If a ball has a pressure of 1 bar inside and another ball has a pressure of 2 bar with the same radius then both ball are deep inside water then which ball will come first outside the water?

Answer 1

Hi Ava,

A fascinating question, b/c you are given incomplete initial conditions and must make assumptions as to what the exact situation is. Making such assumptions, and knowing just what you have assumed that might not be true, is the mark of a careful scientist!

So, the balls are pressurized. 1) Are these gross (absolute) or net (gauge) pressures? This would matter if you were trying to calculate exact forces acting on balls filled with a compressible fluid, such as air. But if you are trying to only compare the forces relatively, then that doesn't matter.

2) Is the filling medium compressible, i.e. does its density increase with increasing pressure?

If so, then the higher-pressure ball has higher mass, thus less buoyancy, and it will rise more slowly. And, of course, all the bets are off if the two balls are filled with different fillings! The problem didn't specify that....

3) However (and this is a big however!) there have been materials made which have the bizarre property of expanding more under pressure (see, for example, http://silver.neep.wisc.edu/~lakes/NegCompr.html ). If that were true in all dimensions for a solid, you could have less mass in the more-pressured ball. My suspicion, however, is that that would only apply for measured pressure in certain special directions in the solid, so the placement of the pressure gauges, and the preparation of the ball fillings, would have to be "rigged" to give this result. But not impossible!

4) Are the balls rigid-surfaced? If not, then the more-pressurized one will expand to a larger radius, and potentially then rise faster (it's the inverse situation to water droplets falling through air: big droplets fall faster, whereas mist-sized droplets drift down slowly. They all have the same density, but the ratio between "buoyancy" and air resistance is more favorable for falling for the larger droplets). The problem didn't specify rigid ball surfaces!

5) We assume that "deep inside the water" means at the same starting depth!

6) And now for an end-run around the whole problem! (Impress your teacher!): "Come outside the water" can be interpreted as "emerge completely from the water". But even beach balls, inside a swimming pool, may not jump completely free of the water when they surface. It depends on the detailed forces on the ball as it first breaks the surface. So you could argue that the mass and size of the balls might prevent the heavier one, in particular, of ever "coming out of the water". In particular, small balls will have a greater burden to achieve "escape velocity". Let me know how that argument sits with your teacher!

Cheers, -- Mr. d.

Answer 2

Both balls will experience the same upward buoyant force , F = ρ V g , where ρ is the density of the liquid (water), V is the volume of the ball and g is the acceleration due to gravity (g = 9.81 m/s^2).

However, the ball with the higher pressure has more mass of air inside, so experiences a greater downward gravitational force. Thus the net force on the lower pressure ball is greater than that for the higher pressure ball.

Consequently, the drift velocity (upward) of the lower pressure ball will be greater than that for the higher pressure ball. For this reason the lower pressure ball will reach the surface first.

Answer 3

If the "balls" do not change radius, then they will essentially come to the surface at about the same time!

The forces involved are the weight of the ball down and the buoyancy force up. The buoyancy force is the weight of the water displaced and if the balls have the same radius, the buoyancy force is the same regardless of the pressure inside the ball.

Because the weight of the 2 bar ball is slightly more with the extra "gas" that results in the 2 bar pressure, it technically would take a "bit" longer to surface with a "slightly" heavier weight, but it should be a very small difference!