Tricky Q

[adrakdavra]

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What happens to a balloon as it descends into the water at a uniform rate?


A. The volume of the balloon decreases linearly.
B. The volume of the balloon does not change.
C. The volume of the balloon changes a little at first, then gradually more as it descends.
D. The volume of the balloon changes considerably at first, then gradually less as it descends.

 

[Captain Sisko]

What happens to a balloon as it descends into the water at a uniform rate?


A. The volume of the balloon decreases linearly.
B. The volume of the balloon does not change.
C. The volume of the balloon changes a little at first, then gradually more as it descends.
D. The volume of the balloon changes considerably at first, then gradually less as it descends.

I get d. Is that right?

 

[Balantidiumcoli]

What happens to a balloon as it descends into the water at a uniform rate?


A. The volume of the balloon decreases linearly.
B. The volume of the balloon does not change.
C. The volume of the balloon changes a little at first, then gradually more as it descends.
D. The volume of the balloon changes considerably at first, then gradually less as it descends.

P = Density*Height*Gravity + Air Pressure
PV = nRT -> P = nRT/V

Density and gravity are constant while only height changes. Since it is descending at a uniform rate, this implies that pressure is increasing linearly.

Looking at P = nRT/V this means that P is proportional to 1/V (assuming T is constant like n and R) and since P is increasing linearly this means that 1/V must also be decreasing linearly as V increases. So the answer is D.

-Edit-

Put another way: V is proportional to 1/P

 

[peacefulheart]

1. It is right that pressure increases linearly. But volume does not .

2. Using graph of PV=nRT will be better to understand. Think about the graph xy=1 ( or any constant). Let us say P=y , V=x, and nRT= constant, you will see from the graph that V (volume) does not change linearly with P. It change considerably at first since the slope of the graph is very deeper and gradually less later since the slope of the graph is flatten.

3. I can understand from a mathematical point of view. It just does not make sense to me how the whole process could happen. There are only buoyant force(upward) and gravity of balloon (downward)..

4.Could anybody explain how could this happen?

thanks a lot

 

[Balantidiumcoli]

3. I can understand from a mathematical point of view. It just does not make sense to me how the whole process could happen. There are only buoyant force(upward) and gravity of balloon (downward)..

You're looking at the wrong things. All that matters is pressure which is F/A. The problem all ready shows that the velocity is constant so you don't have to worry about new forces acting on the balloon (F=MA). As a solid goes deeper in a liquid, the weight on the solid increases (a greater volume of liquid is above the solid), which means increasing force and increasing pressure. You do the algebra all ready demonstrated above and you can derive the conclusion that V is proportional to 1/P or V=K/P where K is a constant. The graph of 1/P shows a large number as P approaches 0 which decreases greatly at first and lessens with increasing pressure.

So the answer is D.

 

[milski]

At a more superficial level: The more dense the air in the balloon becomes, the harder it is to compress it and the less of a change you need to match the pressure outside. For example, if you double the pressure, the volume will change from V to V/2 - that's loss of V/2. If you double it again, the volume is only V/4 or the lost volume now is just V/4. As the volume continues to decrease, the same change, proportionally, becomes smaller and smaller as an absolute value.

 

[Balantidiumcoli]

At a more superficial level: The more dense the air in the balloon becomes, the harder it is to compress it and the less of a change you need to match the pressure outside. For example, if you double the pressure, the volume will change from V to V/2 - that's loss of V/2. If you double it again, the volume is only V/4 or the lost volume now is just V/4. As the volume continues to decrease, the same change, proportionally, becomes smaller and smaller as an absolute value.

Nice. 👍

 

[peacefulheart]

You're looking at the wrong things. All that matters is pressure which is F/A. The problem all ready shows that the velocity is constant so you don't have to worry about new forces acting on the balloon (F=MA). As a solid goes deeper in a liquid, the weight on the solid increases (a greater volume of liquid is above the solid), which means increasing force and increasing pressure. You do the algebra all ready demonstrated above and you can derive the conclusion that V is proportional to 1/P or V=K/P where K is a constant. The graph of 1/P shows a large number as P approaches 0 which decreases greatly at first and lessens with increasing pressure.

So the answer is D.

1. Thanks for the explanation.

2. What is F in F/A. It seems that water exerts a downward force by pressure and upward buoyant force. If like you said, the velocity is constant , then the upward buoyant force should be equal to the downward gravity. When the balloon is descending in the water, it becomes less buoyant since its volume is decreased. However the weight of balloon does not change. This net force keeps the balloon descending. According to my understanding, the uniform rate means constant acceleration.

3. Or at the water surface, the balloon was given a push or some sort of downward force since there is no way for balloon to descending by itself when the air density inside of balloon is less than water density. otherwise, the whole process does not make sense.

4. Could anybody make a clear reasoning .

thanks

 

[milski]

You are seriously overthinking it. The whole point of saying that it descends at constant rate is to say that the pressure outside the balloon increases at a constant rate. It does not matter what exactly is making it descend - it could be just a scuba diver pulling it down by a string.

Uniform rate also means zero acceleration, not constant acceleration. In this case it's a rate of descend (or rate of change of position), which would be the same thing as uniform velocity.

 

[peacefulheart]

At a more superficial level: The more dense the air in the balloon becomes, the harder it is to compress it and the less of a change you need to match the pressure outside. For example, if you double the pressure, the volume will change from V to V/2 - that's loss of V/2. If you double it again, the volume is only V/4 or the lost volume now is just V/4. As the volume continues to decrease, the same change, proportionally, becomes smaller and smaller as an absolute value.

I like this explanation . It makes a lot of sense,. thanks