A U-tube of base length “ $l$ ” filled with same volume of two liquids of densities $ρ$ and 2 $ρ$ is moving with an acceleration “a” on the horizontal plane as shown in the figure. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by $\frac{xa l}{2 g}$ then $x$ will be

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Detailed Solution

Let $ρ$ be the density of liquid in right limb
$\begin{array}{l} ρ_{atm} + 2 ρgh - 2 ρ (a) (\frac{l}{2}) - ρ (a) \frac{l}{2} - ρgh = ρ_{atm} \\ ρgh = 3 ρa \frac{l}{2} \Rightarrow h = \frac{3 al}{2 g} \end{array}$

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