question_answer 48)

                  A thin rod having length \[{{L}_{0}}\] at \[{{0}^{o}}C\]and coefficient of linear expansion \[\alpha \] has its two ends maintained at temperature \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\], respectively. Find its new length.                

Answer:

                  Assume that temperature of the rod varies linearly from one end to another end. Let \[\theta \] be the temperature of the rod at its mid point.                 Now\[\frac{d\theta }{dt}=\,\frac{KA\,({{\theta }_{1}}-\theta )}{{{L}_{0}}/2}\,\,\frac{KA\,(\theta -{{\theta }_{2}})}{{{L}_{0}}/2}\]                 \[\therefore \]\[{{\theta }_{1}}-\theta =\theta -{{\theta }_{2}}\]                 or \[\theta \,=\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]                 Using  \[L={{L}_{0}}(1+\alpha \theta ),\] we get                 \[L=\,{{L}_{0}}\,\left[ 1+\alpha \left( \frac{{{\theta }_{1}}+\,{{\theta }_{2}}}{2} \right) \right]\]