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Home » The depth of an ocean is . The compressibility of water is and density of water is . At the bottom of the ocean, the fractional compression of water will be A)B)C)D)
The depth of an ocean is 2000 \mathrm{~m}. The compressibility of water is 45 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{N} and density of water is 10^{3} \mathrm{~kg} / \mathrm{m}^{3}. At the bottom of the ocean, the fractional compression of water will be \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)
A)6 \times 10^{-3}
B)10^{-3}
C)9 \times 10^{-3}
D)3 \times 10^{-3}
Physics
The depth of an ocean is 2000 \mathrm{~m}. The compressibility of water is 45 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{N} and density of water is 10^{3} \mathrm{~kg} / \mathrm{m}^{3}. At the bottom of the ocean, the fractional compression of water will be \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)
A)6 \times 10^{-3}
B)10^{-3}
C)9 \times 10^{-3}
D)3 \times 10^{-3}

Correct option is C.

compressibility is given as \kappa=\frac{\frac{\Delta V}{V}}{\Delta P}
\Delta \mathrm{V}=\kappa \times \Delta \mathrm{P} \times \mathrm{V}
Substituting values \Delta \mathrm{P}=\rho \operatorname{gh} \mathrm{Pa}, \mathrm{\kappa}=45 \times 10^{-11} \mathrm{~Pa}^{-1}
\begin{array}{l} \frac{\Delta \mathrm{V}}{\mathrm{V}}=45 \times 10^{-11} \times \Delta \mathrm{P} \\ \frac{\Delta \mathrm{V}}{\mathrm{V}}=45 \times 10^{-11} \times 10^{3} \times 10 \times 2000 \\ \frac{\Delta \mathrm{V}}{\mathrm{V}}=9 \times 10^{-3} \end{array}

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