The depth below the surface of the sea to which a rubber ball must be taken so as to decrease its volume by $0.02 \%$ is . . . . . . $m$.
(Take density of sea water $= 10^3 \ kg \ m^{-3}$,Bulk modulus of rubber $= 9 \times 10^8 \ N \ m^{-2}$,and $g = 10 \ m \ s^{-2}$)

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?

$A$ swimming pool has a depth of $22 \ m$ and area $700 \ m^2$. Calculate the fractional change $\frac{\Delta V}{V}$ of water at the bottom of the swimming pool. Given that the bulk modulus of water is $2.2 \times 10^9 \ N \ m^{-2}$, $g = 10 \ m \ s^{-2}$, and the density of water is $1000 \ kg \ m^{-3}$.

The adiabatic bulk modulus of a gas at a pressure $P$ is (where $\gamma$ is the ratio of specific heat capacities of the gas).

$A$ sphere contracts in volume by $0.01 \%$ when taken to the bottom of a sea $1 \,km$ deep. Find the Bulk modulus of the material of the sphere in $N/m^2$.

The volume contraction of a solid copper cube of edge length $10 \ cm$,when subjected to a hydraulic pressure of $7 \times 10^6 \ Pa$,would be . . . . . . $mm^3$. (Given bulk modulus of copper $= 1.4 \times 10^{11} \ Nm^{-2}$)