A man on a boat, with something in his hand 
Possible Duplicate:
Basic buoyancy question: Man in a boat with a stone 

A man is on a boat, in the middle of a quiet lake.
He has something in his hands (anything you like)
He throws the "something" in the water.
Does the water level increases or decreases? 
And if the answer is "it depends", on what it does?
 A: If the object floats: water level stays the same
If the object sinks: water level decreases
Consider the force balance.  The Earth exerts an upward force on the lake.  Anything floating on the water is included in the weight of the lake.  Since water is constant density, the upward force on the lake is a direct function of the water level - a higher level results in a higher pressure on the bottom, and a lower level results in a lower pressure on the bottom.  Thus, no matter how you rearrange items supported by buoyancy on (or in) the lake, the water level will stay the same.  An object that sinks touches the floor of the lake, so the prior argument no longer applies.  The force balance dictates that the weight of the lake is less (as we consider the sunk object to no longer be a part of the hydrostatic body, since it's supported by a separate and unconnected normal force), and thus the water level will be less.
A: The question Basic buoyancy question: Man in a boat with a stone is an exact duplicate. The answer is that if the thing you throw from the boat has a density greater than water the level falls. If the object's density is less than water the level rises.
Oops:
Ah, yes, as Alan points out in his comment, in Basic buoyancy question: Man in a boat with a stone the equation for the volume of water displaced assumes the object is full submerged, and for a density less than water this wouldn't be the case. To briefly reprise the other question, when the object is in the boat the volume of water displaced is:
$$ V_{disp1} = \frac{M + m}{\rho_w} $$
and if the object is thrown from the boat and fully submerged the volume of water displaced is:
$$ V_{disp2} = \frac{M}{\rho_w} + \frac{m}{\rho_r} $$
where $\rho_r$ is the density of the object, so the change in volume displaced is:
$$ V_{disp1} - V_{disp2} = \frac{m}{\rho_w} - \frac{m}{\rho_r} $$
hence if $\rho_r$ > $\rho_w$ then $V_{disp1} > V_{disp2}$ and the water volume falls. But if the object has a density less than water, and therefore floats, the second equation changes to:
$$ V_{disp2} = \frac{M}{\rho_w} + \frac{m}{\rho_w} = \frac{M + m}{\rho_w} $$
so $V_{disp2} = V_{disp1}$ and the water level doesn't change.
