The volume of an air bubblebecomes three times as it rises from the bootom of a lake to its surface. Assuming temperature to be constant and atmospheric pressure to be 75 cm of Hg and the density of water to be `1//10` of the density of the mercury, the depth of the lake is

Answer

4.79 mL

Explanation

The pressure on the bubble at the bottom of the tank = Atmospheric pressure+ Pressure of 20 feet water column
`=75+(20xx12xx2.54)/(13.6)`
`( :. "density of Hg " 13.6 g cm^(-3))`
=119.8 cm of Hg
According to Boyle.s law
`P_1 V_1 = P_2 V_2`
In the present case,
`P_1 119.8 " cm Hg," V_1=3 mL ,P_2=75 " cm Hg, "`
`:. 119.8xx3=75xxV_2 or V_2= 4.79 mL`
Hence, the valume at the surface of water is 4.79 mL.