Question

The approximate depth of an ocean is $2700\text{m}$. The compressibility of water is $50\times {10}^{-11}{\text{Pa}}^{-1}$ and density of water is ${10}^3{\text{kg/m}}^3$. What fractional compression of water will be obtained at the bottom of the ocean?

• A. $1.5\times {10}^{-2}$
• B. $1.35\times {10}^{-2}$
• C. $1.2\times {10}^{-2}$
• D. $1.0\times {10}^{-3}$

Answer 1

The correct option is B $1.35\times {10}^{-2}$

$\text{Compressibility}=\frac{1}{K}$
where, $K=\text{bulk modulus}$
Bulk modulus,
$K=-\frac{P}{\left(\frac{\Delta V}{V}\right)}$
$\text{Fractional compression}\Rightarrow -\frac{\Delta V}{V}=\frac{P}{K}...\left(i\right)$
$-ve$ sign represents decrease in volume due to compression.

$P=\rho gd$ is the pressure at the bottom of the ocean
$\rho =1000{\text{kg/m}}^3$ is density of water
$g=10{\text{m/s}}^2$
$d=2700\text{m}$ is the depth of ocean
$K=\frac{1}{50\times {10}^{-11}}\text{Pa}$

From Eq. $\left(i\right)$,
$-\frac{\Delta V}{V}=\frac{P}{K}=\frac{\rho gd}{K}$

$\Rightarrow -\frac{\Delta V}{V}=\frac{{10}^3\times 10\times 2700}{\left(\frac{1}{50\times {10}^{-11}}\right)}=1.35\times {10}^{-2}$

Hence $1.35\times {10}^{-2}$ is the desired fractional compression of water at the bottom of the ocean.