The bulk modulus of rubber is $9.1 \times 10^8 \, N/m^2$. To what depth $h$ (in $m$) must a rubber ball be taken in a lake so that its volume is decreased by $0.1 \, \%$?

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

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 $= 60 \ GPa)$.

The compressibility of water is $6 \times 10^{-10} \ N^{-1} \ m^2$. If one litre is subjected to a pressure of $4 \times 10^7 \ N \ m^{-2}$,the decrease in its volume is (in $cc$):

When a sphere is taken to the bottom of a sea of depth $1 \ km$,it contracts in volume by $0.01 \%$. The bulk modulus of the material of the sphere is (Acceleration due to gravity $= 10 \ m \ s^{-2}$,density of sea water $= 10^3 \ kg \ m^{-3}$)

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at a temperature of $57\,^{\circ}C$ is drunk. You can take the body (tooth) temperature to be $37\,^{\circ}C$ and $\alpha = 1.7 \times 10^{-5}/^{\circ}C$,bulk modulus for copper $= 140 \times 10^9\, N/m^2$.