Question

Two uniform metal rods of lengths ${{l}_{1}}$ and ${{l}_{2}}$ and linear coefficients of expansion ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ respectively are connected to form a single rod of length $({{l}_{1}}+{{l}_{2}}).$ When the temperature of the combined rod is raised by t ?C, the length of each rod increases by the same amount. Then $\left( \frac{{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}} \right)$ is:

• 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. $\frac{({{l}_{1}}+{{l}_{2}})}{{{l}_{2}}}$

Answer 1

Answer: (C)

$\frac{{{l}_{2}}}{\left( {{l}_{1}}+{{l}_{2}} \right)}$

Answer 2

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 $\Delta {{l}_{1}}=\Delta {{l}_{2}}$ $\therefore$ $\frac{{{l}_{1}}}{{{\alpha }_{1}}t}=\frac{{{l}_{2}}}{{{\alpha }_{2}}t}\Rightarrow \frac{{{l}_{1}}}{{{\alpha }_{1}}}=\frac{{{l}_{2}}}{{{\alpha }_{2}}}$ $\Rightarrow$ $\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\Rightarrow \frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}$ $\Rightarrow$ $\frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}+1=\frac{{{l}_{1}}}{{{l}_{2}}}+1$ $\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{{{\alpha }_{2}}}=\frac{{{l}_{1}}+{{l}_{2}}}{{{l}_{2}}}$ $\Rightarrow$ $\frac{{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}}$