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Bell shrinking due to water pressure

  1. Oct 31, 2009 #1
    A bell of diameter 3m and bulk modulus of 14x1010N/m2 is toss into the ocean. By how much did the diameter decrease as it sank to a depth of 10km? The bell is assumed to be a sphere of diameter 3m.

    Well o what i did was to calculate the pressure at the depth of 10km so P=Po+[tex]\rho[/tex]gh which gives me P=1.0135x105+1000x9.81x10000=9.82x107. Then Bulk modulus=[tex]\frac{P}{V/Vi}[/tex] where V is change in volume and Vi is the initial volume. This gives me change in diameter to be 0.267m however the answer provided is 0.722mm :cry: What is wrong ??
     
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  3. Nov 1, 2009 #2

    rl.bhat

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    Re: Pressure!!

    Show your calculations.
     
  4. Nov 1, 2009 #3
    Re: Pressure!!

    Hello rl! My working is in the post? Okie i guess my post is too messy haha sorry about that.

    Okie here is what i did


    First i calculate the pressure at the depth of 10 km including the atmospheric pressure Po which is taken to be 1.013x105 Pa.

    P=Po + [tex]\rho[/tex]gh
    P=1.013x105 + 1000 x 9.81 x 10000 = 9.82 x 107 Pa

    So using the formula of bulk modulas,

    Bulk modulas = [tex]\frac{P}{V/Vi}[/tex]
    14x1010=[tex]\frac{P}{V/Vi}[/tex]

    Since formula for both volume are the same V/Vi is reduced to r3/ri3 where r is the change in radius and ri is the initial radius.

    Using 14x1010=[tex]\frac{P}{V/Vi}[/tex] and the P i found previously, i got the change in radius which is r to be 0.267m

    Yup basically that's how i did it :biggrin:
     
    Last edited: Nov 1, 2009
  5. Nov 1, 2009 #4

    rl.bhat

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    Re: Pressure!!

    Vi = 4/3*π*R^3
    V' = 4/3*π*( R - ΔR )^3
    V/Vi = (Vi - V')/Vi = [R^3 - (R-ΔR)^3]/R^3
    Expand the brackets and simplify to find V/Vi.
     
  6. Nov 1, 2009 #5
    Re: Pressure!!

    I don't get it. If we are looking for the change in radius why i can't just find the change in volume and from there calculate the change in diameter?
     
  7. Nov 1, 2009 #6

    rl.bhat

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    Re: Pressure!!

    Find the difference in original volume and compressed volume.
    Change in volume = Bulk modulus*Pressure* Original volume.
     
  8. Nov 1, 2009 #7
    Re: Pressure!!

    I am assuming you meant Change in volume = Bulk modulus*Pressure*Original volume? How did you get this? From my working I get Change in volume =[tex]\frac{Pressure x Original volume}{Bulk Modulus}[/tex] ??
     
  9. Nov 1, 2009 #8

    rl.bhat

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    Re: Pressure!!

    Yes. You are right. It is typo.
     
  10. Nov 1, 2009 #9
    Re: Pressure!!

    So you mean that's the way to do it? But my answer is different from the answer provided
     
  11. Nov 1, 2009 #10

    rl.bhat

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    Re: Pressure!!

    Show your calculations instead of answer.
     
  12. Nov 2, 2009 #11
    Re: Pressure!!

    I am afraid i don't get you, this is my calculation and i can't figure out what's wrong with this and why do you keep asking me to post this?

    [tex]\Delta[/tex]r3=[tex]\frac{9.82*10^7*1.5^3}{140*10^9}[/tex]
    [tex]\Delta[/tex]r=0.1333m [tex]\Rightarrow[/tex] [tex]\Delta[/tex]d=0.2666m
     
  13. Nov 2, 2009 #12

    rl.bhat

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    Re: Pressure!!

    Sorry. Your calculation is not correct.
    Vi = 4/3π*r^3
    Vf = 4/3π*(r - Δr)^3
    Chane in volume =ΔV = Vi - Vf = ...........?
    Then ΔV = P*Vi/B gives you Δr.
     
  14. Nov 2, 2009 #13
    Re: Pressure!!

    ΔV = Vi - Vf
    Δr = ri3-rf3
    ΔV = P*Vi/B

    Vi - Vf = (P/B)*Vi
    [tex]\frac{4}{3}[/tex][tex]\Pi[/tex](ri3-rf3)=(P/B)[tex]\frac{4}{3}[/tex][tex]\Pi[/tex]ri3

    (ri3-rf3)=(P/B)[/tex]ri3
    Δr3 = (P/B)ri3
    Δr3 = [tex]
    \frac{9.82*10^7*1.5^3}{140*10^9}
    [/tex]
    Δr = 0.133 [tex]
    \Rightarrow
    [/tex] Δd=0.266

    Whats wrong in my calculation?
     
  15. Nov 2, 2009 #14

    rl.bhat

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    Re: Pressure!!

    Δr = (ri - rf)
    Δr^3 is not equal to ri^3 - rf^3
    In algebra
    (a^3 - b^3) = (a - b)(a^2 + ab + b^2)
     
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