Does the visible part of an iceberg depend on size? Most people heard that we only see some % of an iceberg above water (10% etc). I suspect the exact % depends on the iceberg's shape (e.g. flat disc vs very narrow cylinder). But, assuming a similar, roughly spherical shape, does this % change with the iceberg's size? If you had roughly spherical icebergs, 1m, 10m, 100m & 1000m (that's a big one) in diameter, would you always see the same % above water?
 A: A perfectly spherical ball of some uniform density $\rho_{ball}$ in water of uniform density $\rho_{water}$ would have the exact same percentage above water regardless of the radius of the ball.
However, in reality it's not quite that simple because the water has different density at different depths, therefore a larger ball of let's say $1000$m radius would have a larger percentage visible due to the colder water at the bottom giving a larger pressure gradient than what a smaller ball would feel close to the surface. Also in reality the iceberg does not have uniform density making it even more complex, but on average it's said that roughly 1/10th of an iceberg is above water.
A: Archimedes’ principle tells us that a floating body displaces its own weight of water. This is regardless of its shape. So a solid body with an average density that is $x\%$ that of the water in which it floats will float with $x\%$ of its volume submerged and $100-x\%$ above the water line.
For an iceberg the proportion that is above the water will depend on the average density of the ice compared with the density of seawater - and the biggest factor of variation here will be the salinity and temperature of the seawater. Wikipedia says that a typical iceberg is $90\%$ as dense as the seawater in which it floats, so $10\%$ of the volume of the iceberg will be above the water line.
