$A$ ball is taken to a depth of $200 \ m$ in a lake. The decrease in its volume is $0.1\%$. Calculate the bulk modulus of the material of the ball.

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.

$A$ bottle has an opening of radius $a$ and length $b$. $A$ cork of length $b$ and radius $(a + \Delta a)$,where $(\Delta a << a)$,is compressed to fit into the opening completely (see figure). If the bulk modulus of the cork is $B$ and the frictional coefficient between the bottle and the cork is $\mu$,then the force needed to push the cork into the bottle 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 $\rho = 10^{3} \, \text{kg/m}^3$,$g = 9.8 \, \text{m/s}^2$)

The volume of a material reduces by $2 \%$ when the pressure is increased from $1 \text{ atm}$ to $2 \text{ atm}$. What is its bulk modulus?

Bulk modulus of water is $2 \times 10^9 \ N/m^2$. The pressure required to increase the volume of water by $0.1 \%$ in $N/m^2$ is: