An air bubble of diameter 1 mm is formed at the bottom of a lake and when it rises to the surface of the water, its diameter becomes 2 mm. If the atmospheric pressure is 760 cm of mercury, then find the depth of the lake. (Density of mercury = `13.6g*cm^(-3)`)
Text Solution
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Pressure at the bottom of the lake = `(76xx13.6xx980+hxx1xx980)dyn*cm^(-2)` [h = depth of the lake, p = atmospheric pressure = `76xx13.6xx980dyn*cm^(-2)`] Volume of the bubble at the bottom of the lake `=4/3pi(0.05)^(3)cm^(3)` and volume of the bubble at the surface of the lake = `4/3pi(0.1)^(3)cm^(3)`. If the temperature of the lake water is uniform, then pV = constant. `therefore" "(76xx13.6+h)xx980xx4/3pi(0.05)^(3)` `=76xx13.6xx980xx4/3pi(0.1)^(3)` or, `(76xx13.6+h)(0.05)^(3)=76xx13.6xx(0.1)^(3)` `therefore" "h=7235cm=72.35m`.
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