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 = 2m/ s . Then the ratio of areas A1 & A2 is

Using equation of continuity, we have v2=A1A2v1From Bernoulli’s theorem,P1+ρgh1+12ρv12=P2+ρgh2+12ρv22⇒gh1−h2=12v22−v12⇒60=A12A22−1v12⇒A1A2=41

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 = 2m/ s . Then the ratio of areas A1 & A2 is

Using equation of continuity, we have v2=A1A2v1From Bernoulli’s theorem,P1+ρgh1+12ρv12=P2+ρgh2+12ρv22⇒gh1−h2=12v22−v12⇒60=A12A22−1v12⇒A1A2=41

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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 = 2m/ s . Then the ratio of areas A₁ & 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,
$P_{1} + {ρgh}_{1} + \frac{1}{2} {ρv}_{1}^{2} = P_{2} + {ρgh}_{2} + \frac{1}{2} {ρv}^{2}_{2}$

$\Rightarrow g (h_{1} - h_{2}) = \frac{1}{2} (v_{2}^{2} - v_{1}^{2})$

$\Rightarrow 60 = (\frac{A_{1}^{2}}{A_{2}^{2}} - 1) v_{1}^{2}$

$\Rightarrow \frac{A_{1}}{A_{2}} = \frac{4}{1}$

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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 = 2m/ s . Then the ratio of areas A1 & A2 is

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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 = 2m/ s . Then the ratio of areas A₁ & A₂ is