The mercury in a barometer stands at 75cm by the side of a freshwater lake An air bubble at - Physics - Mechanical Properties Of Fluids

Question

The mercury in a barometer stands at 75cm by the side of a
freshwater lake An air bubble at the bottom of the lake, when rises
to the surface it's volume becomes three times it's volume at the
bottom If specific gravity of mercury is 40/3 then the depth of the
lake is? ( Ans is given 20m but please explain how?)

Vaibhav T. · Accepted Answer

Dear Student,

Let the depth of lake be h
Atmospheric pressure=75 cm of Hgspecific gravity=density of Hgdensity of water403=density of Hg1  
 because density of water=1 g/cm3density of Hg=403 g/cm3Pressure at the bottom of the lake,P1=Atmospheric pressure+pressure due to water column in lakeP1=75×403×g+h×1×gP1=g1000+h 
 (1)Volume of bubble at botton, V1=VVolume of bubble at surface, V2=3VPressure at surface, P2=Atmospheric pressureP2=75×403×g=1000×g 
 (2)From Boyle's law,P1V1=P2V2g1000+h×V=1000×g×3V1000+h=3000h=3000-1000h=2000 cm=20 m