In a horizontal tube with area of cross-section A₁and A₂ as shown in fig (5) , the liquid is flowing with velocities v₁and v₂ respectively. The difference in the levels of the liquid in the two vertical tubes is h. Then

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$v_{2} - v_{1} = \sqrt{(2 gh)}$

the volume of the liquid flowing through the tube in unit time is A₁v₁

the energy per unit mass of the liquid is the same in both sections of the tube

${v_{2}}^{2} - {v_{1}}^{2} = 2 gh$

answer is B.

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Detailed Solution

$\frac{P_{1}}{ρ} + \frac{V_{1}^{2}}{2} = \frac{P_{2}}{ρ} + \frac{V_{2}^{2}}{2} ∴ P_{1} - P_{2} = ρgh = \frac{ρ}{2} (v_{2}^{2} - v_{1}^{2}) or {v_{2}}^{2} - {v_{1}}^{2} = 2 gh$

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