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A thin rod having length $L_0$ at $0\,^oC$ and coefficient of linear expansion $\alpha $ has its two ends maintained at temperatures $\theta _1$ and $\theta _2$, respectively. Find its new length.
Solution
The temperature in rod changed by going linearly from its one end to another end and temperature at midpoint is $\theta$. In thermal steady state heat current $=\frac{d \mathrm{Q}}{d t}=$ constant.
$\therefore \mathrm{KA} \frac{\theta_{1}-\theta}{\left(\mathrm{L}_{0} / 2\right)}=\frac{\mathrm{KA}\left(\theta-\theta_{2}\right)}{\left(\mathrm{L}_{0 / 2}\right)}$
where $\mathrm{K}$ is thermal conductivity,
$\therefore \theta_{1}-\theta=\theta-\theta_{2}$
$\therefore \theta_{1}+\theta_{2}=2 \theta$
$\therefore \theta=\frac{\theta_{1}+\theta_{2}}{2} \text { temperature of midpoint }$
Now, its length increases with increase in temperature,
$\therefore \mathrm{L}=\mathrm{L}_{0}(1+\alpha \theta)$
$\therefore \mathrm{L}=\mathrm{L}_{0}\left[1 \times \alpha\left(\frac{\theta_{1}+\theta_{2}}{2}\right)\right] \text { which is new length. }$
