An open cubical container of side $H$ is filled with a liquid up to height $\frac{H}{4}.$ Mass of liquid inside container is 'm' (neglect atmospheric pressure). When the container is moved with maximum possible acceleration without spilling water, what will be the magnitude of force applied by liquid on left wall parallel to plane of paper?

The correct option is B $2mg$
Volume of liquid inside container is $\frac{H^3}{4}$. When container is moved with maximum possible acceleration:
So,
volume of this triangular portion = $\frac{H^3}{4}$
$\frac{1}{2}H.x\left(H\right)=\frac{H^3}{4}$
$\Rightarrow x=\frac{H}{2}$

We know that, angle $\theta$ is given by:
$\tan \theta =\frac{a}{g}=\frac{H}{x}=2\Rightarrow a=2g$

Force due to left wall is:
$F=Pressure\times Area$
Now this pressure is varying from top to bottom, so we will take its average value from $0$ to $\rho gH$.

$F=\frac{(0+\rho gH)}{2}.H^2=\frac{\rho g{H}^3}{2}[m=\rho (\frac{H^3}{4}\left)\right]F=2g\left(\rho \frac{H^3}{4}\right)=2mg$