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))`, (DeltaV)/(V) = - (Deltap)/(B)`
`m = rhoV = const.
`drho V + dV . Rho = 0 rArr (drho)/(rho) = - (dV)/(V)`
i.e. `(Deltarho)/(rho) = (Deltap)/(B) rArr (Deltarho)/(rho) = (1)/(100)`
`(1)/(100) = (hrhog)/(B)` [assuming `rho = const.`]
`h rho g = (B)/(100) = (1)/(100K) rArr h rho g = (1 xx 10^(5))/(100 xx 50 xx 10^(-5))`
`h = (10^(5))/(5000 xx 10^(-6) xx 1000 xx 10) = (100 xx 10^(3))/(50) = 2km`