A) \[\frac{{{l}_{1}}}{\left( {{l}_{1}}+{{l}_{2}} \right)}\]
B) \[\frac{\left( {{l}_{1}}+{{l}_{2}} \right)}{{{l}_{1}}}\]
C) \[\frac{{{l}_{2}}}{\left( {{l}_{1}}+{{l}_{2}} \right)}\]
D) \[\frac{({{l}_{1}}+{{l}_{2}})}{{{l}_{2}}}\]
Correct Answer: C
Solution :
Initial length of I rod \[={{l}_{1}}\] Initial length of II rod \[={{l}_{2}}\] Linear coefficient of 1 rod \[={{\alpha }_{1}}\] Linear coefficient of II rod\[={{\alpha }_{2}}\] If temperature of combined rod increases by \[t{{\,}^{o}}C,\]then increase in length in I rod \[\Delta {{l}_{2}}=\frac{{{l}_{1}}}{{{\alpha }_{1}}t}\] Increase in length of II rod \[\Delta {{l}_{2}}=\frac{{{l}_{2}}}{{{\alpha }_{2}}t}\] and \[\Delta {{l}_{1}}=\Delta {{l}_{2}}\] \[\therefore \] \[\frac{{{l}_{1}}}{{{\alpha }_{1}}t}=\frac{{{l}_{2}}}{{{\alpha }_{2}}t}\Rightarrow \frac{{{l}_{1}}}{{{\alpha }_{1}}}=\frac{{{l}_{2}}}{{{\alpha }_{2}}}\] \[\Rightarrow \] \[\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\Rightarrow \frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\] \[\Rightarrow \] \[\frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}+1=\frac{{{l}_{1}}}{{{l}_{2}}}+1\] \[\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{{{\alpha }_{2}}}=\frac{{{l}_{1}}+{{l}_{2}}}{{{l}_{2}}}\] \[\Rightarrow \] \[\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}}\]
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