I don’t think it’s the same thing. In free fall, they’d hit at the same time. On a ramp, their angular momentums would increase at the same rate which would translate into a faster solid ball. WAG.
That was my first thought; ask a Dolphin. 
Well, I’ve gone and done a little reading, and looks like I’ve fought my own ignorance. Apparently the speed that a ball rolls down a hill does depend on its mass distribution (which means that a lot of things I thought I knew were wrong–didn’t Galileo do tests rolling balls down a hill to show that objects fell at the same speed–or does that experiment only work if the balls have the same mass distribution? or am I just remembering it entirely wrong?)
You can do Galileo’s test by either making sure things have the same mass distribution (straightforward enough if, as is likely, they’re all solid and uniform density), or by making the rolling portion of the mass a negligible fraction of the total mass (like a big heavy cart on small wheels).
Wasn’t Galileo’s purported test dropping things off the Leaning Tower of Pisa rather than rolling them? I dunno.
Anyway, two balls of the same mass will be pulled down a hill with the same force. However, the one with most of it’s mass at the surface will take longer to spin up due to its higher moment of inertia.
This is, of course, the same phenomenon that allows ice skaters to increase their spin speed as they pull their arms in. The hollow ball is like a skater with her arms out.
As to the spheres, I think I’d just see which one floats.
The classic answer is to roll the balls down a hill. It’s in just about every high school physics textbook (in the angular momentum chapter) i’ve ever seen. Unfortunately, I forget which gets to the bottom first. Pretty sure it’s the mass-near-the-outside one.
Same density means they both float or both sink.
I’m curious about the feasibility of making two spheres as described and if dropped, they’ll both bounce to the same height.
The one filled immaculately with liquid, wouldn’t be able to compress, only deform.
The hollow one, would be able to deform and compress the gaseous (or vacuum) interior.
Even if they had the same density, mass and hardness, wouldn’t the hollow sphere bounce a tad higher?
Another thought: temperature. Freezing or heating them might cause changes in density of the differing interiors. You might be able to bracket their temperatures to see if eventually one floats and one sinks.
BTW, My initial solution was to spin them. Of course, rolling them down a grade is much simpler.
Good question. I think the hardness is measured externally as the total deformation of the sphere under load. Internally, they’d be composed of different materials, with the hollow one being made of the harder kind. But I wonder if you could really do that… make two objects appear to be of the same hardness at all scales/frequencies while making them different internally.
You can have all the prisoners count independently, but it would take a long time.
Number the prisoners from 0 to 9, and have them only leave the light on if their number is equal to the week number mod 10.
Any time a prisoner is sent to the common room and finds the light on, they know exactly which prisoner must have lit it and can cross them off. Whoever finishes crossing everyone off first calls the warden.
Improvement: They can also leave the light on if they already crossed off the person whose week it is.
I am absolutely not following this at all. If the prisoners are randomly selected, how would you be able to designate a leader to monitor the light switch? How and why would there be a pattern to who gets to visit the room? How would you be able to number and time the frequency of visits if the point is only one gets to visit once a month, and the selection is random?
Say prisoner #1 goes in, has his visit, and leaves the light on. #2 comes in, sees it on, turns it off. #3 comes in, sees it off, then what?
#3 does nothing, and waits for his next opportunity to signal to the leader. In the scheme described by Indistinguishable, the rule is that non-leaders turn the light off only if it is the first time they have visited the room and found the light on. Otherwise, they leave it alone. In effect, once someone has switched the light off the signalling mechanism is blocked until the leader next visits the room and can “receive” the signal and turn the light back on.
Whichever prisoner is randomly selected for the first visit assumes the responsibility of leader. The prisoners presumably can keep track of the passing days, so no other prisoners will incorrectly take on the role of leader.
Nice! Anyone want to figure out how long this would take? (And I assume by “week” above, you mean “month”, right?)
Monte Carlo says about 60 months.
double post
I see, so the mechanics of the thing being limited by chance, it’s true that the light might stay off for three years if the leader happens not to be randomly selected (though I understand the probability says they would be, eventually), yaar? I thought there was something I was missing about monitoring the switch. It doesn’t seem like a terribly expeditious solution, but I don’t have a better one.
Monte Carlo may have screwed up. I be back.
I’m not following this either; or, rather, I’m not following how it can possibly be a quicker system. On average, someone’s leaving the light on only every ten months or so, and that bit of information is passed to one other random prisoner. You have to pass along, I’m guessing, something like 60-70 bits of information before any one prisoner fills up his tally sheet, so we’re talking 600ish months.