Question

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

• A. $\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• B. $\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$16$}\right.$
• C. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
• D. $\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Answer 1

The correct option is B

$\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$16$}\right.$

Step1: Given data.

Maximum diameter of the pipe, $d_{max}=6.4cm$

Minimum diameter of the pipe, $d_{min}=4.8cm$

Step2: The ratio of minimum and maximum velocities of fluid in pipe.

We know that

From equation continuity.

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

where $A_1=A_2=Area$ and $V_1=V_2=velocity$

$\Rightarrow \frac{A_{min}}{A_{max}}=\frac{V_{min}}{V_{max}}$ …..$\left(i\right)$

Now,

Minimum area of the pipe, $A_{min}=\frac{\pi d_{min}^2}{4}$

Maximum area of the pipe, $A_{max}=\frac{\pi d_{max}^2}{4}$

Substitute $A_{min}$ and $A_{max}$ in equation $\left(i\right)$ we get.

$\frac{V_{min}}{V_{max}}=\frac{\frac{\pi d_{min}^2}{4}}{\frac{\pi d_{max}^2}{4}}$

$\Rightarrow$$\frac{V_{min}}{V_{max}}=\frac{4.8^2}{6.4^2}$

$\Rightarrow$ $\frac{V_{min}}{V_{max}}=\frac{4.8^2}{6.4^2}$

$\Rightarrow$$\frac{V_{min}}{V_{max}}=\frac{9}{16}$

Hence, option B is correct.