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 first cube leaves a base area of 5 cm^2, so the depth will be .4 * 6/5 = .48 cm.
The second cube leaves a base area of 4 cm^2, so the depth will be .4 * 6/4 = .6 cm.
The third cube leaves a base area of 3 cm^2, so the depth will be .4 * 6/3 = .8 cm.
The fourth cube leaves a base area of 2 cm^2 up to a level of 1 cm, and a clear 6 cm^2 above that. There are .4 * 6 = 2.4 cm^3 of water. The portion below the 1-cm level is 2 cm^3, leaving .4 cm^3 to occupy the full 6-cm^2 higher base. Dividing, .4 / 6 = 1/15, so the top of the water is 1 1/15 cm above the original base. (1.066666... cm)
With 5 cubes in place, there is room for only 1 cm^2 within 1 cm of the base, leaving 1.4 cm^3 for the free-and-clear next level, so the water level extends another 1.4 / 6 = 7/30 above the 1-cm mark, making the level 1 7/30 cm. (1.233333... cm)
The 6th cube would completely fill the bottom layer. The new water level would be at 1.4 cm.
The rest would follow, 1 cm higher than before: 1.48 cm, 1.6 cm, 1.8 cm, 2 1/15 cm, 2 7/30 cm, 2.4 cm.
Posted by Charlie
on 2006-05-15 09:58:35