Ncert Exercises & Solutions

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:
1. $9:4$
2. $3:2$
3. $\sqrt{3}:\sqrt{2}$
4. $\sqrt{2}:\sqrt{3}$

Hint: Use the equation of continuity.

Step: Find the ratio of velocities at the two cross-sections.
Given: The diameter ( $𝑑 _1$ ) at the first cross-section is $2.5~\text{cm}$ , the diameter ( $𝑑 _2$ ) at the second cross-section is $3.75 ~\text{cm}$ The flow of the liquid is shown in the figure below;

As given,
Applying the equation of continuity for cross-sections $A _1$ and $A_2,$
$\Rightarrow A_1v_1 = A_2v_2$
$\Rightarrow \frac{v_1}{v_2} = \frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r^2_1} = \frac{r_2^2}{r^2_1} = \frac{d^2_2}{d^2_1}$
$\Rightarrow \frac{v_1}{v_2}=\frac{(3.75)^2}{(2.5)^2} \approx \frac{3^2}{2^2} = \frac{9}{4}$
Hence, option (1) is the correct answer.

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