Let’s first state our assumptions:

• Very thin membrane, the balloon skin cannot be compressed more than it already is

• Zero trapped air

• Isothermal conditions, water temperature at the bottom of the ocean is the same as the top, that way we don’t need to deal with thermal expansion

• Balloon is a perfect sphere

Calculations:

1. Pressure on the balloon is densitygravitydepth ~= 1000 x 10 x 100 = 1 Mpa or 10 atmospheres

2. The bulk modulus of water is 2.2 Gpa so (change in volume)/(initial volume) = (change in pressure)/(bulk modulus) = 0.0046% decrease in volume

0.0046% decrease in volume is like a 0.015% decrease in radius

So yes the balloon does get a little bit smaller

Liquid Water does not change it’s density much with pressure, but yes.

Water doesn’t compress (for all intents and purposes), so no. I.E. the water pressure of the ocean would indeed push inward against the balloon much more strongly, but the water inside the balloon would push back by the same amount.

Nothing special would happen if you pop the balloon, the rubber would just rapidly contract, basically the same as it would above the surface.

At 100 m the compression would be nil.

If you take it deep enough for there to be measurable compression and then pop it, nothing will happen because it’s already compressed.

Depends on how much air is in there with the water.

Water itself is incompressible so the balloon won’t shrink that much, but there’s some air in there so it would shrink a bit.

The balloon’s popping action comes from being filled and expanded. It’s an elastic, so if something breaks it, it snaps back.

The only way that wouldn’t happen is if there was mostly air in there, and you went to the appropriate depth where that air was compressed down to its original volume.