__Question 9.13__:^{3}kg m

^{–3}?

__Solution__:*h*.

Pressure at the given depth,

*p*= 80.0 atm = 80 × 1.01 × 10

^{5}Pa

Density of water at the surface,

*ρ*

_{1 }= 1.03 × 10

^{3}kg m

^{–3}

Let

*ρ*

_{2}be the density of water at the depth

*h*.

Let

*V*

_{1}be the volume of water of mass

*m*at the surface.

Let

*V*

_{2}be the volume of water of mass

*m*at the depth

*h*.

Let Δ

*V*be the change in volume.

ΔV = V

_{1}- V

_{2}

= m [ (1/

*ρ*

_{1}) - (1/

*ρ*

_{2}) ]

∴ Volumetric strain = ΔV / V

_{1}

= m [ (1/

*ρ*

_{1}) - (1/

*ρ*

_{2}) ] X (

*ρ*

_{1}/ m)

ΔV / V

_{1}= 1 - (ρ

_{1}/

*ρ*

_{2}) ......(i)

Bulk modulus, B = pV

_{1}/ ΔV

ΔV / V

_{1}= p / B

Compressibility of water = (1/B) = 45.8 X 10

^{-11}Pa

^{-1}

∴ ΔV / V

_{1}= 80 X 1.013 X 10

^{5}X 45.8 X 10

^{-11}= 3.71 X 10

^{}

^{-3}....(ii)

For equations (

*i*) and (

*ii*), we get:

1 - (ρ

_{1}/

*ρ*

_{2}) = 3.71 X 10

^{-3}

*ρ*

_{2}= 1.03 X 10

^{3}/ [ 1 - (3.71 X 10

^{-3}) ]

= 1.034 X 10

^{3}kg m

^{-3}

Therefore, the density of water at the given depth (

*h*) is 1.034 × 10

^{3}kg m

^{–3}.

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