What's the acceleration?

First, we realise that the combined acceleration (g + rocket) must be perpendicular to a drawn line between the surface of each side of the water. Now, we also realise that the slope of this line compared to g must in fact represent the exact acceleration necessary applied from the rocket to maintain mentioned position of the water.

We calculate the slope as follows: $\frac{(10\ cm-20\ cm)}{30\ cm} =-\frac{1}{3}$ (since the accelerometer is set at a 45 degree angle, this one should be fairly obvious)

Meaning that in order to change the angle of force to match our new setup, we have to apply a horizontal force that is $\frac{1}{3}$ of the vertical one (g).

Consider this beautiful illustration: (Remember, the accelerations are perpendicular to what the triangle displays, hence g is horizontally drawn) Conclusively, the answer is $\frac{g}{3} = 3.33... \frac{m}{s^2}$

Sorry if this was confusing, done late hours.

As per the illustration above, this problem can be solved geometrically:

1. The water is in equilibrium, therefore the force on it must be perpendicular to the water level, and it is shown as r .
2. g is the force of gravity, of length 10 per definition
3. Therefore a is the force due to the acceleration of the rocket
4. Because the slant of vector r is 3 (goes up three units for each unit going right), a is 10 / 3 = 3.3333.

Denote the mass of the fluid as $3 m$ . Consider the force and acceleration along the tube since the fluid cannot move perpendicular to the tube. Then the net force along the tube is $m g \cos 45^\circ$ . The sum of the force along the tube equals the mass times the acceleration along the tube:

$m g \sin(45^\circ)=3 m a \cos(45^\circ)$

Therefore $a = \frac{g}{3}$

Easiest to do this graphically with vectors. Water always seeks its lowest point; therefore the magnitude of acceleration must be such that the water "thinks" that the 20cm/10cm distribution is level. The rear water level is 10 cm higher than the front one, and 30 cm to the rear. A line connecting the two water levels "feels" to the water like a level surface. The acceleration vector must be perpendicular to this line, so it must extend downward and to the rear with the same ratio of 3:1. The downward component of the acceleration is g (10 m/s), so the (fictitious) rear-ward acceleration experienced by water must be 10/3.

Resolve the two acceleration vectors (g and a) into components down the tubes and perpendicular to the tubes. The left tube has a "down the tube" acceleration of (g - a)/sqrt(2) and the right hand side has a "down the tube" acceleration of (g + a)/sqrt(2).

Pressure at bottom from left tube = Pressure at bottom of right tube

=> (density) x (tube length L) x (area) x "down tube acceleration L" / (area) = (density) x (tube length R) x (area) x "down tube acceleration R" / (area)

=> (density) x (20/sqrt(2)) x A x (10 - a)/(sqrt(2)) x 1/A = = (density) x (10/sqrt(2)) x A x (10 + a)/(sqrt(2) ) x 1/A

=> (20) x (10 - a) = (10) x (10 + a), thus a = 10/3

I'm not sure if this solution is correct but I got the right answer:

Assuming no force other than gravity was acting on the water, the height of each side would have to be equal. In order to keep the same volume while doing this, both sides would end up being 15 cm in height. The force x brings it down by 5, which is 1/3 of 15. 1/3 of the gravitational constant, 10, is 3.3333 m/s^2. Therefore, x has to be 3.3333 m/s^2

Reference from the topic Rigid body motion under fluid mechanics of Mechanical engineering: When a fluid mass in a container is subjected to a motion such that there is no relative motion between the particles, such a motion is known as rigid body motion. The motion can be either translation or rotation at constant acceleration or a combination of both. As in the given question it has mentioned only about translation so I am using the concept of fluid dynamics, the fluid is under pure translation with a uniform acceleration the piezometric head will have a gradient in the direction of motion. Motion is in x direction with a constant acceleration a(x), then:

-(dh/dx) = tan θ = {a(x)/g}

dh = differential height or difference in height

dx = distance covered or difference between two points in a x-direction

θ = angle

a(x) = acceleration in x-direction

g = gravity