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When an air bubble rises from the bottom...

When an air bubble rises from the bottom of a lake to the upper surface , its diameter increases from 1mm to 2mm. IF the atmospheric pressure is 76 cmHg,calculate the depth of the lake. Density of mercury is `13.6g.cm^-3`.

Text Solution

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Let the depth of the lake be h cm
Pressure on the air bubble ,at the bottom of the lake,`p_1`=atmospheric pressure + pressure of water at depth `h=(76 times 13.6 times 981+h times 1 times 981)dyn.cm^-2`, and pressure at the surface of the lake `p_2`= atmospheric pressure =`76 times 13.6 times 981 dyn.cm^-2`
Volume of the bubble :
at the bottom ,`V_1=4/3 pi (1/20)^3 cm^3,`
at the top,`V_2=4/3 pi (1/10)^3 cm^3`
Using Boyle.s law : `p_1V_1=p_2V_2`, we get
`(htimes1times981+76times13.6times981)times4/3pi(1/20)^3=76 times13.6times981times4/3pi(1/10)^3`
or,`(h+76times13.6)times1/6=76times13.6`
or,`h=76times13.6times8-76times13.6`
`=7235.2cm`
`therefore` This lake is 7235.2 cm or 72.352 m deep.
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