An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of the sides, 1/3rd water spilled out. Acceleration was?

Let
L = length of the container
B = breadth of the container
H = height of the container
as container is a cube.
so, L = B = H
We know that angle made by free surface with the horizontal is given by tan ? = a/g
where a = acceleration of the container in the horizontal direction
Now we can see that there are sides of the container when seen from the side. These are front side and rear side.
When the container is accelerating then the water level on the front side will come down and the water level in the rear side will go up.
The container was completely filled with water so the amount of water that will rise in the rear side will go out or will spill out. so. the water level in the rear side will remain constant because it cannot increase more.
tan ? = h/L
where h = fall in water level on the front side.
L = length of the container
h = L tan ?
Now volume of the free space above the water level after spilling =
(1/2) x h x L x B (since from the fig you can understand why i have taken half volume )
This volume is equal to amount of water spilled out.
Total volume of container - L x B x H
Ratio = (h x L x B)/(2 x L x B x H) = h/2H = 1/3
h = H - L tan ? = H( 1 - tan ?)
h/2H = (1 - tan ?)/2 = 1/3
tan ? = 1/3 = a/g
a = g/3 = answer
Thanks & Regards,
Nirmal Singh
Askiitinas faculty