Derive the relation for linear thermal expansion in solids.
OR
Show that L =$ L_{0} (1+ α \triangle T)$?
Difficulty: Medium
Linear thermal expansion in solids:
Consider a metal rod of length $L_{0}$ at a certain temperature $T_{0}$. Let its length on heating to a temperature T becomes L Thus
Increase in length of the rod =$ \triangle L = L - L_{0}$
Increase in temperature = $\triangle T = T - T_{0}$
It is found that the change in length $ \triangle L$ of a solid is directly proportional to its original length$ L_{0}$, and the change in temperature $\triangle T $.
That is;
$ \triangle L \alpha L_{0} \triangle $ T
or $ \triangle L = \alpha L_{0} \triangle T $ ………. (i)
or $ L - L_{0}= \alpha L_{0} \triangle T $
or L =$ L_{0}+ \alpha L_{0} \triangle T$
or L =$ L_{0} (1+\alpha \triangle T)$ ………. (ii)
Where α is called the coefficient of linear thermal expansion of the substance.
From equation (i), we get
$\alpha$ = $ ( \triangle L )/(L_{0} \triangle T )$ (iii)
Coefficient of linear expansion $\alpha$:
We can define the coefficient of linear expansion$ \alpha$ of a substance as the fractional increase in its length per kelvin rise in temperature.
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