# Q13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03 \times 103 Kg$ ?

S Sayak

Water at the surface is under 1 atm pressure.

At the depth, the pressure is 80 atm.

Change in pressure is $\Delta P=79\ atm$

Bulk Modulus of water is $B=2.2\times 10^{9}Nm^{-2}$

$\\B=-\frac{P}{\frac{\Delta V}{V}}\\ \frac{\Delta V}{V}=-\frac{P}{B} \\\frac{\Delta V}{V}=-\frac{79\times 1.013\times 10^{5}}{2.2\times 10^{9}}\\ \frac{\Delta V}{V}=-3.638\times 10^{-3}$

The negative sign signifies that for the same given mass the Volume has decreased

The density of water at the surface $\rho=1.03\times 10^{3}\ kg\ m^{-3}$

Let the density at the given depth be $\rho '$

Let a certain mass occupy V volume at the surface

$\\\rho=\frac{m}{V}\\$

$\\\rho'=\frac{m}{V+\Delta V}\\$

Dividing the numerator and denominator of RHS by V we get

$\\\rho'=\frac{\frac{m}{V}}{1+\frac{\Delta V}{V}}\\ \rho'=\frac{\rho }{1+\frac{\Delta V}{V}}\\ \rho'=\frac{1.03\times 10^{3}}{1+(-3.638\times 10^{-3})}\\ \rho'=1.034\times 10^{3}\ kg\ m^{-3}$

The density of water at a depth where pressure is 80.0 atm is $1.034\times 10^{3}\ kg\ m^{-3}$.

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