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Volume-of-a-bubble-is-3-times-larger-when-it-reaches-the-surface-from-the-bottom-of-the-lake-What-is-the-depth-of-the-lake-A-10-m-D-40-m-B-20-m-E-50-m-C-30-m-




Question Number 13695 by Joel577 last updated on 22/May/17
Volume of a bubble is 3 times larger  when it reaches the surface from the bottom  of the lake.  What is the depth of the lake?    (A) 10 m             (D) 40 m  (B) 20 m             (E) 50 m  (C) 30 m
Volumeofabubbleis3timeslargerwhenitreachesthesurfacefromthebottomofthelake.Whatisthedepthofthelake?(A)10m(D)40m(B)20m(E)50m(C)30m
Commented by Joel577 last updated on 22/May/17
I use  p_(surface)  = 1 atm  p_(bottom)  . V_(bottom)  = p_(surface)  . V_(surface)   p_(bottom)  . V = 1 . 3V  p_(bottom)  = 3 atm    And now I′m stuck
Iusepsurface=1atmpbottom.Vbottom=psurface.Vsurfacepbottom.V=1.3Vpbottom=3atmAndnowImstuck
Commented by mrW1 last updated on 22/May/17
p_(bottom)  −p_(surface)  = 3−1 = 2 atm  =20m water depth
pbottompsurface=31=2atm=20mwaterdepth
Answered by ajfour last updated on 22/May/17
P_0 V_f =(P_0 +ρgh)V_i   h=((P_0 ((V_f /V_i )−1))/(ρg))        and as (V_f /V_i )=3   h =((2P_0 )/(ρg)) =((2(1.013×10^5 N/m^2 ))/((1000kg/m^3 )(9.806m/s^2 )))   h=((202.6)/(9.8))m ≈((29)/(1.4))m ≈ 20m .
P0Vf=(P0+ρgh)Vih=P0(VfVi1)ρgandasVfVi=3h=2P0ρg=2(1.013×105N/m2)(1000kg/m3)(9.806m/s2)h=202.69.8m291.4m20m.
Commented by Joel577 last updated on 22/May/17
thank you very much
thankyouverymuch

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