Question

The coefficient of linear expansion of a crystal in one direction is ${\alpha }_1$ and that in every direction perpendicular to it is ${\alpha }_2$. Then, the coefficient of cubical expansion of the crystal is

• A. ${\alpha }_1+{\alpha }_2$
• B. $2{\alpha }_1+{\alpha }_2$
• C. ${\alpha }_1+2{\alpha }_2$
• D. None of these

Answer 1

The correct option is C ${\alpha }_1+2{\alpha }_2$

Given,
Coefficient of linear expansion along a direction is ${\alpha }_1$.
Coefficient of linear expansion along any direction perpendicular to the direction of ${\alpha }_1$ is ${\alpha }_2$.
We know from the concept of volume expansion,
$V=V_0(1+\gamma \Delta T)......\left(1\right)$
For a small change in temperature, new length obtained along each direction is given by the formula
$L=L_0(1+\alpha \Delta T)$
$\therefore$ For total volume $V$ we can write $\left(1\right)$ as follows
$L^3=\left[L_0\right(1+{\alpha }_1\Delta T\left)\right]\left[L_0^2\right(1+{\alpha }_2\Delta T{)}^2]$
$\Rightarrow L^3=L_0^3(1+{\alpha }_1\Delta T)(1+{\alpha }_2\Delta T{)}^2.........(2)$
Taking, $L_0^3=V_0\text{and}{L}^3=V$ in $\left(2\right)$
$V=V_0(1+{\alpha }_1\Delta T)(1+{\alpha }_2\Delta T{)}^2$
Using $\left(1\right)$, we get
$1+\gamma \Delta T=(1+{\alpha }_1\Delta T)(1+{\alpha }_2\Delta T{)}^2$
Applying binomial expansion and neglecting the higher order powers of $\Delta T$,
$1+\gamma \Delta T\approx (1+{\alpha }_1\Delta T)(1+2{\alpha }_2\Delta T)$
$\Rightarrow 1+\gamma \Delta T\approx (1+{\alpha }_1\Delta T+2{\alpha }_2\Delta T)$
$\Rightarrow \gamma ={\alpha }_1+2{\alpha }_2$
Hence, option (c) is the correct answer.