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Physics Question on mechanical properties of fluid

The approximate depth of an ocean is 2700 m2700\, m. The compressibility of water is 45.4×10−11Pa−145.4 \times 10^{-11} Pa ^{-1} and density of water is 103kg/m3.10^{3} kg / m 3 . What fractional compression of water will be obtained at the bottom of the ocean?

A

1.0×10−21.0 \times 10^{-2}

B

1.2×10−21.2 \times 10^{-2}

C

1.4×10−21.4 \times 10^{-2}

D

1.8×10−21.8 \times 10^{-2}

Answer

1.2×10−21.2 \times 10^{-2}

Explanation

Solution

compressibility is given as K=ΔVVΔPK =\frac{\frac{\Delta V }{ V }}{\Delta P } ΔV=K×ΔP×V\Delta V =K\times \Delta P \times V Substituting values ΔP=ρ gh Pa\Delta P =\rho \,gh \,Pa , K=45×10−11Pa−1 K=45 \times 10^{-11} Pa ^{-1} ΔVV=45×10−11×ΔP\frac{\Delta V }{ V }=45 \times 10^{-11} \times \Delta P ΔVV=45×10−11×103×10×2700\frac{\Delta V }{ V }=45 \times 10^{-11} \times 10^{3} \times 10 \times 2700 ΔVV=1.2×10−2\frac{\Delta V }{ V }=1.2 \times 10^{-2}