An open vessel containing water is given a constant acceleration a in the horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle $\mathrm{\theta }$ given by

From the figure that there are two forces operating on the liquid's free surface.

One is the force brought on by gravity, or the weight of the liquid, and the other is brought on by the water's horizontal acceleration.

Using the trigonometric formula:

$\mathrm{tan\theta }=\frac{\mathrm{perpendicular}}{\mathrm{base}}$

The perpendicular force is now shown in the figure to be caused by horizontal acceleration, whereas the bass force is caused by gravity.

Hence,

$$\begin{gathered} \mathrm{tan\theta }=\frac{\mathrm{ma}}{\mathrm{mg}} \\ \mathrm{tan\theta }=\frac{\mathrm{a}}{\mathrm{g}} \\ \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{a}}{\mathrm{g}}\right) \end{gathered}$$

Hence the correct answer is ${\tan }^{-1}\left(\frac{\mathrm{a}}{\mathrm{g}}\right).$