If the normal density of sea water is `1.00 g//cm^(3)`, what will be its density at a depth of 4km? Given compressibility of water = 0.00005 per atmosphere. 1 atmospheric pressure `=10^(6) dyn e//cm^(2), g=980 cm//s^(2)`.

Answer

`1.0245 g//c c`

Explanation

Here , `rho = 1g//c c ,`
compressibility, k = 0.0005 per atmosphere ` =(0.00005)/(10^6 dyn e //cm^2) = 5xx10^(-11)cm^2dyn e` Depth, `h = 5km = 5xx1000xx 100 cm`
` = 5xx10^5 cm` increase in pressrue at depth 5 km of water is `Delta P = h rho g = (5xx10^5)xx1xx980 dyn e//cm^2` Density of water at depth h is `rho' = rho (1+ k Delta P)`
` =1 [1+(5xx10^(-11))xx(5xx10^5)xx1xx980]`
` =1[1 +0.0245] =1.0245 g//c c`