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

The volume of an air bubblebecomes three times as it rises from the bo

1.	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» Solution :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.