Question
The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_{1}$ to $T_{2}$ resulting in increase of its length by $\Delta L_{1}$. The temperature is further increased from $T_{2}$ to $T_{3}$ such that the increase in its length is $\Delta L_{2}$.
Given $T_{3} + T_{1} = 2{\text{ }T}_{2}$ and $T_{2} - T_{1} = \Delta T$, the value of $\Delta L_{2}$ is $\_\_\_\_$ .
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Solution
Sol. [Key Concept: Successive expansion: $\Delta\ L_{2} = L_{0}\alpha\Delta\ T + \Delta\ L_{1}\alpha\Delta\ T = \Delta\ L_{1}(1 + \alpha\Delta\ T)$.]
$$\Delta L_{1} = L_{0}\alpha\left( {\text{ }T}_{2} - T_{1} \right) $$So final length at $T_{2} = L_{0} + \Delta L_{1}$
$\Delta L_{2} = \left( L_{0} + \Delta L_{1} \right)\alpha\left( T_{3} - T_{2} \right)$ and $T_{3} - T_{2} = T_{2} - T_{1} = \Delta T$
$${\Delta L_{2} = L_{0}\alpha\Delta\text{ }T + \Delta L_{1}\alpha\Delta\text{ }T }{= \Delta L_{1} + \Delta L_{1}\alpha\Delta\text{ }T }{= \Delta L_{1}(1 + \alpha\Delta T) }$$
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