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### Mechanical Properties of solids NCERT Solutions Class 11 Physics - Solved Exercise Question 9.13

Question 9.13:
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?
Solution:
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, ρ1 = 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/ρ1) - (1/ρ2) ]
∴ Volumetric strain = ΔV / V1
= m [ (1/ρ1) - (1/ρ2) ] X (ρ1 / m)
ΔV / V1 = 1 - (ρ1/ρ2)     ......(i)
Bulk modulus, B = pV1 / ΔV
ΔV / V1 = p / B
Compressibility of water = (1/B) = 45.8 X 10-11 Pa-1
∴ ΔV / V1 = 80 X 1.013 X 105 X 45.8 X 10-11  =  3.71 X 10-3    ....(ii)
For equations (i) and (ii), we get:
1 - (ρ1/ρ2)   =   3.71 X 10-3
ρ2 = 1.03 X 103 / [ 1 - (3.71 X 10-3) ]
= 1.034 X 103 kg m-3
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.