An ideal fluid flows in the pipe as shown in the figure. The pressure in the fluid at the bottom `P_(2)` is the same as it is at the top `P_(1)`. If the velocity of the top `v_(1) = 2 m//s`. Then the ratio of areas `A_(1).A_(2)`, is

Options

A. `2:1`
B. `4:1`
C. `8:1`
D. `4:3`

Answer

B

Explanation

Using equation of continuity we have `v_(2)=(A_(1))/(A_(2)v_(1)`
From Bernoulli's theorem
`p_(1) +rhogh_(1)+1/2rhoh_(1)^(2)`
`p_(2)+rhogh_(2)+1/2rhov_(2)^(2)=g(h_(1)-h_(2))=1/2(v_(2)^(2)-v_(1)^(2))`
`implies 60=((A_(1)^(2))/(A_(2)^(2))-1)v_(1)^(2)implies(A_(1))/(A_(2))=4/1`