Question

A cubical solid aluminum ($K=-V\left(\frac{dP}{dV}\right)=70GPa$) 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,

$\begin{array}{rcl}K& =& -V\left(\frac{dP}{dV}\right)\\ & =& 70GPa\end{array}$

Edge length $a=1m$

Depth is,

$\begin{array}{rcl}d& =& 5km\\ & =& 5000m\end{array}$

The density of water, $\rho ={10}^{3}kg{m}^{-3}$

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

Step 2. Formula used

$\begin{array}{rcl}K& =& -V\left(\frac{dP}{dV}\right)\end{array}$

Where, $K=$ bulk modulus

$P=$ pressure

$V=$ initial volume of the substance

Step 2. Finding the change in edge length,

Let $a$ be the cubic block's edge length then the block's volume,

$V=a^3$

Differentiating both sides we get,

$dV==3a^2$

The change in volume with respect to initial volume,

$\frac{dV}{V}=3\times \frac{da}{a}$

Putting this value in bulk modulus relation,we get

$\begin{array}{rcl}-V\left(\frac{dP}{dV}\right)& =& \frac{-V\left(\rho gd\right)}{3\times {\displaystyle \frac{da}{a}}}\end{array}$

$\Rightarrow$ $70\times {10}^9=\frac{-1\left(1000\times 10\times 5000\right)}{3\times {\displaystyle da}}$

$\Rightarrow$ $da=\frac{-1\left(1000\times 10\times 5000\right)}{3\times {\displaystyle 70\times {10}^9}}$

$\Rightarrow$ $da=0.238\times {10}^{-3}m$

$\Rightarrow$ $da\approx 0.24mm$

Hence, the change in the edge length of the block in $mm$ is $\mathbf{0}\mathbf{.}\mathbf{24}$.