IIT JEE , AIEEE Forum - Physics , Mechanics

LAMPARD

When a large bubble rises from of a lake,its radius doubles.The atm. pressure is equal to that of a column of liquid A of height 70cm.The depth of lake is(sp. gravity of liquid A is 15)

computer001

EDITED:
EDITED:
oh im gettin 10.5 too not 5.25 as i said previously

LAMPARD

Sry dude,even i was gettin 10.5,but ans. is sumthing different...73.5m

anandghegde

ya ans is 73.5m

use
P_1V_1 = P_2V_2

subscripts 1 denote the surface at 2 denote the bottom.

as radius doubles, V_1 = 8V_2

$P_2 = P_1 + ho_{water} gh$

where h is the depth of water

Substitute in the eq and solve you get the correct ans

akhil_o

edited

pressure at surface=P0=atmospheric pressure
at the bottom of lake,pressure =8 times, as volume is 1/8
so we have P=8P0
now we have
the difference in pressures due to liquid level=7P0
h*1000*10=0.7 m of liquid A*7
h*10000=0.7*15000*10*7
73 m

akhil_o

anand is right!

vacuumhead_pratyush

the excess pressure in the bubble (when at the bottom) is 2T/r ,while at the top it is 2T/2r = T/r .

so dont u think that the height also depends on "T" ???????

vacuumhead_pratyush

some one plzzzzzz reply................

coolvignesh

yeah.. considering this process as isothermal..
what anand did was completely right....
only thing is that u've to convert everything into its SI units...
PV = constant is asolutely fine and best formula to use here..
by that calculate pressure at depth and at surface...
after that, the difference in pressure equals hpg.. where p is density of liquid, h is depth of lake and g is acceleration due to gravity

coolvignesh

73.5 m is right answer

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