Volume Displacement

Question: If you were in a boat in a small swimming pool (so that you could  
measure the water level) and there were rocks in the boat, What would happen 
to the water level if you threw the rocks into the water?  Would it rise, 
lower, or remain the same level as when the rocks were in the boat?
---------------------------------------
Let us note in passing that since each rock sinks, each rock's 
overall density must be greater than that of water.  When a rock is in the 
boat, Archimedes' principle tells us that the weight of the volume of water 
displaced due to that rock is equal to the weight of the rock.  Since the rock 
is denser than water, this displaced volume of water is larger than the volume 
of the rock (weight = mass density times volume times g, the acceleration due 
to gravity).  But when the rock is on the bottom of the pool, it merely 
displaces a volume of water equal to the rock's volume.  Hence, more water is 
displaced when the rock is in the boat than when the rock is on the pool 
bottom.  Therefore, the water level will drop each time a rock is thrown out 
of the boat into the water.

R.C. Winther
=====================================================