Question

A cubical solid aluminum ($bulkmodulus=-V(dP/dV)=70GPa$) block has an edge length of $1m$ on the surface of the earth. It is kept on the floor of a $5km$ deep ocean. Taking the average density of water and the acceleration due to gravity to be ${10}^{3}kg{m}^{-3}$ and $10m{s}^{-2}$, respectively, the change in the edge length of the block in $mm$ is ________.

Answer 1

Step 1: Given Data:

Bulk modulus $B=-V\left(\frac{dP}{dV}\right)$

$=70GPa$

$=70\times {10}^{9}Pa$

Depth in water $\begin{array}{rcl}H& =& 5km\end{array}$

$\begin{array}{rcl}& =& 5000m\end{array}$

Density of water $\rho ={10}^{3}kg{m}^{-3}$

Acceleration due to gravity $g=10m{s}^{-2}$

side of cube $l=1m$

Step 2: Calculating the pressure acting on the cube:

$\begin{array}{rcl}P& =& \rho gH\\ & =& {10}^3\times 10\times 5000\\ & =& 50\times {10}^{6}Pa\end{array}$

Step 3: Calculating the Change in length:

$\begin{array}{rcl}B& =& -V\left(\frac{dP}{dV}\right)\\ 70\times {10}^9& =& -V\left(\frac{50\times {10}^6}{dV}\right)\\ \frac{dV}{V}& =& -\frac{50\times {10}^6}{70\times {10}^9}\\ \frac{dV}{V}& =& -\frac{5}{7\times {10}^3}\\ \frac{3∆l}{l}& =& -\frac{5}{7000}\\ 3∆l& =& -\frac{5}{7\times {10}^3}\times 1\\ ∆l& =& -\frac{5}{7\times 3}\times {10}^{-3}\\ & =& -0.238\times {10}^{-3}m\\ & =& -0.238mm\end{array}$

A negative sign indicates a decrease in edge length.

Hence, the change in the edge length of the block is 0.238 mm.