Find 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 $\alpha = 1.7 \times 10^{-5} {}^{\circ} C^{-1}$,bulk modulus for copper $B = 140 \times 10^9 \ Nm^{-2}$.

The compressibility of water is $6 \times 10^{-10} \,m^{2}/N$. If one litre of water is subjected to a pressure of $4 \times 10^{7} \,N/m^{2}$, then the decrease in its volume in millilitre will be

$A$ 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 . . . . .

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

The bulk modulus of rubber is $9.1 \times 10^8 \, N/m^2$. To what depth should a rubber ball be taken in a lake so that its volume is decreased by $0.1\%$? (Take density of water $\rho = 10^3 \, kg/m^3$ and $g = 10 \, m/s^2$)

Consider a fluid in a container. Let the density of the fluid at the surface and at depth $H$ be $\rho_0$ and $\rho$ respectively. The bulk modulus of the fluid is $B_0$. If $\rho = \frac{\rho_0}{1 - \frac{\rho g H}{B_0}}$, find the constant $\alpha$ in the expression $\rho = \frac{\rho_0}{1 + \alpha \rho g H}$ (Assume $\frac{\rho - \rho_0}{\rho_0} \ll 1$).