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Physics Question on Pressure

The approximate depth of an ocean is 2700 m2700\, m. The compressibility of water is 45.4?10−11Pa−145.4 ? 10^{-11} Pa^{-1 } and density of water is 103 kg/m310^3\, kg/m^3.What fractional compression of water will be obtained at the bottom of the ocean ?

A

1.2×10−21.2 \times 10^{-2}

B

1.4×10−21.4 \times 10^{-2}

C

0.8×10−20.8 \times 10^{-2}

D

1.0×10−21.0 \times 10^{-2}

Answer

1.2×10−21.2 \times 10^{-2}

Explanation

Solution

compressibility is given as κ=ΔVVΔP\kappa=\frac{\frac{\Delta V }{ V }}{\Delta P }
ΔV=κ×ΔP×V\Delta V =\kappa \times \Delta P \times V
Substituting values ΔP=ρghPa,κ=45×10−11Pa−1\Delta P =\rho g h Pa , \kappa=45 \times 10^{-11} Pa ^{-1}
ΔVV=45×10−11×ΔP ΔVV=45×10−11×103×10×2700 ΔVV=1.2×10−2 \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}