Question

If $\alpha ,\beta$ and $\gamma$ are the coefficient of linear, areal and volume expansion respectively, then which of the following options correctly represent the relation between them?

• A. $\gamma =3\alpha$
• B. $\alpha =3\gamma$
• C. $\beta =3\alpha$
• D. $\gamma =3\beta$

Answer 1

The correct option is A $\gamma =3\alpha$

Let us consider a three dimensional solid having a volume expansion coefficient $\left(\gamma \right)$. Let the original volume of the solid be $\left(V\right)$.
Here we assume, Volume $V=L^3$.
Then by heating the solid by changing the temperature $\left(\Delta T\right)$, the new volume of the solid can be written as,
$V+\Delta V={(L+\Delta L)}^3............\left(1\right)$
where $(L+\Delta L)$ is the new length of the solid along each direction in 3-dimensional space.
Using binomial expansion on $\left(1\right)$, we get
$V+\Delta V=(L^3+{\left(\Delta L\right)}^3+3L^2\Delta L+3L{\left(\Delta L\right)}^2)$
From the definition of linear and volume expansions
$\Delta L=L\alpha \Delta T$ and $\Delta V=V\gamma \Delta T$
$\Rightarrow V+\Delta V=L^3+(L\alpha \Delta T{)}^3+3L^2(L\alpha \Delta T)+3L(L\alpha \Delta T{)}^2$
$\Rightarrow V+V\gamma \Delta T=L^3+L^3({\alpha }^3{\Delta T}^3+3{\alpha }^2{\Delta T}^2+3\alpha \Delta T)$
[Neglecting ${\alpha }^2\text{and}{\alpha }^3$ terms as $\alpha$ is very small and using $V=L^3$]
$\Rightarrow V\gamma \Delta T=L^3\times 3\alpha \Delta T$
we get, $\gamma =3\alpha$
Hence, option (a) is the correct answer.