Question: Show that β ≈ 3α, by calculating the change in volume ΔV of a cube with sides of length L.

We calculate the change in length by the formula for linear expansion in solids and find the new length and we further calculate the change in volume and find the new volume.

The formula for length expansion of solids$\mathrm{\Delta L}=\mathrm{\alpha L\Delta T}$

The formula for volume expansion of solids $\mathrm{\Delta V}=\mathrm{\beta V\Delta T}$

Here,are the coefficient of linear and volume expansion respectively, its original length and volume, and is the temperature change.

New length$=\mathrm{L}+\mathrm{\Delta L}$

New volume$=\mathrm{V}+\mathrm{\Delta V}$

But for the cube,

Volume = (Length)³

$$\begin{gathered} {(\mathrm{L}+\mathrm{\Delta L})}^3=\mathrm{V}+\mathrm{\Delta V} \\ {(\mathrm{L}+\mathrm{\alpha L\Delta T})}^3=\mathrm{V}+\mathrm{\beta V\Delta T} \\ {\mathrm{L}}^3{(1+\mathrm{\alpha \Delta T})}^3=\mathrm{V}(1+\mathrm{\beta \Delta T}) \\ {\mathrm{L}}^3(1+{\mathrm{\alpha }}^3{\mathrm{\Delta T}}^3+3\mathrm{\alpha \Delta T}+3{\mathrm{\alpha }}^2{\mathrm{\Delta T}}^2)=\mathrm{V}(1+\mathrm{\beta \Delta T}) \end{gathered}$$

Since α is very small, second and fourth terms in the bracket on LHS are negligible. Also, we can write volume as,.So the equation becomes

$$\begin{gathered} (1+3\mathrm{\alpha \Delta T})=(1+\mathrm{\beta \Delta T}) \\ 3\mathrm{\alpha \Delta T}=\mathrm{\beta \Delta T} \\ 3\mathrm{\alpha }=\mathrm{\beta } \end{gathered}$$

Therefore, it is proved that$\mathbf{\beta }=\mathbf{3}\mathbf{\alpha }$