@sammygerbil There are a couple of issues with your approach, in my opinion: 1. The problem asks you to find the minimum non-horizontal acceleration. What you have calculated is the minimum horizontal acceleration, but anyway: 2) it doesn't give a correct result, as it is still higher than the minimum required acceleration. Finishing the calculations, you get $a=\frac{(\rho_w-\rho_u)H}{\rho_wL}$. With the data given, this gives $a=g$.

But the minimum possible value for $a$ is $g/\sqrt{2}$ for an angle of $45^\circ$. I will show you my approach shortly. While my solution is completely different from the one in the grading sheet, it does give the correct result.

@Mr.Xcoder can you post your solution because i am interested too?

Your claim the vertical component of applied acceleration $a$ has the same effect on both limbs, so we should make this component zero to minimize the magnitude of $a$ is not necessarily correct, as its vertical component aims to reduce the pressure difference

@Nobodyrecognizeable Sure, I'm writing it down right now.

@Mr.Xcoder can i repost your problem For bookmark purposes?

@Nobodyrecognizeable Done :)

So first of all, let $\phi$ be the angle $a$ forms with the horizontal. Then, he have that $\Delta P=\rho_w (g-a\sin\phi)H-\rho_o (g-a\sin\phi)H=(\rho_a-\rho_o)(g-a\sin\phi)H$.

On the other hand, the horizontal component of the acceleration also creates a pressure difference, given by: $\Delta P'=\rho_w aL\cos\phi$ (proof: $F_x=ma_x=m\rho_w\cdot SL$, where $S$ is the cross-sectional area of the tube, and $\Delta P'=F_x/S=\rho_w La_x$). From the equilibrium of the liquid columns, we deduce that $$\Delta P=\Delta P'\implies a=\dfrac{g}{\sin\phi+\frac{\rho_wL\cos\phi}{(\rho_w-\rho_o)H}}$$

@Mr.Xcoder great!

Wait, something must have gone wrong here.

@Mr.Xcoder hi how do you get the L?

@Nobodyrecognizeable L is given, 5cm.

@Mr.Xcoder you didn't write that on the question.

@Mr.Xcoder in your last calculations you didn't put the value of H .

Yes, I've noticed now

@Mr.Xcoder you are right $\phi = 45°$

@Mr.Xcoder Now to evaluate $\dfrac{da}{d\phi}=0$: Let $A\equiv \dfrac{\rho_wL}{(\rho_w-\rho_o)H}$, then $\dfrac{d}{d\phi}(\sin\phi+A\cos\phi)=0\implies A=1\cot\phi\implies \phi=45^\circ$. Plugging into the initial expression of $a$: $$\boxed{a=\dfrac{g}{\sqrt{2}}}\approx 7m/s^2$$