The density of a metal at normal pressure is $\rho$. Its density when it is subjected to an excess pressure $p$ is $\rho^{\prime}$. If $B$ is the Bulk modulus of the metal,the ratio $\frac{\rho^{\prime}}{\rho}$ is:

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 isothermal bulk modulus of a gas at atmospheric pressure is:

Compute the bulk modulus of water from the following data: Initial volume $= 100.0 \; L$,Pressure increase $= 100.0 \; atm$ $(1 \; atm = 1.013 \times 10^{5} \; Pa)$,Final volume $= 100.5 \; L$. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

$A$ 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 . . . . .

$A$ spherical ball contracts in volume by $0.02 \%$,when subjected to a normal uniform pressure of $50$ atmosphere. The Bulk modulus of its material is ............. $Nm^{-2}$.