A vessel containing water has a constant acceleration of $19.6{\text{m/s}}^2$ along horizontal direction. The free surface of water will be inclined with the horizontal at an angle of : [Take $g=9.8{\text{m/s}}^2$]

The correct option is D ${\sin }^{-1}\left[\frac{2}{\sqrt{5}}\right]$
The free surface of the liquid will get tilted and become perpendicular to the net force (of pseudo force $\&$ weight) on the fluid mass.


For horizontally accelerated fluid,
$\text{tan}\theta =\frac{ma}{mg}=\frac{a}{g}...\left(i\right)$
$\Rightarrow \tan \theta =\frac{19.6}{9.8}=2$
$\therefore \tan \theta =2$


$\Rightarrow \sin \theta =\frac{2}{\sqrt{5}}$
$\therefore \theta ={\sin }^{-1}\left[\frac{2}{\sqrt{5}}\right]$