Question

The approximate depth of an ocean is $2700\text{m}$. The compressibility of water is $45.4\times {10}^{-11}{\text{Pa}}^{-1}$ and density of water is ${10}^3{\text{kg/m}}^3$. Fractional compression of water will be:
[Take $g=10{\text{m/s}}^2$]

• A. $0.8\times {10}^{-2}$
• B. $1.0\times {10}^{-2}$
• C. $1.2\times {10}^{-2}$
• D. $1.4\times {10}^{-2}$

Answer 1

The correct option is C $1.2\times {10}^{-2}$

Depth of ocean $h=2700\text{m}$
Compressibility of water, $z=45.4\times {10}^{-11}{\text{Pa}}^{-1}$
Density of water, $\rho ={10}^3{\text{kg/m}}^3$
Compressibility is defined as,
$z=\frac{1}{\text{Bulk modulus}}$
$\Rightarrow z=\frac{1}{\frac{-\Delta p}{\frac{\Delta V}{V}}}$
$\Rightarrow \frac{\Delta V}{V}=-z\Delta p$
$\Rightarrow \frac{\Delta V}{V}=-z\rho gh$
[$\Delta p=\rho gh$, pressure difference due to depth of water column]
$\Rightarrow \frac{\Delta V}{V}=-45.4\times {10}^{-11}\times 1000\times 10\times 2700$
$\Rightarrow \frac{\Delta V}{V}=-1.2\times {10}^{-2}$
Here, $-ve$ sign shows compression or reduction in volume of water.
$\frac{\Delta V}{V}$ denotes fractional change in volume.
$\Rightarrow$Fractional compression of water will be $1.2\times {10}^{-2}$