In a $U-$ tube, the liquid level stands at same level when it is at rest. When $U-$ tube is accelerated towards right, as shown in figure, the difference $h$ between level of two arms will be
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$\tan \theta=\frac{a}{g}=\frac{h}{L}$
$\therefore \mathrm{h}=\frac{\mathrm{aL}}{\mathrm{g}}$
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