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

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9 : 4

3 : 2

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

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

answer is A.

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Detailed Solution

According to the equation of continuity
$A_{1} v_{1} = A_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{A_{2}}{A_{1}} = \frac{π D_{2}^{2} / 4}{π D_{1}^{2} / 4} = {(\frac{D_{2}}{D_{1}})}^{2}$
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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