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Question: The approx. the depth of an ocean is 2700 m. The compressibility of water is \(45.4 \times {10^{ - 1...

The approx. the depth of an ocean is 2700 m. The compressibility of water is 45.4×10−11Pa−145.4 \times {10^{ - 11}}P{a^{ - 1}}and the density of water is 103kg/m3{10^3}kg/{m^3}. What fractional compression of water will be obtained at the ocean's bottom?
A. 1.0×10−21.0 \times {10^{ - 2}}
B. 1.2×10−21.2 \times {10^{ - 2}}
C. 1.4×10−21.4 \times {10^{ - 2}}
D. 0.8×10−20.8 \times {10^{ - 2}}

Explanation

Solution

Use the expression of compressibility which is given as:
κ=ΔVVΔP\kappa = \dfrac{{\dfrac{{\Delta V}}{V}}}{{\Delta P}}
Take ΔP\Delta Pas ρgh\rho gh (where, ρ\rho is the density of water and hhis depth of an ocean)
Take g=10ms−2g = 10m{s^{ - 2}}

Complete step-by-step solution :The property of reducing to small space by pressure is called compressibility. This property is a consequence of porosity, and the change of bulk comes from the particles being brought closer together by the pressure.
The expression of compressibility is given as:
κ=ΔVVΔP⋯(1)\kappa = \dfrac{{\dfrac{{\Delta V}}{V}}}{{\Delta P}} \cdots (1)
where, κ\kappa is the compressibility
ΔVV\dfrac{{\Delta V}}{V}is the fractional compression
ΔP\Delta Pis change of pressure
From equation (1),
ΔV=κ×ΔP×V ΔVV=κ×ΔP⋯(2)  \Delta V = \kappa \times \Delta P \times V \\\ \dfrac{{\Delta V}}{V} = \kappa \times \Delta P \cdots (2) \\\
And ΔP=ρgh⋯(3)\Delta P = \rho gh \cdots \left( 3 \right)
According to the question, it is given that:
ρ=103kg/m3 g=10ms−2 h=2700m κ=45×10−11Pa−1  \rho = {10^3}kg/{m^3} \\\ g = 10m{s^{ - 2}} \\\ h = 2700m \\\ \kappa = 45 \times {10^{ - 11}}P{a^{ - 1}} \\\
From equation (2) and (3)
ΔVV=κ×ρgh ⇒ΔVV=45×10−11×103×10×2700 ⇒ΔVV=1215000×10−8 ∴ΔVV=1.2×10−2  \dfrac{{\Delta V}}{V} = \kappa \times \rho gh \\\ \Rightarrow \dfrac{{\Delta V}}{V} = 45 \times {10^{ - 11}} \times {10^3} \times 10 \times 2700 \\\ \Rightarrow \dfrac{{\Delta V}}{V} = 1215000 \times {10^{ - 8}} \\\ \therefore \dfrac{{\Delta V}}{V} = 1.2 \times {10^{ - 2}} \\\
So, the option (B) is the correct answer.

Note:- The compressibility of liquid is the calculation of the change in density that is to be produced in the liquid by making a change in pressure specified. Fluids have very low compressibility.
Use the formula of pressure to find out the fractional compression.
Take the value of g≃10ms−2g \simeq 10m{s^{ - 2}}(approx.)
Gas dynamics is the study of the flows in which involves the change in density and temperature