How stuff works

[Fred]

On the other hand, if we all just let our inhibitions run wild.........there'd be a lot of happy campers.  
I'm sorry, I don't know what's wrong with me today.

[flyinlow]

1 & 2 would be even.

3 would be lower assuming the steel ball was large enough to sink the ice cube originally.

[ws23rt]

1 & 2 would be even.

3 would be lower assuming the steel ball was large enough to sink the ice cube originally.

It is true the extra weight of the steel ball would hold the ice lower until it fell to the bottom. The ice would raise and water level drop accordingly.
The other two examples would not change as the weight of the ice would not change

[Lighthorseman]

Would anything change if one of the ice cubes was painted orange with a 01 on the side? 

[Fred]

The shit that goes on in peoples heads is astounding.  

No wonder we can solve every problem when we all put our heads together.  

[Mopar Nut]

Two heads are better than one.

[Nacho-RT74]

I didn't want to read all replies just in case the final answer was there, but this is my conclusion after use my brain.

every one will be lower than originally when freezed BUT the one with iron ball will be higher than the rest.

the one with the air buble will be the lower of all of course.

reason ? since water expands when freezed, with all glasses freezed at same level ( so same total freezed volume ) the only section of the total volume will never change is the one with the steel ball. The air bubble will dissapear, and the only water freezed will keep that area filled.

metal contracts when freezed too, same as air, but not the same rate than the water expands.

of course everything will be more noticeable or not according of the size of the air bubble and steel ball

[.]

*snip* reason ? since water expands when freezed, with all glasses freezed at same level ( so same total freezed volume ) the only section of the total volume will never change is the one with the steel ball. The air bubble will dissapear, and the only water freezed will keep that area filled. *snip*

I'm afraid you're wrong, here's the logic:

Cup 1: solid ice.
Since water expands when it freezes the ice cubes are lower in density. The part of the ice cube that floats equals the part of volume that would disappear as the ice would melt - because water is more compact then ice.
-> no change in level.

Cup 2: Ice with air
The volume of the ice cube is larger then the actual amount of ice (since the air takes up place aswell) - however the same principal applies here: The air contributes to the overall density of the ice cube - making it even less dense. Hence the ice cube will float even more then its solid counterpart. So the cube is less deep inside the water, making cup 2 filled with more water then cup 1. As the ice melts the air leaves the ice cube and the ice melts. There is less ice to melt then in cup 1, but there needed to be more water in it from the stat to fill the cup to the edge
-> no change in level.

Cup 3: Ball inside the ice
The steel ball increases the density of the overall ice cube, making it sink. as the ice melts it will detach from the ball and go towards the surface. However since the ice was completely under water when you filled the cup in the beginning of the experiment the complete sum of water in your cup is less because the submerged ice took in place for twice the amount of water it would result in when it melts
-> water level drops

[stripedelete]

Would anything change if one of the ice cubes was painted orange with a 01 on the side? 

Not sure.  But I can tell you the one full of Scotch is empty.