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 \(A _1\)
\(\Rightarrow A_1v_1 = A_2v_2\)
\(\Rightarrow \frac{v_1}{v_2} = \frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r^2_1} = \frac{r_2^2}{r^2_1} = \frac{d^2_2}{d^2_1}\)
\(\Rightarrow \frac{v_1}{v_2}=\frac{(3.75)^2}{(2.5)^2} \approx \frac{3^2}{2^2} = \frac{9}{4}\)
Hence, option (1) is the correct answer.
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