Quiz - Physics - Minimum Amount of Water That Must Be Displaced

[Pedro Burrito]

From the Physics Section of the PADI Dive Theory Exam:

A 175 kg/385 lb anchor that displaces 115 l/4cf of water lies on the bottom in 14m/46ft of salt water. What is the minimum amount of water that must be displaced from a lifting device to bring the anchor to the surface?

a. 54.9 litres / 2 cubic feet

b. 60 litres / 2.16 cubic feet

c. 65 litres / 2.34 cubic feet

d. 110 litres / 4 cubic feet​

I will post a daily question from my exams to help newer divers and to encourage more experienced divers to interact gracefully and helpfully with the newer divers.

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Thank you for your patience while we try to give people something to discuss other than Covid-19 and/or Politics. I will post the answer covered by the spoiler tag later today.

 

[saxman242]

Damn, hit the wrong button. None of these seem quite right though.

 

[ScubaWithTurk]

You take the weight of the object and divide by 1.03 for saltwater. This gives you 169.9 L. (we divide to convert weight to volume and multiply to convert volume to weight)

next is 169.9L - 115L = 54.9L

 

[saxman242]

Yah, but 1.03 is way high for surface level salt water densities, with an average sitting around 1.025 kg/l. Puts the actual answer about 56l for most places. I saw 65 and thought it was the 56 I was looking for.

Given the choices, and using 3 significant figures, I agree that 54.9 is right for the rounding error... And as such, I hit the wrong button.

 

[Angelo Farina]

Salt water density is around 1025 kg/m3. The negative buoyancy thrust of the anchor is 175 - 115*1.025 = 57.125 kg.
This requires a displacement of 57.125/1.025 = 55.7 liters for being overcome.
Hence 54.9 liters is less than sufficient. The anchor will not move, you need at least 55.7 liters.
So the next available answer is 60 liters, and I did choose this one, as this is enough for lifting.

 

[Ukmc]

Salt water density is around 1025 kg/m3. The negative buoyancy thrust of the anchor is 175 - 115*1.025 = 57.125 kg.
This requires a displacement of 57.125/1.025 = 55.7 liters for being overcome.
Hence 54.9 liters is less than sufficient. The anchor will not move, you need at least 55.7 liters.
So the next available answer is 60 liters, and I did choose this one, as this is enough for lifting.

This was exactly my thought process as well, except I divided 175 by 1.025 to make calculation easier. Then, I am all in Liters: 175/1.025-115 for a minimum force which is 55.7

 

[BlueTrin]

Salt water density is around 1025 kg/m3. The negative buoyancy thrust of the anchor is 175 - 115*1.025 = 57.125 kg.
This requires a displacement of 57.125/1.025 = 55.7 liters for being overcome.
Hence 54.9 liters is less than sufficient. The anchor will not move, you need at least 55.7 liters.
So the next available answer is 60 liters, and I did choose this one, as this is enough for lifting.

I suspect the expected answer is A) even though you are correct.

I think PADI usually uses 1.03 for their quizzes in seawater.

 

[Scraps]

I used Angelo's thought process.
My calculated solution was between two of the answers provided.
You can't go with the closer answer. You have to choose the answer that will lift the object.

 

[rhwestfall]

Spoiler: Imperial Solution

(385/64) - 4 = 2.02 (yep, splitting hairs, but 2.0 won't cut it)

 

[scubadada]

Currently, neck and neck