Estimate the change in the density of water in ocean at a depth of $400$ m below the surface. The density of water at the surface $= 1030$ kg $m^{- 3}$ and the bulk modulus of water $= 2 \times 10^{9}$ N $m^{- 2}$ .

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Solution

We know bulk modulus $k = \frac{V d p}{- d V} = \frac{ρ d p}{d ρ}$ where $d ρ$ = change in density $d p$ =change in pressure and $ρ$ =density of the
fluid.
Given $ρ_{w a t e r}$ =1030 kg/ $m^{3}$
Now pressure at surface of the ocean $P_{1}$ =1 atm=101325 Pa
and pressure at a height of 400 m below the surface of water $P_{2} = ρ g h$ where g=acceleration due to gravity and h= height =400 m
Thus putting the above values we get $P_{2}$ =1030 $\times$ 9.81 $\times$ 400=4041720 Pa
Thus $d p$ =4041720-101325 =39.40395 $\times 10^{5}$ Pa
Hence $d ρ$ = $\frac{ρ d p}{k}$
$= \frac{1030 (39.40 \times 10^{5)}}{2 \times 10^{9}}$
$d ρ = 2.029 k g / m^{3}$

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