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Cubes in a tin (Posted on 2006-05-15) Difficulty: 2 of 5
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.

See The Solution Submitted by Jer    
Rating: 2.6667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: THE REAL SOLUTION 2 | Comment 20 of 23 |
(In reply to THE REAL SOLUTION 2 by JAMES)

Initially there is 2 * 3 * .4 = 2.4 cm^3 of water, as the base is 2 * 3 = 6 cm^2 and the height of the water is .4 cm.

If 1 cm^2 of the base is overlain by the first cube and the water has risen .1666666 cm, then the base of the water volume is reduced to 5 cm^2, and its height is now, per your solution, .4 + .166666 = .566666 cm, then the volume would be 5 * .566666 = 2.833333 cm^3.  There wasn't enough water initially to fill this volume, so the water cannot have gone this high with the first cube.  Each successive cube will raise the height further than the one before, as a smaller base is being used.

Looked at another way, remember, the first cube does not fully enter the water.  Only a portion is under water and available to displace some upwards.  As more cubes are added, the rising level allows more and more of previously placed cubes to affect the height of the water.

On the other subject: the problem with placing the 6th cube is that a 1 cm^3 volume of water is trapped as soon as you start to place it, with no way of exiting to above the newly placed cube.  That's just a logistics problem, not counted in the "theoretical" solution to this problem.

Edited on May 19, 2006, 11:30 am
  Posted by Charlie on 2006-05-19 11:21:07

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