Question

A cubical solid aluminium (bulk modulus=$(-V)\dfrac{dP}{dV}=70\,GPa$) block has an edge length of on the surface of the earth. It is kept on the floor of a $5\,km$ deep ocean. Taking the average density of water and the acceleration due to gravity to be $10^3\,kg/m^3$ and $10\,m/s^2$, respectively, the change in the edge length of the block in mm is?

Answer 1

Hint:A substance's bulk modulus is a measurement of how resistant it is to compression. It's the ratio of the infinitesimal pressure rise to the volume's resultant relative drop. Other moduli explain the material's response (strain) to other types of stress: the shear modulus represents shear, and Young's modulus describes linear stress.

Complete step by step answer:
Only the bulk modulus is relevant for a fluid. These three moduli do not provide enough information to describe the behaviour of a complex anisotropic solid like wood or paper, hence the entire generalised Hooke's law must be used. Isothermal compressibility is defined as the reciprocal of the bulk modulus at a constant temperature.The bulk modulus $K$ can be defined as:
$K = - V\dfrac{{dP}}{{dV}}$
When \[P\]means pressure, \[V\]signifies the starting volume of the substance, and $\dfrac{{dP}}{{dV}}$ denotes the pressure-volume derivative.

Because volume and density are inversely proportional, it follows that
$K = \rho \dfrac{{dP}}{{d\rho }}$
where $\dfrac{{dP}}{{d\rho }}$represents the derivative of pressure with respect to density and $\rho $ is the initial density (i.e. pressure rate of change with volume). The compressibility of a substance is determined by the inverse of the bulk modulus. The bulk modulus is usually described as the isothermal bulk modulus at constant temperature, but it can also be defined as the adiabatic bulk modulus at constant entropy.Now let us solve the problem: Let $a$ be the cubic block's edge length then the block's volume,$V = {a^3}$.

Differentiating both sides we get,
$dV = 3{a^2}$
$ \Rightarrow \dfrac{{dV}}{V} = 3\dfrac{{da}}{a}\,\,\,\,...(1)$
Bulk modulus of block, $B = - V\dfrac{{dP}}{{dV}}$
Here putting $p = \rho gh$ and $\dfrac{{dV}}{V} = \dfrac{{3a}}{a}\,$ [from eq $1$]
$B = - \dfrac{{\rho gha}}{{3da}}$
Now, as given in the problem
$B = 70 \times {10^9}\,Pa$
$\Rightarrow \rho = {10^3}\,kg/{m^3},g = 10\,m/{s^2},h = 5000\,m,a = 1\,m$

Now putting the value in :
$B = - \dfrac{{\rho gha}}{{3da}}$
$70 \times {10^9} = \dfrac{{ - ({{10}^3} \times 10 \times 5000 \times 1)}}{{(3da)}}$
$ \Rightarrow da = \dfrac{{(5 \times {{10}^7})}}{{(3 \times 70 \times {{10}^9})}} \\
\Rightarrow da = \dfrac{5}{{21}} \times {10^3}$
$ \Rightarrow da = 0.238 \\
\therefore da\approx 0.24\,mm$

Hence,the change in the edge length of the block in mm is 0.24 mm.

Note:The bulk modulus of elasticity is one of the measures of the mechanical properties of solids. Other elastic modules include Young’s modulus and Shear modulus. In any case, the bulk elastic properties of a material are used to determine how much it will compress under a given amount of external pressure.