Water displacement in volume

Question

I am having a hard time understanding the following problem

There exists water in a tank whose
base is 2 feet by 6 feet. If a rectangular solid whose
dimensions are 1 foot by 1 foot by 2 feet is totally
immersed in the water, how many inches will the
water rise?

Can somebody explain this to me nicely?

Answer 1

Well, The volume of the rectangular solid is : $$ V_{solid} = 1 \times 1 \times 2 = 2 \text{ cubic feet} $$ The volume of the tank is : $$ V_{tank} = 6 \times 2 \times h = 12h \text{ cubic feet} $$ Where h is the height of the tank.

Now you know that the new volume of the tank will be : $$ V_{tank_{after}} = V_{tank} + V_{solid}$$ Let's call $\delta$ the difference between the height $h$ before adding the volume and the height after adding it :

$$12(\delta+h) = 12h + 2$$ $$\iff 12\delta = 2$$

Answer 2

There is a water tank, which implicitly is assumed to be formed like a cuboid (all sides are rectangles), and one of its sides (called the base) is assumed to be horizontal (so in equilibrium the water surface is always parallel to that surface, that is, also the water-filled volume has the shape of a cuboid). This water tank is filled to an unspecified height with water.

Now a rectangular solid is immersed in the water. This solid displaces the same volume of water, therefore the water level rises, so that the base area times the height of the water equals the sum of the volumes of the water and the immersed solid.

What you are asked is how much the water rises, that is, the difference between the height of the water before and after immersing the solid.