MCQ
An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5cm and 3.75cm. The ratio of the velocities in the two pipes is:
- ✓$9:4$
- B$3:2$
- C$\sqrt{3}:\sqrt{2}$
- D$\sqrt{2}:\sqrt{3}$
✓
Answer
Correct option: A.
$9:4$
According to continuity equation (Law of Conservation of mass)
$\text{a}_1\text{v}_1=\text{a}_2\text{v}_2$
$\frac{\text{v}_1}{\text{v}_2}=\frac{\text{a}_2}{\text{a}_1}=\frac{\pi\big(\frac{\text{d}_2}{2}\big)^2}{\pi\big(\frac{\text{d}_1}{2}\big)^2}=\frac{\text{d}^2_2}{4}\cdot\frac{4}{\text{d}^2_1}=\frac{\text{d}^2_2}{\text{d}^2_1}$
$\frac{\text{v}_1}{\text{v}_2}=\frac{(3.75)^2}{(2.50)^2}=\Big[\frac{3}{2}\Big]^2$
$\frac{\text{v}_1}{\text{v}_2}=\frac{9}{4}$
or ${\text{v}_1}:{\text{v}_2}:9:4$
$\text{a}_1\text{v}_1=\text{a}_2\text{v}_2$
$\frac{\text{v}_1}{\text{v}_2}=\frac{\text{a}_2}{\text{a}_1}=\frac{\pi\big(\frac{\text{d}_2}{2}\big)^2}{\pi\big(\frac{\text{d}_1}{2}\big)^2}=\frac{\text{d}^2_2}{4}\cdot\frac{4}{\text{d}^2_1}=\frac{\text{d}^2_2}{\text{d}^2_1}$
$\frac{\text{v}_1}{\text{v}_2}=\frac{(3.75)^2}{(2.50)^2}=\Big[\frac{3}{2}\Big]^2$
$\frac{\text{v}_1}{\text{v}_2}=\frac{9}{4}$
or ${\text{v}_1}:{\text{v}_2}:9:4$
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