Compare the densities of water at the surface and bottom of a lake 100 metre deep , given that the compressibility is `10^(-3)//22` per atmosphere and 1 atmosphere = `1.015 xx 10^(5)Pa`.

Explanation

Compressibility = `1/B=(10^(-3))/(22) = (1)/922 xx 10^(3) `per atm
`B = 22 xx 10^(3) atm = 22 xx 10^(3) xx 1.015 xx 10^(5) Pa`
Let V be the volume of 1 kg water at surface and `(V-Delta V)` be its volume at the bottom of lake 100m deep.
Increase in pressure `p = h rho g`
`=100 xx 10^(3) xx 9.8 Nm^(-2)`
`(Delta V)/(V) = p/B = (100 xx 10^(3) xx 9.8 )/(22 xx 10^(3) xx (1.015 xx 10^(5)))`
`("Density of water at the surface")/("density of water at the bottom") = (1//V)/(11//(V-Delta V))`
`=(V-Delta V)/(V) = 1-(Delta V)/(V) = 1 -(9.8)/(22xx1.015 xx 10^(3))`
`=0.99956`.