The increase in pressure required to decrease the $200 \, L$ volume of a liquid by $0.008 \, \%$ in $kPa$ is (Bulk modulus of the liquid $= 2100 \, MPa$)

If a brass sphere of radius $36 \ cm$ is submerged in a lake at a depth where the pressure is $10^7 \ Pa$, then the change in the radius of the sphere is $($Bulk modulus of brass $= 60 \ GPa)$.

$A$ rubber ball is taken into the deep sea such that its volume decreases by $x \%$. The bulk modulus of rubber is $K$ and the density of sea water is $\rho$. The depth $h$ to which the rubber ball is taken is proportional to $(g = \text{acceleration due to gravity})$:

$A$ ball falling in a lake of depth $200 \ m$ shows $0.1\%$ decrease in its volume at the bottom. What is the bulk modulus of the material of the ball?

$A$ cubical solid aluminium (bulk modulus $B = -V \frac{dP}{dV} = 70 \text{ GPa}$) block has an edge length of $1 \text{ m}$ on the surface of the earth. It is kept on the floor of a $5 \text{ km}$ deep ocean. Taking the average density of water $\rho = 10^3 \text{ kg m}^{-3}$ and the acceleration due to gravity $g = 10 \text{ m s}^{-2}$,the change in the edge length of the block in $\text{mm}$ is . . . . .

When a rubber ball is taken to a depth of $h$ meters in deep sea,its volume decreases by $0.5\, \%$. Calculate the depth $h$. (Given: Bulk modulus of rubber $B = 9.8 \times 10^{8} \, \text{N/m}^2$,Density of sea water $\rho = 10^{3} \, \text{kg/m}^3$,$g = 9.8 \, \text{m/s}^2$)