Question

The approximate depth of an ocean is $2700m$. The compressibility of water is $45.4\times {10}^{-11}P{a}^{-1}$ and density of water is ${10}^{3}kg/m^3$. What fractional compression of water will be obtained at the bottom of the ocean?(Take $g=10m{s}^{-2}$)

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

Answer 1

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

compressibility is given as $\kappa =\frac{\frac{\Delta V}{V}}{\Delta P}$

$\Delta V=\kappa \times \Delta P\times V$

Substituting values $\Delta P=\rho ghPa$ , $\kappa =45\times {10}^{-11}P{a}^{-1}$

$\frac{\Delta V}{V}=45\times {10}^{-11}\times \Delta P$

$\frac{\Delta V}{V}=45\times {10}^{-11}\times {10}^3\times 10\times 2700$

$\frac{\Delta V}{V}=1.2\times {10}^{-2}$