• # question_answer The coefficients of linear expansions of brass and steel are ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ respectively. When we take a brass rod of length ${{l}_{1}}$ and a steel rod of length ${{l}_{2}}$ at ${{0}^{o}}C,$ then the difference in their lengths $({{l}_{2}}-{{l}_{1}})$ will remain the same at all temperatures if:               [AIPMT 1999] A) ${{\alpha }_{1}}{{l}_{1}}={{\alpha }_{2}}{{l}_{2}}$ B)       ${{\alpha }_{1}}{{l}_{2}}={{\alpha }_{2}}{{l}_{1}}$ C)  $\alpha _{1}^{2}{{l}_{2}}=\alpha _{2}^{2}{{l}_{1}}$ D)                   ${{\alpha }_{1}}l_{2}^{2}={{\alpha }_{2}}l_{1}^{2}$

 Linear expansion coefficient $=\frac{\text{change in length}}{\text{original length }\times \text{ rise temperature}}$ or         $\alpha =\frac{\Delta l}{lt}$ or         $\Delta l=l\,\alpha \,t$ For brass rod, $\Delta {{l}_{1}}={{l}_{1}}{{\alpha }_{1}}t$ For steel rod, $\Delta {{l}_{2}}={{l}_{2}}{{\alpha }_{2}}t$ Since, ${{l}_{2}}-{{l}_{1}}=$ constant (give) So,       $\Delta {{l}_{2}}-\Delta {{l}_{1}}=0$ or         $\Delta {{l}_{2}}=\Delta {{l}_{1}}$ $\therefore$      ${{l}_{2}}{{\alpha }_{2}}t={{l}_{1}}{{\alpha }_{1}}t$ As $t\ne 0,$ hence ${{l}_{2}}{{\alpha }_{2}}={{l}_{1}}{{\alpha }_{1}}$