Water is flowing through a horizontal tube having cross-sectional areas of its two ends being A and A^ such that the ratio A / A^ is 5 . If the pressure difference of water between the two ends is 3 10^5 ~N ~m ⁻² , the velocity of water with which it enters the tube will be (neglect gravity effects)

Step-by-step solution

According to Bernoulli's theorem P₁+ 1 2 v₁^2=P₂+ 1 2 v₂^2 ( i ) From question, P₁-P₂=3 10^5, A₁ A₂ =5 According to equation of continuity aligned & A₁ v₁=A₂ v₂ & or, A₁ A₂ = v₂ v₁ =5 & v₂=5 v₁ aligned From equation (i) aligned & P₁-P₂= 1 2 (v₂^2-v₁^2 ) & or 3 10^5= 1 2 1000 (5 v₁^2-v₁^2 . & 600=6 v₁ 4 v₁ & v₁^2=25 & v₁=5 ~m / s aligned