Question

The approximate depth of an ocean is $2700\, m$. The compressibility of water is $45.4 ? 10^{-11} Pa^{-1 }$ and density of water is $10^3\, kg/m^3$.What fractional compression of water will be obtained at the bottom of the ocean ?

• 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

compressibility is given as $\kappa=\frac{\frac{\Delta V }{ V }}{\Delta P }$
$\Delta V =\kappa \times \Delta P \times V$
Substituting values $\Delta P =\rho g h Pa , \kappa=45 \times 10^{-11} Pa ^{-1}$
$
\begin{array}{l}
\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}
\end{array}
$