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# if normal density of sea water is 1g/cm-3,what is the density at a depth of 3km.[compressibility of water =0.00005 per atm, 1 atm=10^6dyne /cm2. please give step by step explanations

## 1 Answers

11 months ago

Here, ρ=1.00 g/cm^3

Compressibility "k" = 0.00005 per atm

= 0.00005 / 10^6 dyne/cm^2

= 5 × 10^−11 cm^2/dyne

Depth "h" = 3 km = 3 × 1000 × 100 cm

= 3 ×10^5 cm

Increases in pressure at depth 4 km of water is

ΔP = hρg = ( 3 x 10^5) × 1 × 980 dyne/cm^2

If ρ' is density at depth h, then

ρ' = ρ (1 + kΔP) = 1 [1 + (5×10^−11) × (3 × 10^5 × 980)]

ρ' = 1 +  (5 × 10^−11 x 2940 x 10^5

ρ' =  1 + 14,700 x 10^-11+5

ρ' = 1 + 14,700 x 10^-6

ρ' = 1 +1.4 x 10^4-6

ρ' = 1 + 1.4 x 10^-2

ρ' = 1 + 0.0014 = 1.014 g /cm^3

Thus the density of the sea water at 3 km depth is 1.014 g /cm^3

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