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 $v_1=2m/s$. Then the ratio of areas $A_1:A_2$ is:

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

Answer 1

The correct option is B $4:1$

Using equation of continuity, we have $v_2=\frac{A_1}{A_2}{v}_1$

From Bernoulli's theorem,

$p_1=\rho g{h}_1+\frac{1}{2}\rho v_1^2$

$p_1+\rho g{h}_2+\frac{1}{2}\rho v_2^2\Rightarrow g(h_1-h_2)=\frac{1}{2}(v_2^2-v_1^2)$

$\Rightarrow 60=(\frac{A_1^2}{A_2^2}-1)v_1^2\Rightarrow \frac{A_1}{A_2}=\frac{4}{1}$