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

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 $\rho = 10^{3} \, \text{kg/m}^3$,$g = 9.8 \, \text{m/s}^2$)

$A$ spherical ball of volume $2000 \text{ cm}^3$ is subjected to a hydraulic pressure of $15 \text{ atm}$. If the change in volume is $5 \times 10^{-2} \text{ cm}^3$,the bulk modulus of the material of the spherical ball is (Given: $1 \text{ atm} = 10^5 \text{ Nm}^{-2}$)

An external pressure $P$ is applied on a cube at $0^o C$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

The density of a metal at normal pressure $P$ is $\varrho$. When it is subjected to an excess pressure $p$,the density becomes $\varrho^{\prime}$. If $K$ is the bulk modulus of the metal,then the ratio $\frac{\varrho^{\prime}}{\varrho}$ is

The adiabatic elasticity of a gas is equal to