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 $k g m^{- 3}$ and the bulk modulus of water = $2 \times 10^{9} N m^{- 2}$ .

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Solution

B = $2 \times 10^{9} N m^{2}$

$P_{0} = \frac{m}{V_{0} P_{0}} = \frac{m}{V_{d}}$

So, $\frac{P_{d}}{P_{0}} = \frac{V_{0}}{V_{d}}$ ....(i)

d= 400 m

$P_{0} = 1030 k g / m^{3}$

vol. strain = $\frac{V_{0} V_{d}}{V_{0}}$

$B = (\frac{P_{0} - V_{d}}{v_{0}})$

$\Rightarrow 1 - \frac{V_{d}}{V_{0}} = \frac{P_{0} g h}{B}$

$\Rightarrow \frac{V_{d}}{V_{0}} = (1 - p_{0} \frac{g h}{B})$ ....(ii)

Putting value of (ii) in equation (i), we get

$\frac{P_{d}}{P_{0}} = \frac{1}{(1 - p_{0} \frac{g h}{B})}$

$p_{d} = \frac{1}{(1 - p_{0} \frac{g h}{B})} P_{0}$

= $1032 k g / m^{3}$

change in density = $p_{d} - p_{0}$

= $2 k g / m^{2}$

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