Question

An ideal fluid is flowing in a pipe in a streamline flow. The pipe has a maximum and minimum diameter of $6.4\text{cm}$ and $4.8\text{cm}$ respectively. Find out the ratio of minimum to maximum velocity.

• A. $\frac{81}{256}$
• B. $\frac{9}{16}$
• C. $\frac{3}{4}$
• D. $\frac{3}{16}$

Answer 1

The correct option is B $\frac{9}{16}$

Given:
Maximum diameter of cylindrical pipe $\left(d_1\right)=6.4\text{cm}$
Minimum diameter of cylindrical pipe
$\left(d_2\right)=4.8\text{cm}$
From equation of continuity, we can say that minimum velocity $v_{min}$ is where Area $A_1$ is maximum and maximum velocity $v_{max}$ is where area $A_1$ is minimum
So, $A_1{v}_{min}=A_2{v}_{max}$
$\therefore \frac{v_{min}}{v_{max}}=\frac{A_2}{A_1}=\frac{\pi {d_2}^2}{\pi {d_1}^2}={\left(\frac{4.8}{6.4}\right)}^2=\frac{9}{16}$