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Mechanical Properties of Solids

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Question : 13 of 21
Marks: +1, -0
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−31.03 \times 10^{3} \, \text{kg}\, \text{m}^{-3}? Compressibility of water is 45.8×10−11 Pa−1.45.8 \times 10^{-11} \, \text{Pa}^{-1}.
Solution:  
Here, P=80.0 atm.P=80.0 \, \text{atm}.
=80.0×1.013×105 Pa=80.0 \times 1.013 \times 10^{5} \, \text{Pa}
Compressibility, 1B=45.8×10−11 Pa−1\frac{1}{B}=45.8 \times 10^{-11} \, \text{Pa}^{-1}
Density of water at surface, ρ=1.03×103 kg m−3\rho=1.03 \times 10^{3} \, \text{kg} \, \text{m}^{-3}
Let ρ′ be the density of water at the given depth. If V and V′ are volume of certain mass M of ocean water at surface and at a given depth, then
V=MρV=\frac{M}{\rho} and V′′=Mρ′′V''=\frac{M}{\rho''}
∴\therefore Change in volume, ΔV=V−V′′=M(1ρ−1ρ′′)\Delta V=V-V''=M\left(\frac{1}{\rho}-\frac{1}{\rho''}\right)
∴\therefore Volumetric strain, ΔVV=M(1ρ−1ρ′)×ρM=1−ρρ′\frac{\Delta V}{V}=M\left(\frac{1}{\rho}-\frac{1}{\rho'}\right)\times \frac{\rho}{M}=1-\frac{\rho}{\rho'}
or ΔVV=1−1.03×103ρ′… (i)\frac{\Delta V}{V}=1-\frac{1.03 \times 10^{3}}{\rho'}\ldots\ (i)
As, Bulk modulus B=PVΔVB=\frac{PV}{\Delta V} or ΔVV=PB\frac{\Delta V}{V}=\frac{P}{B}
∴ΔVV=(80.0×1.013×105)×45.8×10−11\therefore \frac{\Delta V}{V}= (80.0 \times 1.013 \times 10^{5}) \times 45.8 \times 10^{-11}
=3.712×10−3=3.712 \times 10^{-3}
Putting this value in (i) we get
1−1.30×103ρ′=3.712×10−31-\frac{1.30 \times 10^{3}}{\rho'}=3.712 \times 10^{-3}
or ρ′=1.30×1031−3.712×10−3\rho'=\frac{1.30 \times 10^{3}}{1-3.712 \times 10^{-3}}
=1.034×103 kg m−3=1.034 \times 10^{3} \, \text{kg} \, \text{m}^{-3}
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