Question

A non-isotropic solid metal cube has coefficient of linear expansion as $5\times {10}^{-5}{/}^{\circ }C$ along the x-axis and $5\times {10}^{-6}{/}^{\circ }C$ along y-axis and z-axis. If the coefficient of volumetric expansion of the solid is $C\times {10}^{-6}{/}^{\circ }C$ then the value of $C$ is

Answer 1

• Let the side of the cube taken as $l_1,l_2,l_3$ in x, y, z directions.
• Hence the volume$V=l_1\times l_2\times l_3$ .....(1)
• Let the change in volume be$\Delta v$
• The coefficient of volume expansion can be calculated as:

Volume expansion coefficient=$\frac{\Delta v}{V\Delta t}$

• ​Take log of (1) on both sides

$lnV=ln{l}_1+ln{l}_2+ln{l}_3$

• Differentiating the equation above.

$\frac{\Delta v}{V}=\frac{\Delta l_1}{l_1}+\frac{\Delta l_2}{l_2}+\frac{\Delta l_3}{l_3}$

• Multilply the eqaution by $\frac{1}{\Delta t}$
$\frac{1}{V}\frac{\Delta v}{\Delta t}=\frac{1}{l_1}\frac{\Delta l_1}{\Delta t}+\frac{1}{l_2}\frac{\Delta l_2}{\Delta t}+\frac{1}{l_3}\frac{\Delta l_3}{\Delta t}$
• Given that $\frac{1}{l_1}\frac{\Delta l_1}{\Delta t}=5\times {10}^{-5}{\text{/}}^{o}C$, $\frac{1}{l_2}\frac{\Delta l_2}{\Delta t}=5\times {10}^{-6}{\text{/}}^{o}C$
• Then,​

  $\frac{1}{V}\frac{\Delta v}{\Delta t}=5\times {10}^{-5}+5\times {10}^{-6}$

  $=6\times {10}^{-6}$

  $6\times {10}^{-6}=C\times {10}^{-6}$

  $C=6$