$A$ rail track made of steel having length $10\,m$ is clamped on a railway line at its two ends as shown in the figure. On a summer day,due to a rise in temperature by $20\,^oC$,it is deformed as shown in the figure. Find $x$ (displacement of the centre) if $\alpha_{steel} = 1.2 \times 10^{-5} \,^oC^{-1}$.

(N/A) By using the Pythagoras theorem for the given right-angled triangle:
$\left(\frac{L+\Delta L}{2}\right)^{2} = \left(\frac{L}{2}\right)^{2} + x^{2}$
$x = \sqrt{\left(\frac{L+\Delta L}{2}\right)^{2} - \left(\frac{L}{2}\right)^{2}}$
$x = \frac{1}{2} \sqrt{(L+\Delta L)^{2} - L^{2}}$
$x = \frac{1}{2} \sqrt{L^{2} + 2L\Delta L + (\Delta L)^{2} - L^{2}}$
$x = \frac{1}{2} \sqrt{2L\Delta L + (\Delta L)^{2}}$
Since $\Delta L$ is very small,$(\Delta L)^{2}$ can be neglected.
$x \approx \frac{1}{2} \sqrt{2L\Delta L}$
Given $\Delta L = L \alpha \Delta T$,we substitute this into the equation:
$x = \frac{1}{2} \sqrt{2L(L \alpha \Delta T)} = \frac{L}{2} \sqrt{2 \alpha \Delta T}$
Substituting the values $L = 10\,m$,$\alpha = 1.2 \times 10^{-5} \,^oC^{-1}$,and $\Delta T = 20\,^oC$:
$x = \frac{10}{2} \sqrt{2 \times 1.2 \times 10^{-5} \times 20}$
$x = 5 \sqrt{48 \times 10^{-5}} = 5 \sqrt{4.8 \times 10^{-4}}$
$x = 5 \times 2.19 \times 10^{-2} \approx 0.1095\,m \approx 11\,cm$.