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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\

Vedclass · Accepted Answer

$1.36 	imes 10^{-2}$

Vedclass · Answer

(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 $g = 10\; m s^{-2}$)

Similar Questions

The average depth of the Indian Ocean is about $3000 \,m$. The value of fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at the bottom of the ocean is (given that the bulk modulus of water is $2.2 \times 10^9 \,N/m^2$,$g = 9.8 \,m/s^2$,$\rho_{H_2O} = 1000 \,kg/m^3$):

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An increase in pressure required to decrease the $200 \ L$ volume of a liquid by $0.004\%$ in a container is .......... $kPa$ (Bulk modulus of the liquid $= 2100 \ MPa$).

The pressure required to decrease the volume of $4000 \ cc$ water by $0.05 \%$ is (Bulk modulus of water $= 2.2 \times 10^9 \ N/m^2$)

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