Bulk Modulus Problem: Sphere of Brass Diameter Change

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SUMMARY

The discussion revolves around calculating the diameter change of a solid brass sphere (bulk modulus of 14.0 x 1010 N/m2) when submerged to a depth of 1.0 km in the ocean. The gauge pressure at this depth is calculated as 9.81 x 106 N/m2. The initial volume of the sphere is determined to be 14.14 m3, leading to a volume change of 9.906 x 10-4 m3. The correct approach to find the new radius involves adjusting the volume calculation to account for the change, ultimately yielding a diameter decrease of 0.0721 mm instead of the incorrectly calculated 12.36 cm.

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  • Familiarity with volume formulas for spheres (V = 4/3πr3).
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Homework Statement



A solid sphere of brass (bulk modulus of 14.0*10^10 N/m^2) with a diameter of 3.00 m is thrown into the ocean. By how much does the diameter of the sphere decrease as it sings to a depth of 1.0 km?


Homework Equations



Gauge pressure = density(water)*g*h
Bulk Modulus = -(ΔP/ΔV)*V


The Attempt at a Solution



I tried solving for ΔP = gauge pressure = 1000*9.81*1000=9.81*10^6.

I then isolated ΔV = -ΔP*V/B

V = 4/3*pi*r^3 = 14.14 m^3

I plugged in and got 9.906*10^-4 m^3.

I then solved for r in v=4/3*pi*r^3 (where I took V to be equal to 9.906*10^-4).

The answer I am getting for r is 6.18cm, which would be a difference of 12.36 cm when considering the diameter. But the answer is 0.0721 mm.

Any help would be appreciated. Thanks!
 
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Your error is here: "I then solved for r in v=4/3*pi*r^3 (where I took V to be equal to 9.906*10^-4)". You should have used v = 14.14 - 9.906*10^-4, and you need to figure out how to get the change in r without roundoff error.

Chet
 

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