The average depth of an oil well is $2000 \, m$. If the bulk modulus of oil is $8 \times 10^8 \, N/m^2$ and the density of oil is $1500 \, kg/m^3$, the fractional compression at the bottom of the well is (take $g = 10 \, m/s^2$): (in $\%$)

The compressibility of water is $4 \times 10^{-5}$ per unit atmospheric pressure. The decrease in volume of $100 \ cm^3$ of water under a pressure of $100 \ atm$ will be ......... $cm^3$.

$A$ sample of a liquid is kept at $1 \ atm$. It is compressed to $5 \ atm$,which leads to a change of volume of $0.8 \ cm^3$. If the bulk modulus of the liquid is $2 \ GPa$,the initial volume of the liquid was . . . . . . litre. (Take $1 \ atm = 10^5 \ Pa$)

If a rubber ball is taken to a depth of $200 \ m$ in a pool,its volume decreases by $0.1 \%$. If the density of the water is $1 \times 10^3 \ kg/m^3$ and $g = 10 \ m/s^2$,then the bulk modulus (volume elasticity) in $N/m^2$ will be:

$A$ 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 $(g = \text{acceleration due to gravity})$:

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: