TPR MCAT Review Series Physics (Buoyancy)

[Synapsis]

Spoiler alert if you plan on taking this test.

A question says that a balloon has a buoyant force F at a distance d beneath the surface of water. If submerged to a distance 4d, how does that force change?

A) .25F
B) .5F
C) F
D) 4F

Answer: A

Here's how I did it. I started with Archimedes' Principle, saying that the buoyant force equals the weight of the fluid displaced by the object. So F = pVg. Pretend p is rho. Then I said that volume is just the surface area times the depth, and changed the equation to: F = pAdg.

So if the depth increases by a factor of 4 I get F = 4(pAdg) and answer D. Why exactly is this wrong?

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[tuhtles]

My first instinct was what you did. But then I thought if you put the balloon at a deeper depth, then its volume would decrease at the higher pressure so it would displace less water, thereby making the buoyant force smaller by decreasing the volume. But I'm not quite sure how to mathematically arrive to the answer in this case because I can't seem to find a formula that quantifies the relationship between pressure and volume.

If I remember correctly from TBR, the liquid will also be more dense in deeper depths because the liquid molecules will also be impacted by the increase in pressure, making the liquid molecules closer together. I don't think this would be a significant effect though. A denser liquid would also mean greater buoyant force, which goes against the answer.

What did the explanation say?

 

[sakabato93]

If I remember correctly from TBR, the liquid will also be more dense in deeper depths because the liquid molecules will also be impacted by the increase in pressure, making the liquid molecules closer together. I don't think this would be a significant effect though. A denser liquid would also mean greater buoyant force, which goes against the answer.

I don't know about that. Generally, liquids are said to be incompressible. If the fluid were a gas this would be true, but liquids usually have uniform density. But pressure does increase as depth increases, but not due to increased density, but the weight of the water increases.

 

[Jumbo]

My first instinct was what you did. But then I thought if you put the balloon at a deeper depth, then its volume would decrease at the higher pressure so it would displace less water, thereby making the buoyant force smaller by decreasing the volume. But I'm not quite sure how to mathematically arrive to the answer in this case because I can't seem to find a formula that quantifies the relationship between pressure and volume.

If I remember correctly from TBR, the liquid will also be more dense in deeper depths because the liquid molecules will also be impacted by the increase in pressure, making the liquid molecules closer together. I don't think this would be a significant effect though. A denser liquid would also mean greater buoyant force, which goes against the answer.

What did the explanation say?

You're over-thinking it. Unless the balloon is submerged very deep in an ocean, the pressure difference between d and 4d would be negligible.

Spoiler alert if you plan on taking this test.

A question says that a balloon has a buoyant force F at a distance d beneath the surface of water. If submerged to a distance 4d, how does that force change?

A) .25F
B) .5F
C) F
D) 4F

Answer: A

Here's how I did it. I started with Archimedes' Principle, saying that the buoyant force equals the weight of the fluid displaced by the object. So F = pVg. Pretend p is rho. Then I said that volume is just the surface area times the depth, and changed the equation to: F = pAdg.

So if the depth increases by a factor of 4 I get F = 4(pAdg) and answer D. Why exactly is this wrong?

Shouldn't the answer be C? The formula for buoyancy refers to the volume of the displaced water by the object. It doesn't matter where the balloon is, it has the same volume everywhere so it will displace the same volume of water everywhere, and the buoyant force on the balloon will be the same everywhere. You're confusing the volume of the displaced water with the volume of the column of water above the balloon.

 

[RickP]

You're over-thinking it. Unless the balloon is submerged very deep in an ocean, the pressure difference between d and 4d would be negligible.



Shouldn't the answer be C? The formula for buoyancy refers to the volume of the displaced water by the object. It doesn't matter where the balloon is, it has the same volume everywhere so it will displace the same volume of water everywhere, and the buoyant force on the balloon will be the same everywhere. You're confusing the volume of the displaced water with the volume of the column of water above the balloon.

IIRC

change in pressure = g*density*change in hight

with Pressure = Force/Area.

so with 2 directly being proportional shouldnt it be 4d?

 

[Synapsis]

You're over-thinking it. Unless the balloon is submerged very deep in an ocean, the pressure difference between d and 4d would be negligible.



Shouldn't the answer be C? The formula for buoyancy refers to the volume of the displaced water by the object. It doesn't matter where the balloon is, it has the same volume everywhere so it will displace the same volume of water everywhere, and the buoyant force on the balloon will be the same everywhere. You're confusing the volume of the displaced water with the volume of the column of water above the balloon.

Nope, it's A.

The explanation says gauge pressure is proportional to depth by P = pgD. This makes sense. Then if you quadruple D, you get 4 times the pressure. Since pressure and volume are inversely related, increasing the pressure by a factor of four decreases the volume by a factor of 4. Then by F = 0.25 (pVg), this gives 0.25F.

I posted a question because I was hoping there was a better way to look at this.

 

[sakabato93]

Nope, it's A.

The explanation says gauge pressure is proportional to depth by P = pgD. This makes sense. Then if you quadruple D, you get 4 times the pressure. Since pressure and volume are inversely related, increasing the pressure by a factor of four decreases the volume by a factor of 4. Then by F = 0.25 (pVg), this gives 0.25F.

I posted a question because I was hoping there was a better way to look at this.

Pressure and volume are inversely related for liquids??? That's pretty unintuitive... especially since liquids are, generally, incompressible.

 

[RickP]

Is it just me or does the question seem a little vague. When I answered above I scanned the question and just read force and took that as the force applied onto the balloon.

So if Ρ = F/A Δ Ρ = ρgΔh is it correct to assume then that the force applied onto the balloon is 4x greater and then as op said, because p inversely related to v the boyant force is 0.25?

 

[wjs010]

This is a confusing question. When I first thought about it, I thought it was 4f. Now I see some of you are thinking about the gas laws. This one is tought. The buoyant force should be equal to the weight of the water displaced by the balloon. So which is it?! Weight of water displaced is pVh right? .

 

[tuhtles]

I think this question is tricky because what's being submerged is a balloon that has gas in it and is therefore compressible at lower depths due to the increased pressure. If the object being submerged was an incompressible solid, then the answer would be a bit more straightforward.