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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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