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 solution #6
With the help of Excel and its functions, the solution I have calculated for the cylinder (lying on it side) raises the water level to approximately 4.0016820055 cm in depth.
The following equations were used to calculate this depth (height). The approximate depth was determined by finding where the height of the water equaled the height of the cylindrical segment.
(VOLwater+VOLcyl. seg.)/(WIDTHpot*LENGTHpot) = HEIGHTwater
VOLwater = 300 cm3; WIDTHpot = 10 cm; LENGTHpot = 10 cm
VOLcyl. seg.= L*r2*cos-1((r-h)/r) - (r-h)*L*SQRT((2*r*h) - h2)
where L = LENGTHcyl.; r = RADIUScyl.; h = HEIGHTcyl. seg.
The cylindrical segment is the solid portion of a cylinder below a cutting plane oriented parallel to the cylinder's axis of symmetry.
The definition and equation for the volume of a Cylindrical Segment here is taken from the CRC Concise Encyclopedia of Mathematics by Eric W. Weisstein.
Edited on June 30, 2006, 3:37 pm
Posted by Dej Mar
on 2006-06-29 23:28:20