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The approximate depth of an ocean is 2700 m2700\text{ m}. The compressibility of water is 45.4×1011 Pa145.4 \times 10^{-11}\text{ Pa}^{-1} and 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

0.8×1020.8 \times 10^{-2}

B

1.0×1021.0 \times 10^{-2}

C

1.2×1021.2 \times 10^{-2}

D

1.4×1021.4 \times 10^{-2}

Step-by-Step Solution

  1. Identify the Formula: Fractional compression (ΔVV\frac{\Delta V}{V}) is related to the change in pressure (ΔP\Delta P) and compressibility (kk) by the definition of Bulk Modulus (BB). Compressibility is the reciprocal of Bulk Modulus (k=1/Bk = 1/B). k=1B=ΔV/VΔP    ΔVV=k×ΔPk = \frac{1}{B} = \frac{\Delta V/V}{\Delta P} \implies \frac{\Delta V}{V} = k \times \Delta P

  2. Calculate Pressure Change: The pressure at the bottom of the ocean due to the water column is given by hydrostatic pressure formula: ΔP=ρgh\Delta P = \rho g h where ρ=103 kg/m3\rho = 10^3\text{ kg/m}^3, g10 m/s2g \approx 10\text{ m/s}^2 (or 9.8 m/s29.8\text{ m/s}^2), and h=2700 mh = 2700\text{ m}. ΔP=103×10×2700=2.7×107 Pa\Delta P = 10^3 \times 10 \times 2700 = 2.7 \times 10^7\text{ Pa}

  3. Calculate Fractional Compression: ΔVV=(45.4×1011 Pa1)×(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=122.58×104\frac{\Delta V}{V} = 122.58 \times 10^{-4} ΔVV=1.2258×102\frac{\Delta V}{V} = 1.2258 \times 10^{-2} Rounding to one decimal place matches the option 1.2×1021.2 \times 10^{-2}.

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