Question

If the normal density of sea water is 1g/cm^3.What will be its density at a depth of 3km? Given, compressibility of water is 0.0005per atm(1 atm= 10^6 dyne/cm^2)​

Answer 1

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

Explanation:

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

Answer 2

The density at depth is 1.147 g/cm³.

Explanation:

Given that,

Density of sea water$\rho=1\ g/cm^3$

Depth = 3 km

$h = 3\times10^{5}\ m$

Compressibility $k = 0.0005\ atm$

$k=\dfrac{0.0005}{10^6}\ cm^2/dyne$

$k=5\times10^{-10}\ cm^2/dyne$

We need to calculate the increases in pressure at depth 3 km of water

Using formula of changing pressure

$\Delta P=h\rho g$

Put the value into the formula

$\Delta P=3\times10^{5}\times1\times980$

$\Delta P=294000000$

We need to calculate the density at depth

Using formula of density

$\rho'=\rho(1+k\Delta P)$

$\rho'=1(1+5\times10^{-10}\times294000000)$

$\rho'=1.147\ g/cm^3$

Hence, The density at depth is 1.147 g/cm³.

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Topic : density

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