$A$ sphere contracts in volume by $0.01 \%$ when taken to the bottom of a sea $1 \,km$ deep. Find the Bulk modulus of the material of the sphere in $N/m^2$.

The volume contraction of a solid copper cube of edge length $10 \ cm$,when subjected to a hydraulic pressure of $7 \times 10^6 \ Pa$,would be . . . . . . $mm^3$. (Given bulk modulus of copper $= 1.4 \times 10^{11} \ Nm^{-2}$)

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 $\%$)

$A$ glass slab is subjected to a pressure of $10 \, atm$. The fractional change in its volume is (Bulk modulus of glass $= 37 \times 10^9 \, N \, m^{-2}$,$1 \, atm = 1 \times 10^5 \, N \, m^{-2}$)

$1\, m^3$ of water is brought inside a lake up to a depth of $200\, m$ from the surface. What will be the change in volume if the bulk modulus of elasticity of water is $22000\, \text{atm}$? (Given: density of water $\rho = 1\times10^3\, kg/m^3$, atmospheric pressure $P_0 = 10^5\, N/m^2$, and $g = 10\, m/s^2$)

$A$ 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: