An ideal fluid of density flows steadily through a horizontal pipe. The pipe narrows from a radius of 2r at section 1 to a radius of r at section 2. If the velocity of the fluid at section 1 is v , the pressure difference between section 1 and section 2 ( P₁ - P₂ ) is:

Step-by-step solution

According to the equation of continuity, A₁ v₁ = A₂ v₂ . Given r₁ = 2r and r₂ = r , the areas are A₁ = (2r)^2 = 4 r^2 and A₂ = r^2 . Substituting the values: (4 r^2) v = ( r^2) v₂ v₂ = 4v Applying Bernoulli's principle for a horizontal pipe ( h₁ = h₂ ): P₁ + 1 2 v₁^2 = P₂ + 1 2 v₂^2 Substituting v₁ = v and v₂ = 4v : P₁ + 1 2 v^2 = P₂ + 1 2 (4v)^2 P₁ - P₂ = 1 2 (16v^2) - 1 2 v^2 P₁ - P₂ = 15 2 v^2 Answer: 15 2 v^2