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The volume of an air bubble becomes 8 times the original volume in rising from the bottom of a lake to its surface. If the barometric height is 0.76 m of Hg (density of Hg is 13.6g/cm3 and g=9.8 m/s2), what is the depth of the lake?Aylinder hand |
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Answer» 72.35 mExplanation:SINCE volume of bubble become 8 times on reaching the surface.Therefore pressure on surface of lake 1 / 8 times pf pressure at the bottom.By BOYLE's law pressure at bottom of the lake = 8 P [ P = pressure at surface. ] = > Pressure due to column of liquid = 8 P - P = 7 PP = 7 × 0.76 × 13.6 × 10³ × 9.8 N / m²Now we know : P = ρ h gDensity of mercury ρ = 13.6 × 10³ kg / m³= > h = ( 7 × 0.76 × 13.6 × 10³ × 9.8 ) / ρ G= > h = ( 7 × 0.76 × 13.6 × 10³ × 9.8 ) / 10³ × 9.8= > h = 72.35 mTherefore , depth of the lake is 72.35 m. |
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