An ideal fluid flows in the pipe as shown in the figure. The pressure in the fluid at the bottom P₂ is the same as it is at the top P₁. If the velocity of the top v₁ = 2 m/s. Then the ratio of areas A₁ and A₂ is

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8:1

4:3

4:1

2:1

answer is B.

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Detailed Solution

Using equation of continuity, we have $v_{2} = \frac{A_{1}}{A_{2}} v_{1}$

From Bernoulli's theorem,

$\begin{array}{l} p_{1} + {ρgh}_{1} + \frac{1}{2} {ρv}_{1}^{2} = p_{2} + {ρgh}_{2} + \frac{1}{2} {ρv}_{2}^{2} \\ \Rightarrow g (h_{1} - h_{2}) = \frac{1}{2} (v_{2}^{2} - v_{1}^{2}) \\ \Rightarrow 60 = (\frac{A_{1}^{2}}{A_{2}^{2}} - 1) v_{1}^{2} \Rightarrow \frac{A_{1}}{A_{2}} = \frac{4}{1} \end{array}$