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 is $v_1=2\text{m/s}$. Then the ratio of areas $\frac{A_1}{A_2}$ is

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

Answer 1

The correct option is B $4:1$

Given,
$P_1=P_2$
$v_1=2m/s$
By continuity equation,
mass flow rate is same
$A_1{v}_1=A_2{v}_2$ ($\because {\rho }_1={\rho }_2$)
$⟹\frac{A_1}{A_2}=\frac{v_2}{v_1}$----- (1)
Now by using bernoulli's equation,
$PV+\frac{1}{2}m{v}^2+mgh=cont.$
For per unit mass, equation becomes
$\frac{PV}{m}+\frac{1}{2}\frac{m{v}^2}{m}+\frac{mgh}{m}=cont.$
Now applying equation at inlet and outlet,
$\frac{P_1}{\rho }+\frac{v_1^2}{2}+g{h}_1=\frac{P_2}{\rho }+\frac{v_2^2}{2}+g{h}_2$
as, $P_1=P_2$ and $h_1-h_2=3$

$⟹\frac{2^2}{2}+10\times 3=\frac{v_2^2}{2}$
$⟹4+60=v_2^2$
$v_2=8m/s$
put $v_1$ and $v_2$ in equation (1) we get,
$\frac{A_1}{A_2}=\frac{8}{2}=\frac{4}{1}$