Solution:
The percentage change in length is given by \(\frac{\Delta L}{L} \times 100 = \alpha \Delta T \times 100 = (2 \times 10^{-5}) \times 100 \times 100 = 0.2%\).
The percentage change in length of 1 m iron rod if its temperature changes by 100ºC is (\(\alpha\) for iron is \(2 \times 10^{-5}/\text{ºC}\))
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