There is a rectangular tin 2cm by 3cm at the base, 2cm deep and open at the top.
It is filled with water to a depth of .4cm.
Twelve 1cm steel cubes are placed in the tin one by one. The first six form a single layer at the bottom. The next six form a second layer.
To what height does the water level rise after the placement of each cube?
Assume the cubes fit together tightly, leaving no spaces.
The volume of water is 2*3*.4 = 2.4cc
Cube one reduces the bottom area to 5scm (square centimeters) so the depth is not 2.4/5 = .48cm
Cube two raises the depth to 2.4/4 = .6cm
Cube three raises it to 2.4/3 = .8cm
Cube four will bring the depth above the first layer. The bottom contains the first 2cc, above them will be the remaining .4cc over the full cross section. .4/6=.06667 so the full depth is 1.06667cm
Cube five leaves 1cc at the bottom. 1 + 1.4/6 = 1.23333cm
Cube six fills the bottom layer entirely so the water rises a full cm, to 1.4cm.
Cubes seven, eight and nine are just 1cm more than one, two and three. 1.48, 1.6, 1.8.
Cubes ten and eleven will overflow the tin, the water can't rise above 2cm.
Cube twelve forces out the rest of the water, so there is no water level to speak of.
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