Question

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 $A_{1} \cdot A_{2}$ is

• A. $2: 1$
• B. $4: 1$
• C. $8: 1$
• D. $4: 3$

Answer 1

Answer: (B)

Using equation of continuity, we have $v_{2}=\frac{A_{1}}{A_{2}} v_{1}$
From Bernoulli's theorem, $p_{1}+\rho g h_{1}+\frac{1}{2} \rho v_{1}^{2}$
$p_{2}+\rho g h_{2}+\frac{1}{2} \rho v_{2}^{2}$
$\Rightarrow g\left(h_{1}-h_{2}\right)=\frac{1}{2}\left(v_{2}^{2}-v_{1}^{2}\right)$
$\Rightarrow 60=\left(\frac{A_{1}^{2}}{A_{2}^{2}}-1\right) v_{1}^{2}$
$\Rightarrow \frac{A_{1}}{A_{2}}=\frac{4}{1}$