question_answer 44)

                    We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If \[{{\alpha }_{iron}}=\,1.2\times \,{{10}^{-5}}/K\] and \[{{\alpha }_{brass}}=\,1.8\times \,{{10}^{-5}}/K,\] what should be take as length of each strip?                

Answer:

                  \[{{L}_{i}}-{{L}_{b}}=10\,cm\]                                        ?.. (i)                 Now \[{{L}_{i}}={{L}_{io}}(1+{{\alpha }_{i}}\,\Delta \theta )\]                 \[{{L}_{b}}={{L}_{bo}}\,(1+\,{{\alpha }_{b}}\Delta \theta )\]                 \[\therefore \] \[{{L}_{i}}-{{L}_{b}}=({{L}_{io}}-{{L}_{ba}})+\,\Delta \theta \,({{L}_{io}}{{\alpha }_{i}}-{{L}_{bo}}{{\alpha }_{b}})\]                 Since \[{{L}_{io}}-{{L}_{bo}}=\] constant.                 \[\therefore \] \[{{L}_{io}}\,{{\alpha }_{i}}-{{L}_{bo}}\,{{\alpha }_{b}}=0\] or \[{{L}_{io}}\,{{\alpha }_{i}}\,={{L}_{bo}}\,{{\alpha }_{b}}\]                 or \[\frac{{{L}_{i}}}{{{L}_{b}}}\,=\frac{{{\alpha }_{b}}}{{{\alpha }_{i}}}\,=\frac{1.8\,\times \,{{10}^{-5}}}{1.2\,\times \,{{10}^{-5}}}=\frac{3}{2}\]                 or \[{{L}_{i}}=\,\frac{3}{2}\,{{L}_{b}}=\,1.5\,{{L}_{b}}\]                           ?. (ii)                 Put this value of eqn. (i), we get                 \[0.5\,{{L}_{b}}=10\,cm\] or \[{{L}_{b}}=20\,cm\]                 and from eqn. (ii), \[{{L}_{i}}=1.5\times 20=30\,cm\]