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A rail track made of steel having length $10\,m$ is clamped on a railway line at its two ends as shown in figure. On a summer day due to rise in temperature by $20\,^oC$ , it is deformed as shown in figure. Find $x$ (displacement of the centre) if $\alpha _{steel} = 1.2 \times 10^{-5} \,^oC^{-1}$
Solution
By using Pythagoras theorem, for given right angle triangle,
$\left(\frac{\mathrm{L}+\Delta \mathrm{L}}{2}\right)^{2}=\left(\frac{\mathrm{L}}{2}\right)^{2}+x^{2}$
$x=\sqrt{\left(\frac{\mathrm{L}+\Delta \mathrm{L}}{2}\right)^{2}-\left(\frac{\mathrm{L}}{2}\right)^{2}}$
$=\frac{1}{2}\left(\sqrt{(\mathrm{L}+\Delta \mathrm{L})^{2}-\mathrm{L}^{2}}\right)$
$=\frac{1}{2}\left[\sqrt{\mathrm{L}^{2}+2 \mathrm{~L} \Delta \mathrm{L}+\Delta \mathrm{L}^{2}-\mathrm{L}^{2}}\right]$
$=\frac{1}{2}\left[\sqrt{2 \mathrm{~L} \Delta \mathrm{L}+\Delta \mathrm{L}^{2}}\right]$
As $\Delta \mathrm{L}$ is very small, $\Delta \mathrm{L}^{2}$ is neglected.
$x=\frac{1}{2} \sqrt{2 \mathrm{~L} \Delta \mathrm{L}}$
But $\Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T}$
$\therefore x=\frac{1}{2} \sqrt{2 \mathrm{~L} \times \mathrm{L} \alpha \Delta \mathrm{T}}$
$=\frac{1}{2} \mathrm{~L} \sqrt{20 \Delta \mathrm{T}}$
$\therefore x=\frac{10}{2} \times \sqrt{2 \times 1.2 \times 10^{-5} \times 20}$
$=5 \times \sqrt{4.8 \times 10^{-4}}$
$=5 \times 2 \times 1.1 \times 10^{-2}$
$=11 \times 10^{-2}$
$=0.11 \mathrm{~m}=11 \mathrm{~cm}$
Here, the increase in length of mercury is neglected as it is very small.
