Question

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters $2.5$cm and $3.75$ cm. The ratio of the velocities in the two pipes is according to the equation of continuity

Answer 1

Step 1: Given Data

Given: Diameters of pipe through a cross-section of pipes are $2.5$cm and $3.75$cm.

Let $A_1,A_2$are the areas of the cross sections and $v_1,v_2$ are velocities in two pipes.

Step 2: Formula Used

According to the equation of continuity

$$\begin{gathered} A_1{v}_1=A_2{v}_2 \\ \Rightarrow \frac{v_1}{v_2}=\frac{A_2}{A_1}------\left(1\right) \end{gathered}$$

Step 3: Calculate the Areas

Area of a circle$=\pi {\left(\frac{d}{2}\right)}^2$
$$\begin{gathered} \Rightarrow A_1=\mathrm{\pi }{\left(\frac{{\mathrm{d}}_1}{2}\right)}^2 \\ \Rightarrow {\mathrm{A}}_1=\mathrm{\pi }{\left(\frac{2.5}{2}\right)}^2 \end{gathered}$$

and $A_2=\mathrm{\pi }{\left(\frac{{\mathrm{d}}_2}{2}\right)}^2$

$\Rightarrow A_2=\mathrm{\pi }{\left(\frac{3.75}{2}\right)}^2$

Step 4: Calculate the Ratio

From equation $\left(1\right)$

$$\begin{gathered} \Rightarrow \frac{v_1}{v_2}=\frac{\mathrm{\pi }{\left({\displaystyle \frac{3.75}{2}}\right)}^2}{\mathrm{\pi }{\left({\displaystyle \frac{2.5}{2}}\right)}^2} \\ \Rightarrow \frac{v_1}{v_2}={\left(\frac{3.75}{2.5}\right)}^2 \\ \Rightarrow \frac{v_1}{v_2}={\left(\frac{3}{2}\right)}^2 \\ \Rightarrow \frac{v_1}{v_2}=\frac{9}{4} \end{gathered}$$

Hence, the ratio of the velocities in the two pipes is $\frac{9}{4}$.