2 ρ v 2 2 ⇒ ρ g h 1 - h 2 + v 1 2 - v 2 2 2 = P 2 - P 1 ⇒ ρ 10 × 0 . 1 + 2 - 8 2 = - 100 ⇒ ρ = 50   kg   m - 3