Which type of elastic modulus is there in liquids and gases?

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 $, m$)

If the average depth of an ocean is $4000 \ m$ and the bulk modulus of water is $2 \times 10^9 \ N m^{-2}$,then the fractional compression $\frac{\Delta V}{V}$ of water at the bottom of the ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is . . . . . . (Given,$g=10 \ m s^{-2}, \rho=1000 \ kg m^{-3}$)

If a rubber ball is taken to a depth of $200 \ m$ in a pool,its volume decreases by $0.1 \%$. If the density of the water is $1 \times 10^3 \ kg/m^3$ and $g = 10 \ m/s^2$,then the bulk modulus (volume elasticity) in $N/m^2$ will be:

An external pressure $P$ is applied on a cube at $0^o C$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

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 $B = 1.6 \times 10^{11} \; N/m^{2}$)