First slide

Fluid dynamics

Question

An ideal fluid flows through a pipe of circular cross section made of two sections with diameter 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is

Easy

A

9 :4
B

3 : 2
C

$\sqrt{3} : \sqrt{2}$
D

$\sqrt{2} : \sqrt{3}$

Solution

According to equation of continuity $A_{1} v_{1} = A_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{A_{2}}{A_{1}} = \frac{\frac{{πD}_{2}^{2}}{4}}{\frac{{πD}_{1}^{2}}{4}} = {(\frac{D_{2}}{D_{1}})}^{2}$
Here $D_{1} = 2.5 cm, D_{2} = 3.75 cm$
$∴$ $\frac{v_{1}}{v_{2}} = {(\frac{3.75}{2.5})}^{2} = {(\frac{3}{2})}^{2} = \frac{9}{4}$

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