A liquid flows in the tube from left to right as shown in figure. $A_{1}$ and $A_{2}$ are the area of cross - sections of the portions of the tube as shown. If speed of liquid in these portions be $V_{1} & V_{2}$ respectively . The ratio of speeds $\frac{V_{1}}{V_{2}}$ will be

\frac{A_{1}}{A_{2}}

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\frac{A_{2}}{A_{1}}

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\sqrt{\frac{A_{2}}{A_{1}}}

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\sqrt{\frac{A_{1}}{A_{2}}}

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Solution

The correct option is B $\frac{A_{2}}{A_{1}}$
From the equation of continuity,
Flow rate is constant
$V_{1} A_{1} = V_{2} A_{2}$
So, $\frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}}$

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