Calculate the apprroximate change in density of water in a lake at a depth of `400 m ` below the surface. The density of water at the surface id `1030 kg//m^(3)` and bulk modulus of water is `2xx10^(9) N//m^(2)`.

Answer

B, C

Explanation

`B=2xx10^9N/m^2`
` P_0=m/(V_0P_0)-m/V_d`
So, `p_d/p_0=V_0/V_d`…..i
`d=400m`
`P_0=1030kg/m^3`
vol. strain =`(V_0V_d)/V_0`
`B=((P_0-V_d)/v_0)`
`1-V_d/V_0=(P_0gh)/B`
`rarr V_d/V_0=(1-p_0(gd)/B)``…………ii`
Putting value of ii in equatioin i we get
`P_d/P_0=1/((1-p_0(gh)/B))`
`p_d=1/((1-p_0(gh)/B))P_0`
`=1.32 kg/m^3`
change in density `= p_d-p_0`
`=2kg/m^2`