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 pressure of a medium is changed from $1.01 \times 10^5 \ Pa$ to $1.165 \times 10^5 \ Pa$ and the change in volume is $10\%$ keeping the temperature constant. The Bulk modulus of the medium is:

$A$ spherical ball contracts in volume by $0.02 \%$,when subjected to a normal uniform pressure of $50$ atmosphere. The Bulk modulus of its material is ............. $Nm^{-2}$.

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:

$A$ solid sphere of radius $r$ made of a soft material of bulk modulus $K$ is surrounded by a liquid in a cylindrical container. $A$ massless piston of area $a$ floats on the surface of the liquid,covering the entire cross-section of the cylindrical container. When a mass $m$ is placed on the surface of the piston to compress the liquid,the fractional decrement in the radius of the sphere,$\left( \frac{dr}{r} \right)$ is:

The density of a metal at normal pressure is $\rho$. Its density when it is subjected to an excess pressure $p$ is $\rho^{\prime}$. If $B$ is the Bulk modulus of the metal,the ratio $\frac{\rho^{\prime}}{\rho}$ is: