When at rest, a liquid stands at the same level in the tubes as shown in the figure. But as indicated, a height difference $h$ occurs when the system is given an acceleration $a$ towards the right. Then $h$ is equal to

\frac{a L}{2 g}

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\frac{g L}{2 a}

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\frac{g L}{a}

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\frac{a L}{g}

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Solution

The correct option is D $\frac{a L}{g}$
$h ρ g = L . ρ a$
$a = \frac{h g}{L}$
Or $\Rightarrow h = \frac{a L}{g}$

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