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 $g = 10 \ m \ s^{-2}$)

If the bulk modulus of water is $2 \times 10^9 \ N/m^2$, then the required pressure to reduce the given volume of water by $2 \%$ is

Determine the volume contraction of a solid copper cube,$10 \; cm$ on an edge,when subjected to a hydraulic pressure of $7.0 \times 10^{6} \; Pa$.

$A$ spherical ball contracts in volume by $0.02 \%$, when subjected to a normal uniform pressure of $50 \text{ atm}$. The Bulk modulus of its material is ............. $Nm^{-2}$. (Take $1 \text{ atm} = 1.01 \times 10^5 \text{ Pa}$)

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.

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