# Physics Help

Discussion in 'MCAT Discussions' started by Machine33, Sep 8, 2011.

1. ### Machine33 OMS-4

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If you look at the attachment (don't mind my awesome paint skills ), which container experiences the greatest pressure at the bottom of the fluid? 1, 2, 3, or all 3 experience the same pressure at the bottom.

Thanks so much!

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3. ### Donald Juan

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1, because pressure is force/area. I'm guessing they all have the same volume of liquid, so the forces are equal on all of them, but 1 has the smallest area.

4. ### Machine33 OMS-4

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So pgh doesnt apply here? But instead F/A?

5. ### Machine33 OMS-4

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Bueller? Bueller??

6. ### Chrome19

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It applies. P=F/A = mg/A. But m=density (p)*volume = density*crosssectional area (A)*height(h).

So P=pgAh/A

If area is uniform, then P = pgh and things are straightforward.

Going back to P = pgAh/A we can write areas as pgA1h/A2 where A1 is the cross-sectional of the vessels at any height and A2 is the crosssectional of the base at the same height. Height appears to be the same in the figure and pg is constant. So the solution of this problem requires the determination of the ratio of A1/A2 which is tricky to do from the problem. Casual inspection suggests the first figure has the highest ratio of A1/A2 and thus will have the highest pressure. This also requires the assumption that the mass of water is the same in all vessels.

7. ### MShopes

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Yea as other posters mentioned, it has to do with P=F/A, least area means highest pressure so it is the first container.

You could try to use P=pgh but well apparently all values (pgh) are the same here. That is, the density p is the same since it is the same fluid and highest is the same and g is always the same so this equation does not help much in distinguishing between pressures.

8. ### Temperature101 Banned Banned

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I think they have the same pressure regardless of the shape of the containers (assuming the fluid densities are the same) since the depth are the same. P = density*gravity*height. There was a similar example in TBR.

9. ### MShopes

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Well maybe the same overall pressure but the question asked about the bottom of the container only. In the overall, the area would be the same for all three and the height will be the same as well so the pressure will be the same. But for just the bottom, the area is obviously different and thus choice A wins.

10. ### Temperature101 Banned Banned

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You can't apply P = F/A in this case. The pressure at the bottom of each container is the same. Assuming the height of the fluid is 10cm...if the pressure is the same at 5cm depth for each container, why would it be different at 10cm depth(which is the bottom of each container)?

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11. ### MShopes

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I don't see why not, can you explain?

12. ### Temperature101 Banned Banned

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I guess it will work but requires more thinking. Are you saying P = F/A or P= mg/A... P = Vol*density*g/A...Remember that these vessels do not have the same volume of fluid...so the ones with less surface area will hold less volume of fluid...Therefore, the area is proportional with the volume occupied...We will end up with the same pressure. This formula will work as well but requires more thinking. Hope it makes sense.

13. ### BerkReviewTeach Company Rep & Bad Singer Exhibitor

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Temperature101 is correct here.

Think about it this way. If you were to swim to the bottom of the deep end in a pool, no matter what the size or shape of the pool, you would feel the same pressure from the column of water above you. The pool may get wider, but at a given depth you are beneath a column of water that weighs so much per square meter no matter where you are in the pool or what shape the pool may be.

14. ### DaLeon

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The pressure is the same at the bottom of all, assuming the water column is of equal height (which I think it's safe to assume).

15. ### Chrome19

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This seems to be a tricky problem. I was thinking of the problem assuming all containers had the same volume of fluid.

Ok, let's assume the volumes are different. If the bottom was infinitely small (think of pointed six-inch heels of shoes), are you saying the pressure would still be the same at the base? And what if we had a gas in the containers and we measured pressure as collisions against the walls of the vessels--how would pressure change?

17. ### Chrome19

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OK. I think this makes sense now. The force exerted on the walls will vary from vessel to vessel, but this variation correlates with the surface area so that F/A is the same. Pretty simple.

Anyways, my MCAT is over.

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18. ### indianjatt

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Wouldn't 1 have the highest pressure? At the top of container 1, the liquid is open to the atmosphere so it is at 1 ATM. The other two are vacuums. Since they change by the same amount going down a depth h, wouldn't 1 have the highest pressure by a value of 1 atmosphere?

19. ### BerkReviewTeach Company Rep & Bad Singer Exhibitor

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Chrome19:

Feels great to be done, doesn't it? Here's hoping you have at least two teens when scores come back.

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20. ### KindofBlue Member

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I believe the pressure is all the same too.
But..I think you should erase the image, you know...

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21. ### medceg Lifelong Student

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Here is my take on this tricky figure:

If the height of the water column level in each figure was the same with respect to the bottom of the container, then the pressure would be the same.

BUT

If the volume of the water in each figure was the same, I would think the left most figure would have the greatest amount of pressure. Why? Because the height of the water column would increase in the left most figure since there is less cross sectional area at the bottom of the flask. The left most figure would have the highest water column. The right most figure would have less water column height than the left most figure.. And the middle figure would have the shortest water column height. Since there is more area at the bottom of the flask in the middle figure, the water column would be lower than the other two figures because there is more space for water to occupy.

So both Donald Juan and Temperature101 are correct, because you both assumed different constants. You all are right and the figure is ambiguous.

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This.

23. ### Oncoloman Member

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I agree. One has to consider the cross-sectional area which applies to both P=F/A and P= pgh. The density of that "slice" of water is equivalent to all of the others ( obviously), yet when considering gravity it can be seen that the the cross sectional area of the contaner which experiences that similar force is much less than the others. Hence the force of water is applied to a much smaller cross-sectional area. So pgh is true, but in this case we have varying shapes of contaners so P= F/A must be applied. When it is we can see that the pressure is greater in the first.

24. ### Chrome19

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Thanks mate. Sure feels good. And I'm expecting at least 10's in the sciences. If I score 9 or higher in VR, I'll most likely score over 30. I'm just hoping my science scores really reflect my practice exams and my knowledge. If not, I'll retake and re-apply next year.

25. ### Chrome19

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This makes sense.

26. ### junaid hassan

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the length of shortest air column closed at one end tht resonates with a vibrating tunning fork of frequency 520 vib/sec will be

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