question_answer 48)

                  An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by DT. Find change in the angle ABC. [Coeff. Of linear expansion for Cu is \[{{\alpha }_{1}}\], Coeff. of linear expansion for 2 Al is \[{{\alpha }_{2}}\]]                

Answer:

                  From figure \[\cos \theta =\frac{l_{1}^{2}+l_{2}^{2}-l_{3}^{2}}{2{{l}_{1}}{{l}_{2}}}\]                 or \[l_{1}^{2}+l_{2}^{2}-l_{3}^{2}=2{{l}_{1}}{{l}_{1}}\cos \theta \]             Differentiating both sides, we get                 \[2{{l}_{1}}d{{l}_{1}}+2{{l}_{2}}d{{l}_{2}}-2{{l}_{3}}d{{l}_{3}}\]                 \[=2{{l}_{2}}d{{l}_{1}}\cos \theta +2{{l}_{1}}d{{l}_{2}}\cos \theta -2{{l}_{1}}{{l}_{2}}\sin \theta d\theta \]                 or \[{{l}_{1}}d{{l}_{1}}+{{l}_{2}}d{{l}_{2}}-{{l}_{3}}d{{l}_{3}}=({{l}_{2}}d{{l}_{1}}+{{l}_{1}}d{{l}_{2}})\]                 \[\cos \theta -{{l}_{1}}{{l}_{2}}\sin \theta \,\cos d\theta \]                 When rods are heated, their lengths are increased. The change in the lengths of the rods is given by                 \[d{{l}_{1}}={{l}_{1}}{{\alpha }_{1}}\,\Delta T,\,d{{l}_{2}}={{l}_{2}}{{\alpha }_{1}}\Delta T\] and \[d{{l}_{3}}\]\[={{l}_{3}}{{\alpha }_{2}}\Delta t\]                 Also, \[\theta ={{60}^{o}}\]                 Put these values in eqn. (i), we get                 \[l_{1}^{2}{{\alpha }_{1}}\Delta T+l_{2}^{2}{{\alpha }_{1}}\Delta T-l_{3}^{2}{{\alpha }_{2}}\Delta T\]             \[=(l_{2}^{2}{{\alpha }_{1}}\Delta T+l_{1}^{2}{{\alpha }_{1}}\Delta T)\cos {{60}^{o}}-{{l}_{1}}{{l}_{2}}\sin {{60}^{o}}d\theta \]                 Also \[{{l}_{1}}={{l}_{2}}={{l}_{3}}=l\]                 \[\therefore \]\[2{{\alpha }_{1}}{{l}^{2}}\Delta T-{{\alpha }_{2}}{{l}^{2}}\Delta T={{l}^{2}}{{\alpha }_{1}}\Delta T-\frac{{{l}^{2}}\sqrt{3}}{2}d\theta \]                 or \[d\theta =\frac{2({{\alpha }_{2}}-{{\alpha }_{1}})\Delta T}{\sqrt{3}}\]