Question

The approximate depth of an ocean is 2700m. 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 bottom of the ocean? $(Take,g=10m{{s}^{-2}})$
$\begin{aligned} & (A)1.2\times {{10}^{-2}} \\\ & (B)1.4\times {{10}^{-2}} \\\ & (C)0.8\times {{10}^{-2}} \\\ & (D)1.0\times {{10}^{-2}} \\\ \end{aligned}$

Answer 1

We will use the formula for Compressibility of a liquid as water is also a liquid. The formula states that, the compressibility of any liquid is equal to, the relative change in its volume(that is fractional compression of the liquid) divided by the difference in pressure. We shall use this to calculate the fractional compression of the liquid in terms of other known terms.

Complete answer:
Let us first define some terms that are to be used in our equations at later stages.
Let the compressibility factor of water be denoted by $\kappa$. Then, the value of $\kappa$is equal to:
$\Rightarrow k=45.4\times {{10}^{-11}}P{{a}^{-1}}$
Let the volume of water in the ocean be denoted by $V$ and the change in volume due to compression be given by $\vartriangle V$ .
Then, the fraction of compressed water will be written as:
$=\dfrac{\vartriangle V}{V}$
Now, the difference in pressure (say $\vartriangle P$) at the surface of the ocean and the bottom will be written as:
$\Rightarrow \Delta P=\rho gh$
Here,
$\begin{aligned} & \Rightarrow \rho =1000kg{{m}^{-3}} \\\ & \Rightarrow g=10m{{s}^{-2}} \\\ & \Rightarrow h=2700m \\\ \end{aligned}$
Now, the compressibility of a liquid is given by the following formula:
$\Rightarrow \kappa =\dfrac{\left( \dfrac{\vartriangle V}{V} \right)}{\vartriangle P}$
On rearranging terms, we get:
$\Rightarrow \dfrac{\vartriangle V}{V}=\kappa \times \vartriangle P$
Putting the values of these terms in the right-hand side of the equation, we get:
$\begin{aligned} & \Rightarrow \dfrac{\vartriangle V}{V}=45.4\times {{10}^{-11}}\times 1000\times 10\times 2700 \\\ & \Rightarrow \dfrac{\vartriangle V}{V}=1.22\times {{10}^{-2}} \\\ & \Rightarrow \dfrac{\vartriangle V}{V}\approx 1.2\times {{10}^{-2}} \\\ \end{aligned}$
Hence, the fractional compression of water that will be obtained at the bottom of the ocean is equal to $1.2\times {{10}^{-2}}$.

Hence, option (A) is the correct option.

Note:
The compressibility factor of a liquid is generally very low. This is because of the very strong intermolecular force of attraction in liquids. Even in the above problem, the fractional compression was just about 1.2 percent at such a large depth of 2700 meters. This shows that liquids are highly incompressible.