### Brain Teasers

# Underwater Balloon

You have a rubber balloon with a string attached to it, and a weight attached to the other end of the string. You put just the right amount of air in the balloon such that if you submerge the balloon and weight under water to a depth of 30 feet, the balloon will not rise or sink (its buoyancy force is exactly balanced by gravity).

You then pull the balloon and weight down another 30 feet. If you let the balloon go at this depth, will it rise or sink?

Assume that the water temperature is the same at both depths.

You then pull the balloon and weight down another 30 feet. If you let the balloon go at this depth, will it rise or sink?

Assume that the water temperature is the same at both depths.

### Hint

The overall density of the water/weight system at 30 feet is exactly the same as the density of water. How does the density of the system change at 60 feet?### Answer

The balloon will sink. At a depth of 60 feet, the water pressure is greater than it is at 30 feet (by about 15 psi). Because of this, the balloon will shrink, causing the balloon/weight system to increase in density (the total volume is smaller while the total mass stays the same). Since the system is now more dense than it was at 30 feet, it will sink.Also important is the fact that since water is pretty much incompressible, the water density at both depths will be about the same.

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## Comments

Whoa. That was harrrrrrrrrrrrrrrrd!

Excellent teaser. I first thought that the additional pressure would push it up but then I thought about what would happen to the balloon. BTW keep up your informative explanations on some of the other peoples teasers. They are often much better than the original answers.

This is a good teaser.

I am trying to understand this teaser, but I am having difficulty. Sounds a bit in-depth.

Great problem

I got this one!

Thought so!

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