The float on a frozen lake We have a bit of a debate going on at work.
If a basketball (or any round float - let's assume one firm enough to not change size with temperature) was floating on a lake, and overnight the lake froze, let's say solid for the sake of argument. Would more, less, or the same amount of basketball be below the surface line of the frozen lake than when it was liqid.
That is to ask, does the volume of x increase or decrease when the lake freezes solid.

 A: You have not specified a frictionless float, but it doesn't matter. In any case (assuming the air temperature remains constant), the volume of ball above the ice will remain constant. As the ice gets thicker, the volume below the ice will get less and less. 
A lake or any stationary body of water freezes from the top down. This creates a sheet of ice with a hole in it, and the ball will remain resting on that hole. Even better, if the ball is not frictionless, ice will bond to surface irregularities in the ball (water makes an excellent glue when it freezes), and the ball will be fixed in place wrt the upper surface of the ice. Further ice formation below the top layer will not pull the ball down.
The process is seen in reverse, when freezing farm fields causes subsurface rocks to rise to the surface. In this case, as the upper soil freezes it grabs the top of the rock, and further freezing causes the ice in the soil to expand, lifting the rock. With a void under the rock, soil will fall in and prevent the rock from dropping back to its original depth when a thaw occurs. Repetition of the freeze/thaw process will gradually lift the rock an appreciable amount.
In principle, a frictionless, flexible ball (not, for instance, a ball made of styrofoam), exposed to an extremely fast drop to cryogenic temperatures, might contract enough to fall through the hole and sink. The continued low temperatures might then freeze the surface of the water before the air in the ball warmed up, trapping the ball under the ice, but this seems like a pretty unlikely scenario.
A: Lets assume for a general fluid which is 
1. a perfect thermal conductor (opposite to water)
2. its solid state is more dense than the liquid state (opposite to water)
3. in its liquid phase its density continuously increases with decrease in temperature (same happens with water upto 277 K then the reverse happens)
In such a fluid, the Archimedes principle holds upto the Freezing point and so starting from some initial temperature of the fluid and then going down to the freezing point, the basketball comes out of the fluid more and more with decrease in temperature since the density of fluid increases. Then at freezing point the solid state is achieved and so the ball remains stuck at whatever position it was just before freezing (assuming perfect thermal conductor, so the whole system is in thermal equilibrium).
But water has different characteristics from the above mentioned fluid. The actual conditions affecting the stated problem are many in number so many simplifying assumptions are taken. A naive answer can be found taking ice as an insulator, so it doesn't allow the heat of the lake to escape to the environment and as a result only a  layer of ice (assuming it is of continuous depth) is formed on the lake. Now lets also assume that the basketball has very high pressure inside of it and thus it doesn't shrink with small change in temperature. Lets also assume that the contact surface between the basketball and ice is frictionless as well as the ice doesn't exert any sort of normal force on the basketball, i.e the basketball can sink through the ice. Now the Archimedes principle says that the weight of water displaced is equal to the weight of the floating object. So earlier some water only was displaced, but now water is displaced as well as there is an extra layer of ice formed on top of it.
Now let us bring some more details. The answer depends on


*

*how thick is the ice layer

*what is the average temperature of water before and after

*what happens when ice is not frictionless


Case 1. Ice is not frictionless - Refer to details of the answer given by WhatRoughBeast. A nice description is given there.
Case 2. Ice is frictionless but the layer is very thin - The density of water at the average temperature before freezing and after freezing will dictate the level. If the density before freezing is more than density at freezing, more of the basketball goes in and vice versa.
Case 3. Ice is frictionless but the layer is somewhat thick - Now the answer is not straightforward. The sum of contributions from the density of water at the average temperature before freezing and after freezing as well as the thickness of ice will dictate the level. The initial conditions need to be clearly specified in order to go to any conclusion.
Case 4. The real world case - Ice has friction, the layer is somewhat thick and average temperatures are different. - A combination of all these factors is needed in order to properly account for the rise. 
Please note that effects such as forces due to surface tension, pressure differences inside ball etc. are not considered as they give a smaller contribution than the above mentioned factors. Strain on ice due to shear is not included as it is very difficult, especially in case of thin ice layer near the breaking point in the stress- strain curve.
A: The ball acts as an insulator - meaning that the water right below the ball will be freezing more slowly than the water to the side. As ice forms to either side, it will then try to "grow" into the space below the ball. This creates a lateral force, and I suspect this force will push the ball (slightly) up.
I expect that the indentation below the ball will be quite a bit smaller than it was when the water was still liquid.
