Question

Find the depth of lake at which density of water is `1%` greater than that at the surface. Given compressibility `k= 50 xx10^(-6) atm^(-1)`.

Answer 1

`B=(DeltaP)/(-DeltaV//V)implies(DeltaV)/(V)=-(DeltaP)/(B)`
We know `P=P_(atm)+hrhog` and `=rhoV`=constant
`drhoV+dVrho=0implies(drho)/(rho)=-(dV)/(V)i.e., (Deltarho)/(rho)=(DeltaP)/(B)implies(Deltarho)/(rho)=(1)/(100)=(hrhog)/(B)`
[assuming `rho=` constant], `hrhog=(B)/(100)=(1)/(100K)implieshrhog=(1xx1xx10^(5))/(100xx50xx10^(-6))`
`h=(10^(5))/(5000xx10^(-6)xx1000xx10)=(100xx10^(3))/(50)m=2km`