Question

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are $6.4\text{cm}$ and $4.8\text{cm}$, respectively. The ratio of the minimum and maximum velocities of fluid in this pipe is:

• A. $\frac{9}{16}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{3}{4}$
• D. $\frac{81}{256}$

Answer 1

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

From the equation of the continuity,

$A_1{v}_1=A_2{v}_2$

Here, $v_1$ and $v_2$ are the velocities at the areas of cross-section $A_1$ and $A_2$ of the pipe.

As, $v\propto \frac{1}{A}$

$v$ will be minimum at $A_{max}$ and vice versa.

$\therefore v_{min}{A}_{max}=v_{max}{A}_{min}$

$\Rightarrow \frac{v_{min}}{v_{max}}=\frac{A_{min}}{A_{max}}$

$\Rightarrow \frac{v_{min}}{v_{max}}=\frac{\pi {\left(\frac{4.8}{2}\right)}^2}{\pi {\left(\frac{6.4}{2}\right)}^2}=\frac{9}{16}$

Hence, option $\left(\text{A}\right)$ is correct.