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A tube U-shaped has a uniform cross-section with arm lengths $l_{1}$ and $l_{2} (l_{2} > l_{1})$ . Tube has a liquid of density $ρ_{1}$ filled to a height h. Another liquid of density $ρ_{_{2}} = \frac{ρ_{1}}{2}$ is poured in arm A. Both liquid are immiscible. The length of the second liquid that should be poured A so that first overtone of A is in unison with fundamental frequency of B is.

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$\frac{3}{2} (l_{1} - 3 l_{2} - 2 h)$

$\frac{2}{3} (l_{1} + 3 l_{2} + 2 h)$

$\frac{3}{2} (l_{1} - 3 l_{2} + 2 h)$

$\frac{2}{3} (l_{1} - 3 l_{2} + 2 h)$

answer is D.

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Detailed Solution

Let y be the length of second liquid poured in A let the first liquid comes down by a level x is arm A and rites by x in arm B

$ρ_{1} g y = ρ_{2} g (2 x) x = \frac{ρ_{2}}{2 ρ_{1}}, y = \frac{y}{4} l_{A} = (l_{1} - h) - (y - \frac{y}{4}) l_{B} = l_{2} - h + \frac{y}{4} 3 \frac{V}{4 l_{A}} = \frac{V}{4 l_{B}} ∴ y = \frac{2}{3} (l_{1} - 3 l_{2} + 2 h)$

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