
The approx. the depth of an ocean is 2700 m. The compressibility of water is $45.4 \times {10^{ - 11}}P{a^{ - 1}}$and the density of water is ${10^3}kg/{m^3}$. What fractional compression of water will be obtained at the ocean's bottom?
A. $1.0 \times {10^{ - 2}}$
B. $1.2 \times {10^{ - 2}}$
C. $1.4 \times {10^{ - 2}}$
D. $0.8 \times {10^{ - 2}}$
Answer
585.6k+ views
Hint: Use the expression of compressibility which is given as:
$\kappa = \dfrac{{\dfrac{{\Delta V}}{V}}}{{\Delta P}}$
Take $\Delta P$as $\rho gh$ (where, $\rho $is the density of water and $h$is depth of an ocean)
Take $g = 10m{s^{ - 2}}$
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:
$\kappa = \dfrac{{\dfrac{{\Delta V}}{V}}}{{\Delta P}} \cdots (1)$
where, $\kappa $is the compressibility
$\dfrac{{\Delta V}}{V}$is the fractional compression
$\Delta P$is change of pressure
From equation (1),
$
\Delta V = \kappa \times \Delta P \times V \\
\dfrac{{\Delta V}}{V} = \kappa \times \Delta P \cdots (2) \\
$
And $\Delta P = \rho gh \cdots \left( 3 \right)$
According to the question, it is given that:
$
\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)
$
\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 $g \simeq 10m{s^{ - 2}}$(approx.)
Gas dynamics is the study of the flows in which involves the change in density and temperature
$\kappa = \dfrac{{\dfrac{{\Delta V}}{V}}}{{\Delta P}}$
Take $\Delta P$as $\rho gh$ (where, $\rho $is the density of water and $h$is depth of an ocean)
Take $g = 10m{s^{ - 2}}$
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:
$\kappa = \dfrac{{\dfrac{{\Delta V}}{V}}}{{\Delta P}} \cdots (1)$
where, $\kappa $is the compressibility
$\dfrac{{\Delta V}}{V}$is the fractional compression
$\Delta P$is change of pressure
From equation (1),
$
\Delta V = \kappa \times \Delta P \times V \\
\dfrac{{\Delta V}}{V} = \kappa \times \Delta P \cdots (2) \\
$
And $\Delta P = \rho gh \cdots \left( 3 \right)$
According to the question, it is given that:
$
\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)
$
\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 $g \simeq 10m{s^{ - 2}}$(approx.)
Gas dynamics is the study of the flows in which involves the change in density and temperature
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