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This comes from a brain teaser but I'm not sure I can solve it:

You are in a rowing boat on a lake. A large heavy rock is also in the boat. You heave the rock overboard. It sinks to the bottom of the lake. What happens to the water level in the lake? Does it rise, fall or stay the same?

I'd say that the water level drops, because when you drop the stone in the lake, the water level rises according to the volume of the stone, BUT the water level decreases by a volume of water weighting the stone's weight, when you take it off the boat.

Is that correct?

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I think your own answer is nicer than any of the listed ones so far. Like you say, on the boat, the rock displaces its weight in water. In the lake, the rock displaces its volume in water. The latter is less displaced water than the former, because the rock is denser than water. –  Steve B Jun 18 '12 at 21:50
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up vote 10 down vote accepted

This diagram is my attempt to show the situation first when the rock is in the boat and secondly when you've chucked the rock over the side.

Boat

The mass of the boat of $M$ and the mass of the rock is $m$. The density of water is $\rho_w$ and the density of the rock is $\rho_r$. In the first case Archimedes' principle tells us that the volume of water displaced is:

$$ V_{disp1} = \frac{M + m}{\rho_w} $$

In the second case the volume of water displaced is:

$$ V_{disp2} = \frac{M}{\rho_w} + \frac{m}{\rho_r} $$

where the second term is just the volume of the rock. If we take the difference of these two we get:

$$ V_{disp1} - V_{disp2} = \frac{m}{\rho_w} - \frac{m}{\rho_r} $$

I think it's safe to assume that $\rho_r$ > $\rho_w$, i.e. the rock sinks in water, and in that case $ V_{disp1} - V_{disp2}$ is positive i.e. more water is displaced when the rock is in the boat, so the water level falls when you chuck the rock overboard.

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