# 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 × 103 Kg M–3? - Physics

What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?

#### Solution 1

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa

Density of water at the surface, ρ= 1.03 × 103 kg m–3

Let ρ2 be the density of water at the depth h.

Let V1 be the volume of water of mass m at the surface.

Let V2 be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = V1 - V2

=m(1/rho_1 - 1/rho_2)

:."Volumetric strain" = (triangle V)/V_1

= m(1/rho_1 - 1/rho_2) xx rho_1/m

:.triangleV/V_1 = 1 - rho_1/rho_2   ....(i)

Bulk modulus, B = (rhoV_1)/(triangleV)

(triangleV)/V_1 = rho/B

Compressibity of water = 1/B = 45.8 xx 10^(-11) Pa^(-1)

:.(triangleV)/V_1 = 80xx 1.013 xx 10^5 xx 45.8 xx 10^11 = 3.71 xx 10^(-3)   ...(ii)

For equation i and ii we get

1 - rho_1/rho_2 = 3.71 xx 10^(-3)

rho_2 = (1.03 xx 10^3)/(1-(3.71xx10^(-3)))

=1.034 xx 10^3 kg m^(-3)

Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.

#### Solution 2

Compressibility of water k = 1/B = 45.8 xx 10^11 Pa^(-1)

Change in pressure, trianglep = 80 atm - 1 atm

= 79 atm = 79 xx 1.013 xx 10^5 Pa

Density of water at the surface

rho = 1.03 xx 10^3 kg m^(-3)

As  B= (trianglep.V)/(triangleV) or (triangleV)/V = (trianglep)/B = trianglep xx 1/B = trianglep xx k

or triangleV/V = 79 xx 1.013 xx 10^5 xx 45.8 xx 10^(-11) = 3.665 xx 10^(-5)

Now (triangleV)/V = ((M/rho)-(M/(rho')))/(M/rho) = 1 - rho/rho'

or rho/(rho') = 1 - (triangleV)/V

or rho' = rho/(1-(triangleV/V))

or rho' = (1.03 xx 10^3)/(1-3.665 xx 10^(-3)) = (1.03 xx 10^3)/0.996

= 1.034 xx 10^3 "kg/m"^3

Concept: Elastic Moduli - Bulk Modulus
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#### APPEARS IN

NCERT Class 11 Physics
Chapter 9 Mechanical Properties of Solids
Q 13 | Page 244