The approximate depth of an ocean is 2700 , m | vedclass

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(C) Given:Depth of ocean,$d = 2700 \, m$Density of water,$ho = 10^3 \, kg/m^3$Compressibility of water,$K = 45.4 	imes 10^{-11} \, Pa^{-1}$Accelera

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$1.2 	imes 10^{-2}$

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(C) Given:Depth of ocean,$d = 2700 \, m$Density of water,$ho = 10^3 \, kg/m^3$Compressibility of water,$K = 45.4 	imes 10^{-11} \, Pa^{-1}$Acceleration due to gravity,$g = 10 \, m/s^2$The excess pressure at the bottom of the ocean is given by $\Delta P = ho gd$.$\Delta P = 10^3 	imes 10 	imes 2700 = 27 	imes 10^6 \, Pa$.Compressibility $K$ is defined as the reciprocal of the Bulk Modulus $B$,i.e.,$K = \frac{1}{B}$.Since $B = \frac{\Delta P}{(\Delta V/V)}$,we have $\frac{\Delta V}{V} = K 	imes \Delta P$.Substituting the values:$\frac{\Delta V}{V} = (45.4 	imes 10^{-11} \, Pa^{-1}) 	imes (27 	imes 10^6 \, Pa)$$\frac{\Delta V}{V} = 1225.8 	imes 10^{-5} = 1.2258 	imes 10^{-2} \approx 1.2 	imes 10^{-2}$.

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

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