Fluids question help!

[cardioonc]

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A SCUBA diver decides to bring a helium (He) balloon along with her on a diving trip to the Red Sea (specific gravity of 1013 at sea level). She takes her balloon down with her to a depth of 10 meters. At this depth, she calculates the surface area of the balloon to be 0.5 m2. What is the force being exerted by the He on the inside surface of the balloon?

A. 50.7 kN
B. 101.3 kN (correct answer)
C. 152.0 kN
D. 202.6 kN

 

[kupa]

A SCUBA diver decides to bring a helium (He) balloon along with her on a diving trip to the Red Sea (specific gravity of 1013 at sea level). She takes her balloon down with her to a depth of 10 meters. At this depth, she calculates the surface area of the balloon to be 0.5 m2. What is the force being exerted by the He on the inside surface of the balloon?

A. 50.7 kN
B. 101.3 kN (correct answer)
C. 152.0 kN
D. 202.6 kN

this is actually pretty simple.. just rough calculation without using accurate numbers:

density of water (rough estimate) = 1000kg/m^3
gravity = 10 m/s^2
depth = 10m

atmospheric pressure = 1atm = 100kPa

The pressure the helium atoms would exert on the inner side of the balloon is equal to the pressure the water exerts on the outer side of the balloon.

Therefore, adding the pressure due to atmosphere and due to the weight of the water above the balloon gives: 100000 Pa + 1000 kg/m^3 * 10 m/s^2 * 10m = 200000 Pa.

Since pressure = F/A, and the surface area of the balloon is given as 0.5m^2,
200000 Pa * 0.5m^2 = F = 100 kN

(If you wanted to really solve for the actual answer 101.3kN just use figures of 1013kg/m^3 for the density of water and 101.3kPa for the atmospheric pressure...)


Basically this problem required you to know that at the sea level, the atmospheric pressure is 100kPa (well.. to be precise 101.3kPa).

 

[ilovemcat]

An easier and faster way is just realizing that every 10m depth in water = 1 atm. Therefore Ptot = 2 atm == ~200kPa. This times Area yields ~100kN. If I had to take a guess, I'd say that the OP is just trying to get someone to do his homework.

By the way, no way the specific gravity of the Red Sea is 1,013. More like the Red Sea has a density of 1,013 kg/m^3 and a specific gravity of ~1.

 

[BerkReviewTeach]

An easier and faster way is just realizing that every 10m depth in water = 1 atm. Therefore Ptot = 2 atm == ~200kPa. This times Area yields ~100kN. If I had to take a guess, I'd say that the OP is just trying to get someone to do his homework.

By the way, no way the specific gravity of the Red Sea is 1,013. More like the Red Sea has a density of 1,013 kg/m^3 and a specific gravity of ~1.

Excellent work IloveMCAT!!!

That is a much easier way, which is the name of the game on a timed exam. You're going to kick the MCAT's butt.

 

[cardioonc]

Thanks for the replies. That helps a lot.

ilovemcat, thanks for your reasoning. Actually, this is not a homework question, this is a PR physics problem.