Consider the following problem. A glass contains water that reaches some height $h$ above the ground. We put an ice cube on top of it and wonder what the resulting change in height, $\Delta h$, will be. In order to do that one can apply Archimedes' principle to find out what the volume, call it $V$, of the displaced water will be (which is in fact the same as the volume of ice below surface level).

Now, in order to calculate the increase in height, one has to divide by the cross section of the glass, $\pi r^2$, where r is the radius of the glass. Thus the answer would be: $$\Delta h = \frac{V}{\pi r^2}$$

However, I would have intuitively thought that the area we should take to calculate this would be the cross section of the glass, minus the cross section of the ice cube, say $A$, because the water below the surface level gets displaced to the outer parts of the glass, not the region where the ice cube is now located. So, why is it that the correct answer uses $\pi r^2$ as opposed to $\pi r^2-A$?

  • $\begingroup$ Think about what happens to the water level when the ice melts. $\endgroup$
    – John Darby
    Jan 9, 2023 at 21:29

1 Answer 1


For a floating cube buoyancy and weight forces will be in equilibrium : $$ \rho g V = m_cg \tag 1,$$

expressing displaced volume of water in terms of water column height increase $\Delta h$ :

$$ \rho g ~\pi r^2 \Delta h = m_cg \tag 2, $$

then solving for $\Delta h$ gives :

$$ \Delta h = \frac {m_c}{\rho ~\pi r^2} \tag 3, $$

where $m_c$ is ice cube mass.

Or if you want to express (3-rd) equation in terms of ratio of ice cube/water masses, then substitute water density as $\rho = m_w/\left(\pi r^2 h\right)$ into that equation and with a bit of simplification you'll get :

$$ \Delta h = h \frac {m_c}{m_w} \tag 4,$$

where $h$ initial water height, $m_c$ mass of ice cube and $m_w$ mass of water. To verify this equation imagine that you want to double the water content in the glass, then per (4) ratio of added mass vs initial mass will be 1, and consequently - $h$ will double too.


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