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(A) The pressure $p$ exerted by a $3000\; m$ column of water at the bottom of the ocean is given by the formula $p = h \rho g$.Given: depth $h = 3000\

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$1.36 	imes 10^{-2}$

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(A) The pressure $p$ exerted by a $3000\; m$ column of water at the bottom of the ocean is given by the formula $p = h ho g$.Given: depth $h = 3000\; m$,density of water $ho = 1000\; kg\; m^{-3}$,and acceleration due to gravity $g = 10\; m s^{-2}$.$p = 3000\; m 	imes 1000\; kg\; m^{-3} 	imes 10\; m s^{-2} = 3 	imes 10^{7}\; N m^{-2}$.The bulk modulus $B$ is defined as $B = -\frac{p}{\Delta V / V}$,so the fractional compression is $\frac{\Delta V}{V} = \frac{p}{B}$.Substituting the values: $\frac{\Delta V}{V} = \frac{3 	imes 10^{7}\; N m^{-2}}{2.2 	imes 10^{9}\; N m^{-2}} \approx 1.36 	imes 10^{-2}$.Thus,the fractional compression is $1.36 	imes 10^{-2}$.

The average depth of the Indian Ocean is about $3000\; m$. Calculate the fractional compression,$\Delta V / V,$ of water at the bottom of the ocean,given that the bulk modulus of water is $2.2 \times 10^{9}\; N m^{-2}$. (Take $g = 10\; m s^{-2}$)

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