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 Bernoulli’s and Equation of continuity
` rArr A _ 1 v_ 1 = A _ 2 v_2 rArr v_2 = ((A_1v_1)/(A_2)) `
` P_1 + rho gh _1 + (1)/(2) rho v_1 ^2 + P_2 + rho gh _2 + (1)/(2) rho v _ 2 ^2 `
` g (h_1 - h_2 ) = (1)/(2) (v_2^2 - v_ 1 ^ 2) , 60 = ((A_1^2 ) /(A_2^2)- 1 ) v_1 ^2 rArr (A_1 )/(A_2) = (4)/(1) `