Question

The minimum horizontal acceleration (in ${\text{ms}}^{-2}$) of the container filled with water upto height of $2\text{m}$ as shown, so that the pressure at point $A$ of the container becomes atmospheric is (the tank is of sufficient height)

Answer 1

As the container accelerates, free surface of liquid tilts as shown below
Volume equality gives
$2\times 3=\frac{1}{2}\times h\times 3\Rightarrow h=4\text{m}$

$\therefore \tan \theta =\frac{4}{3}=\frac{a}{g}\Rightarrow a=\frac{4}{3}g=13.33{\text{ms}}^{-2}$