A) \[{{L}_{1}}{{\alpha }_{1}}={{L}_{2}}{{\alpha }_{2}}\]
B) \[{{L}_{1}}\alpha _{1}^{2}={{L}_{2}}\alpha _{2}^{2}\]
C) \[\frac{{{L}_{1}}}{{{\alpha }_{1}}}=\,\,\frac{{{L}_{2}}}{{{\alpha }_{2}}}\]
D) None of these
Correct Answer: A
Solution :
If \[\operatorname{L}_{1}^{'}\,and\,\,L_{2}^{'}\] are their length at some different temperature T then, \[L_{1}^{'}=L_{2}^{'}\left( 1+{{\alpha }_{1}}T \right)\] \[L_{2}^{'}=L_{2}^{{}}\left( 1+{{\alpha }_{2}}T \right)\] \[{{L}_{1}}'-{{L}_{2}}'\,\,=\,\,{{L}_{1}}-{{L}_{2}}\,\,+\,\,T\,({{L}_{1}}{{\alpha }_{1}}-{{L}_{2}}{{\alpha }_{2}})\] For this difference to remain same \[{{L}_{1}}{{\alpha }_{1}}-{{L}_{2}}{{\alpha }_{2}}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{L}_{1}}{{\alpha }_{1}}-{{L}_{2}}{{\alpha }_{2}}\]
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