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 $= 1.4 \times 10^{11} \ Nm^{-2}$)

What is the density of water at a depth where pressure is $80.0 \; atm$, given that its density at the surface is $1.03 \times 10^{3} \; kg \; m^{-3}$? (Take bulk modulus of water $B = 2.2 \times 10^{9} \; Pa$ or compressibility $45.8 \times 10^{-11} \; Pa^{-1}$)

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 fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at a depth of $2.5 \ km$ below the sea level is . . . . . . $\%$. Given: the Bulk modulus of water $B = 2 \times 10^9 \ N m^{-2}$,density of water $\rho = 10^3 \ kg m^{-3}$,and acceleration due to gravity $g = 10 \ m s^{-2}$.

The only elastic modulus that applies to fluids is

If the compressibility of water is $\sigma$ per unit atmospheric pressure,then the decrease in volume $V$ due to $P$ atmospheric pressure will be