Question

The depth of an ocean is \(2000\,\text{m}\). The compressibility of water is \(45 \times 10^{-11}\,\text{m}^2/\text{N}\) and density of water is \(10^3\,\text{kg m}^{-3}\). At the bottom of the ocean, the fractional compression of water will be \((g = 10\,\text{m s}^{-2})\)

• A. \(6 \times 10^{-3}\)
• B. \(10^{-3}\)
• C. \(9 \times 10^{-3}\)
• D. \(3 \times 10^{-3}\)

Hint:

Fractional compression equals compressibility multiplied by applied pressure.

Answer 1

The Correct Option is C

Step 1: Pressure at depth \(h\).
\[ P = \rho g h \]
Step 2: Substitute values.
\[ P = 10^3 \times 10 \times 2000 = 2 \times 10^{7}\,\text{Pa} \]
Step 3: Use compressibility relation.
Fractional compression is given by:
\[ \frac{\Delta V}{V} = \beta P \]
Step 4: Substitute values.
\[ \frac{\Delta V}{V} = 45 \times 10^{-11} \times 2 \times 10^{7} \] \[ \frac{\Delta V}{V} = 9 \times 10^{-3} \]
Step 5: Conclusion.
The fractional compression of water is \(9 \times 10^{-3}\).