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

A

0.8×10−20.8 \times 10^{-2}

B

1.0×10−21.0 \times 10^{-2}

C

1.2×10−21.2 \times 10^{-2}

D

1.4×10−21.4 \times 10^{-2}

Step-by-Step Solution

  1. Identify the Goal: We need to find the fractional compression, which is the ratio of change in volume to original volume (ΔV/V\Delta V / V).
  2. Relate Compressibility and Pressure: Compressibility (kk) is the reciprocal of the Bulk Modulus (BB). The definition of Bulk Modulus is B=ΔPΔV/VB = \frac{\Delta P}{\Delta V / V}. Therefore, fractional compression is given by: ΔVV=ΔPB=kΔP\frac{\Delta V}{V} = \frac{\Delta P}{B} = k \Delta P where ΔP\Delta P is the change in pressure .
  3. Calculate Pressure at Depth: The pressure exerted by a water column of height hh is given by the hydrostatic formula: P=hρgP = h \rho g Given:
  • Depth (hh) = 2700 m2700 \text{ m}
  • Density (ρ\rho) = 103 kg/m310^3 \text{ kg/m}^3
  • Acceleration due to gravity (gg) ≈10 m/s2\approx 10 \text{ m/s}^2 P=2700×103×10=27×106 Pa=2.7×107 PaP = 2700 \times 10^3 \times 10 = 27 \times 10^6 \text{ Pa} = 2.7 \times 10^7 \text{ Pa}
  1. Calculate Fractional Compression: Substitute the pressure and given compressibility (k=45.4×10−11 Pa−1k = 45.4 \times 10^{-11} \text{ Pa}^{-1}) into the equation: ΔVV=(45.4×10−11 Pa−1)×(2.7×107 Pa)\frac{\Delta V}{V} = (45.4 \times 10^{-11} \text{ Pa}^{-1}) \times (2.7 \times 10^7 \text{ Pa}) ΔVV=122.58×10−4\frac{\Delta V}{V} = 122.58 \times 10^{-4} ΔVV≈1.2×10−2\frac{\Delta V}{V} \approx 1.2 \times 10^{-2}
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