A U-tube containing a liquid is subjected to a constant horizontal...

In a frame accelerating horizontally with acceleration $a_0$, the free surface of the liquid aligns itself perpendicular to the effective gravity.

The angle $\theta$ made by the free surface with the horizontal is given by $\tan \theta = \dfrac{a_0}{g}$.

From the geometry of the U-tube, if the separation between the vertical limbs is $l$, the difference in heights $h$ is related to $l$ by $\tan \theta = \dfrac{h}{l}$.

Equating the two expressions for $\tan \theta$, we get $\dfrac{h}{l} = \dfrac{a_0}{g}$.

Therefore, $h = \dfrac{a_0 l}{g}$.