A horizontal tube has different cross sections at points A and B. The areas of cross section are $a_{1}$ and $a_{2}$ respectively, and pressures at these points are $p_{1} = ρ g h_{1} and p_{2} = ρ g h_{2},$ where $ρ$ is the density of liquid flowing in the tube and $h_{1}$ and $h_{2}$ are heights of liquid columns in vertical tubes connected at A and B. If $h_{1} - h_{2} = h,$ then the flow rate of the liquid in the horizontal tube is

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$a_{1} a_{2} \sqrt{\frac{2 g h}{a_{1}^{2} - a_{2}^{2}}}$

$a_{1} a_{2} \sqrt{\frac{2 g}{h (a_{1}^{2} - a_{2}^{2})}}$

$a_{1} a_{2} \sqrt{\frac{(a_{1}^{2} - a_{2}^{2}) h}{2 (a_{1}^{2} - a_{2}^{2})}}$

$\frac{2 a_{1} a_{2} gh}{\sqrt{a_{1}^{2} - a_{2}^{2}}}$

answer is A.

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Detailed Solution

Let $a_{1}$ and $a_{2}$ be cross sectional areas. $v_{1}$ and $v_{2}$ be the velocities of liquid flow at A and B. If $p_{1}$ and $p_{2}$ are the pressures of liquid recorded by manometer

$p_{1} + \frac{1}{2} {ρv}_{1}^{2} = p_{2} + \frac{1}{2} {ρv}_{2}^{2}$

$p_{1} - p_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$

$p_{1} - p_{2} = \frac{1}{2} {ρv}_{1}^{2} (\frac{v_{2}^{2}}{v_{1}^{2}} - 1)$

By equation of continuity $a_{1} v_{1} = a_{2} v_{2}$

also, $p_{1} - p_{2} = h ρ g$

$v_{1} = \sqrt{\frac{2 gh}{\frac{a_{1}^{2}}{a_{2}^{2}} - 1}}$

Rate of flow of liquid is $a_{1} v_{1} = a_{1} a_{2} \sqrt{\frac{2 g h}{a_{1}^{2} - a_{2}^{2}}}$

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