Q.A solid copper sphere of volume $500\ \text{cm}^3$ is lowered into the sea to a depth where the pressure on it increases by $1.1\times10^{7}\ \text{Pa}$. Taking the bulk modulus of copper as $K = 1.4\times10^{11}\ \text{Pa}$, find

Bulk modulus $K$ is the modulus of elasticity associated with a change in the volume of a body under an applied deforming force — the elasticity of volume. Unlike Young's modulus and the modulus of rigidity, which apply only to solids, bulk modulus is a property shared by solids, liquids and gases, since all three states of matter have a definite volume.

For a body of original volume $V$ subjected to a volume stress $dP$ that produces a volume change $dV$, the volume strain is defined as $-dV/V$ (the minus sign ensures the strain is reported as a positive quantity even though $dV$ itself is negative for a pressure increase), and the bulk modulus is

Its SI unit is $\text{N/m}^2$, dimensional formula $[L^{-1}M^1T^{-2}]$. Its reciprocal, $1/K$, is called compressibility — the fractional decrease in volume per unit increase in pressure — measured in $\text{Pa}^{-1}$.

A material with a small bulk modulus (and hence a large compressibility, like water at $2.18\times10^8$ Pa) is easy to compress, while one with a large bulk modulus (like gold or steel, both above $10^{11}$ Pa) strongly resists any change in its volume. This is why a high-pressure balloon or gas cylinder can burst once the internal volume stress exceeds the container material's bulk-elastic limit, and why a submarine hull deep underwater must be engineered to withstand a substantial volume stress from the surrounding water.