Water flows through a horizontal tube as shown in the figure. If the difference of heights of water column in the vertical tubes is h=0.02 m, and the areas of cross section at A and B are $4 \times 10^{- 4} m^{2}$ and $2 \times 10^{- 4} m^{2},$ respectively, then the rate of flow of water across any section is

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$1.46 \times 10^{- 4} m^{3} / s$

$1.30 \times 10^{- 4} m^{3} / s$

$1.60 \times 10^{- 4} m^{3} / s$

$1.70 \times 10^{- 4} m^{3} / s$

answer is B.

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Detailed Solution

$v_{A} a_{A} = v_{B} \times a_{B} \Rightarrow v_{A} \times 4 = v_{B} \times 2 \Rightarrow v_{B} = 2 v_{A},,,,, (i)$

$Again, \frac{1}{2} {ρv}_{A}^{2} + {ρgh}_{A} + p_{A} = \frac{1}{2} {ρv}_{B}^{2} + {ρgh}_{B} + p_{B}$

$\Rightarrow \frac{1}{2} ρ v_{A}^{2} + p_{A} = \frac{1}{2} ρ v_{B}^{2} + p_{B} ({ash}_{A} = h_{B})$

$\Rightarrow p_{A} - p_{B} = \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) = \frac{1}{2} \times 1 \times (4 v_{A}^{2} - v_{A}^{2})$

$\Rightarrow 2 \times 1 \times 1000 = \frac{1}{2} \times 1 \times 3 v_{A}^{2}$

$(p_{A} - p_{B} = 2 cm$ of water column $= 2 \times 1 \times 1000 dyn / {cm}^{2})$

$∴ v_{A} = \sqrt{\frac{4000}{3}} = 36.51 cm / s$

So the rate of flow $= v_{A} a_{A}$

$= 36.51 \times 4 = 146 {cm}^{3} / s$

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