Question

735 Pascal of pressure exerted on a block which is at a depth of 7metre inside
a ocean water. find density.​

Answer 1

Answer:

Explanation: Here `p = 80.0 atm = 80.0 xx 1.013 xx 10^(5)Pa`, Compressibilty , `1/B = 45.8 xx 10^(-11) Pa^(-1)`.

Density of water at surface , `rho = 1.03 xx 10^(3) kg m^(-3)`

Let `rho'` be the density of water at given depth. If V and V' are volumes of certain mass M of ocean water at surface and at a given depth, then `V = (M)/(rho) and V' =(M)/(rho')`

`:. `Change in volume , `Delta V = V - V' = M (1/(rho)-(1)/(rho'))`

`:.` Volumetric strain, `(Delta V)/(V) = M (1/(rho)-(1)/(rho')) xx (rho)/(M) = 1 -(rho)/(rho')` or `(Delta V)/(V) = 1-(1.03 xx 10^(3))/(rho') `...(i)

As, bulk modulus `B = (pV)/(Delta V)` or `(Delta V)/(V) = p/B`

`(Delta V)/(V)= (80.0 xx 1.013 xx 10^(5)) xx 45.8 xx 10^(-11) = 3.712 xx 10^(-3)`

Putting this value in (i) we get

`1-(1.30 xx 10^(3))/(rho') = 3.712 xx 10^(-3)` or `rho' = (1.30 xx 10^(3))/(1-3.712 z 10^(-3)) = 1.304 xx 10^(3) kg m^(-3)`.