Question

A solid copper cube of edge $10\text{cm}$ is subjected to a hydraulic pressure of $7\times {10}^6\text{Pa}$. If compressibility of copper is $\frac{1}{140}{\text{GPa}}^{-1}$, then contraction in its volume will be

• A. $5\times {10}^{-8}{\text{m}}^3$
• B. $4\times {10}^{-8}{\text{m}}^3$
• C. $2\times {10}^{-8}{\text{m}}^3$
• D. ${10}^8{\text{m}}^3$

Answer 1

The correct option is A $5\times {10}^{-8}{\text{m}}^3$

$\text{Bulk Modulus (B)}=\frac{1}{\text{Compressibility}}...\left(i\right)$
$\Rightarrow B=\frac{1}{\frac{1}{140}{\text{GPa}}^{-1}}$

$\therefore B=140\text{GPa}=140\times {10}^9\text{Pa}$
Now from definition of $B$,
$B=\frac{-P}{\left(\frac{\Delta V}{V}\right)}$
$\Delta V=-\frac{PV}{B}...\left(ii\right)$
Substituting the values $P=7\times {10}^6\text{Pa}$
Volume of cube, $V={10}^3=1000{\text{cm}}^3$
$\because 1{\text{cm}}^3={10}^{-6}{\text{m}}^3$
$V=1000\times {10}^{-6}={10}^{-3}{\text{m}}^3$
From Eq $\left(ii\right)$,
$\Delta V=-\frac{7\times {10}^6\times {10}^{-3}}{140\times {10}^9}$

$\therefore \Delta V=-\frac{700}{140}\times {10}^{-8}=-5\times {10}^{-8}{\text{m}}^3$
$-ve$ sign of $\Delta V$ represents contraction in volume