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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-

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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
$$\mathrm{Volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{bubble}\:\mathrm{is}\:\mathrm{3}\:\mathrm{times}\:\mathrm{larger} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bottom} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{lake}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{depth}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lake}? \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:\mathrm{10}\:\mathrm{m}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{40}\:\mathrm{m} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{20}\:\mathrm{m}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:\mathrm{50}\:\mathrm{m} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{30}\:\mathrm{m} \\ $$
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
$$\mathrm{I}\:\mathrm{use}\:\:{p}_{\mathrm{surface}} \:=\:\mathrm{1}\:\mathrm{atm} \\ $$$${p}_{\mathrm{bottom}} \:.\:{V}_{\mathrm{bottom}} \:=\:{p}_{\mathrm{surface}} \:.\:{V}_{\mathrm{surface}} \\ $$$${p}_{\mathrm{bottom}} \:.\:{V}\:=\:\mathrm{1}\:.\:\mathrm{3}{V} \\ $$$${p}_{\mathrm{bottom}} \:=\:\mathrm{3}\:\mathrm{atm} \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{now}\:\mathrm{I}'\mathrm{m}\:\mathrm{stuck} \\ $$
Commented by mrW1 last updated on 22/May/17
p_(bottom) −p_(surface) = 3−1 = 2 atm =20m water depth
$${p}_{\mathrm{bottom}} \:−{p}_{\mathrm{surface}} \:=\:\mathrm{3}−\mathrm{1}\:=\:\mathrm{2}\:\mathrm{atm} \\ $$$$=\mathrm{20}{m}\:{water}\:{depth} \\ $$
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 .
$${P}_{\mathrm{0}} {V}_{{f}} =\left({P}_{\mathrm{0}} +\rho{gh}\right){V}_{{i}} \\ $$$${h}=\frac{{P}_{\mathrm{0}} \left(\frac{{V}_{{f}} }{{V}_{{i}} }−\mathrm{1}\right)}{\rho{g}}\:\:\:\:\:\:\:\:{and}\:{as}\:\frac{{V}_{{f}} }{{V}_{{i}} }=\mathrm{3} \\ $$$$\:{h}\:=\frac{\mathrm{2}{P}_{\mathrm{0}} }{\rho{g}}\:=\frac{\mathrm{2}\left(\mathrm{1}.\mathrm{013}×\mathrm{10}^{\mathrm{5}} {N}/{m}^{\mathrm{2}} \right)}{\left(\mathrm{1000}{kg}/{m}^{\mathrm{3}} \right)\left(\mathrm{9}.\mathrm{806}{m}/{s}^{\mathrm{2}} \right)} \\ $$$$\:{h}=\frac{\mathrm{202}.\mathrm{6}}{\mathrm{9}.\mathrm{8}}{m}\:\approx\frac{\mathrm{29}}{\mathrm{1}.\mathrm{4}}{m}\:\approx\:\mathrm{20}{m}\:. \\ $$
Commented by Joel577 last updated on 22/May/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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