question_answer 45)

                  We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron \[({{\beta }_{ubrass}}=6\times {{10}^{-5}}/K\] and \[{{\beta }_{urion}}=3.55\,\,\times \,{{10}^{-5}}/K)\] to create a volume of 100 cc. How do you think you can achieve this?                

Answer:

                  Let \[{{V}_{io}}=\] initial volume of iron vessel                 \[{{V}_{bo}}\,=\,\] initial volume of brass rod                 Now \[{{V}_{io}}\,-{{V}_{bo}}=100\,\,cc\,\,={{V}_{i}}-{{V}_{b}}\]           ?? (i)                 Now, \[{{V}_{i}}={{V}_{io}}(1+{{\beta }_{i}}\,\Delta \theta )\]                 \[{{V}_{b}}=\,{{V}_{b0}}(1+\,{{\beta }_{b}}\Delta \theta )\]                 \[\therefore \] \[{{V}_{i}}-\,{{V}_{b}}=\,({{V}_{io}}-{{V}_{bo}})\,+\,\Delta \theta \,({{V}_{io}}{{\beta }_{i}}-{{V}_{bo}}{{\beta }_{b}})\]                 Since \[{{V}_{i}}-\,{{V}_{b}}=\] constant                 \[\therefore \] \[{{V}_{io}}\,{{\beta }_{i}}={{V}_{bo}}\,{{\beta }_{b}}\] or \[\frac{{{V}_{io}}}{{{V}_{bo}}}\,=\frac{{{\beta }_{b}}}{{{\beta }_{i}}}\,=\frac{6}{3.55}\]  ?.. (ii)                 From eqns. (i) and (ii)                 \[{{V}_{b}}=\,144.9\,cc\] and \[{{V}_{i}}=\,244.9\,\,cc\]