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

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1\,\%$. (The bulk modulus of rubber is $9.8 \times 10^8 \, N/m^2$ and the density of sea water is $10^3 \, kg/m^3$.) (in $, m$)

An object is located at $2 \, km$ beneath the surface of the water. If the fractional compression $\frac{\Delta V}{V}$ is $1.36 \, \%$,the ratio of hydraulic stress to the corresponding hydraulic strain will be ......... . [Given: density of water is $1000 \, kg \, m^{-3}$ and $g = 9.8 \, m \, s^{-2}$.]

$A$ 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 bulk modulus of copper is $1.4 \times 10^{11} \ Pa$ and the coefficient of linear expansion is $1.7 \times 10^{-5} (^{\circ}C)^{-1}$. What hydrostatic pressure is necessary to prevent a copper block from expanding when its temperature is increased from $20^{\circ}C$ to $30^{\circ}C$?

The Mariana Trench is located in the Pacific Ocean,and at one place it is nearly $11 \; km$ beneath the surface of the water. The water pressure at the bottom of the trench is about $1.1 \times 10^{8} \; Pa$. $A$ steel ball of initial volume $0.32 \; m^{3}$ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches the bottom? (Bulk modulus of steel $B = 1.6 \times 10^{11} \; N/m^{2}$)