You have rectangular pot of water 10cm by 10cm at the bottom and 6 cm deep. It is filled to a depth of 3cm.
You have 7 solid steel shapes in front of you. The question is to find the new level of the water after each shape is put in the pot in the orientation described. The previous shape is removed before adding the next.
1. A cube 5cm on a side.
2. A prism in the shape of a right triangle with legs 5cm long. It is 5cm high but is placed on one of its legs with its hypotenuse sloping out of the water.
3. Another prism, this one having an equilateral triangle of sides 5cm and height 5cm. It is to be placed on its side with two faces sloping up out of the water.
4. A regular hexagonal prism. Each edge of the hexagon is 4cm and the height is 5cm. It is to be placed on its side.
5. A right square pyramid. Its base is 6cm on a side. It is 5cm high. It is to be placed base down.
6. A right cylinder of radius 3cm and length 5cm. It is to be placed on its side.
7a. A right cone of radius 3cm and height 5cm. It is to be placed base down.
7b. The same cone as 7a. This time placed on its side.
(In reply to answers (spoiler)
I have removed my earlier comments to make this clearer.
The base of the cube covers 25 cm², the tank has a surface area of 100 cm² thus the displaced water will cover 75 cm².
After 1cm immersion of the cube 25cc will be occupy 25/75 cm of additional height. After 3cm of immersion this will equate to 1cm.
This means that the cube is still 1cm above the floor of the tank.
As each 1cm immersion is creating a rise of 1/3 cm the new height will be 4/3cm above original surface.
As I noted in comments which I editted out, this is not a simple Archimedes exercise where you can fully immerse a volume. In this case the rise of water is lineal because the cube is uniform upon its entry, but it must also be noted that the depth of immersion of the object must be read against the rising height of the water level and not the base of the tank.
Edited on June 27, 2006, 10:11 pm
Posted by brianjn
on 2006-06-27 20:35:31