Question

The approximate depth of an ocean is $3500\text{m}$. The compressibility of water is $50\times {10}^{-11}{\text{Pa}}^{-1}$ and density of water is ${10}^3{\text{kg/m}}^3$. Find the fractional compression of water that will be obtained at the bottom of the ocean. (Take $g=10{\text{m/s}}^2$)

• A. $5.25\times {10}^{-2}$
• B. $10.5\times {10}^{-2}$
• C. $1.75\times {10}^{-2}$
• D. $2.625\times {10}^{-2}$

Answer 1

The correct option is C $1.75\times {10}^{-2}$

Bulk modulus $\left(K\right),K=\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}$
$\Rightarrow \frac{\Delta V}{V}=\frac{\Delta P}{K}$
where $\Delta P$ is the pressure at the bottom of the ocean.
Now,
$\Delta P=\rho gh$
$h$ is depth of ocean and $\rho$ is density of water
$\frac{\Delta V}{V}=$ Fractional compresssion of water.
$\Rightarrow \frac{\Delta V}{V}=\frac{\rho gh}{K}=\frac{\rho gh}{\left(\frac{1}{C}\right)}$
[$\because$ Given $C=50\times {10}^{-11}{\text{Pa}}^{-1}$]

$\frac{\Delta V}{V}=\frac{{10}^3\times 10\times 3500}{\left(\frac{1}{50\times {10}^{-11}}\right)}=1.75\times {10}^{-2}$
So, fractional compression of water at the bottom of the ocean is $1.75\times {10}^{-2}$