A rail tack made of steel having length l0 m is clamped on a railway line at its two ends (figure). On a summer day due to rise in temperature by 20°C. It is deformed as shown in figure. Find x (displacement of the centre) if 

                      ${\mathrm{\alpha }}_{\mathrm{steel}}= 1.2 \times {10}^{-5}/{ }^0\mathrm{C}$

 

Consider the diagram.

Applying Pythagoras theorem in right angled triangle in figure

${\left(\frac{\mathrm{L}+∆\mathrm{L}}{2}\right)}^2 = {\left(\frac{\mathrm{L}}{2}\right)}^2 +{\mathrm{x}}^2$$\Rightarrow \mathrm{x} = \sqrt{{\left(\frac{\mathrm{L}+∆\mathrm{L}}{2}\right)}^2-{\left(\frac{\mathrm{L}}{2}\right)}^2} = \frac{1}{2}\sqrt{{(\mathrm{L}+∆\mathrm{L})}^2-{\mathrm{L}}^2}$

$\Rightarrow \frac{1}{2}\sqrt{(∆{\mathrm{L}}^2+2\mathrm{L}∆\mathrm{L})}$

As increase in length $∆\mathrm{L}$ is very small, therefore, neglecting (${(∆\mathrm{L})}^2$), we get

$\mathrm{x} = \frac{1}{2}\times \sqrt{2\mathrm{L}∆\mathrm{L}} ---------\left(\mathrm{i}\right)$

But $∆\mathrm{L} = \mathrm{L} \mathrm{\alpha }∆\mathrm{t}------\left(\mathrm{ii}\right)$

Substituting value of $∆\mathrm{L}$ in Eq.(i) from Eq.(ii)

$\mathrm{x}= \frac{1}{2}\sqrt{2\mathrm{L}\times \mathrm{L\alpha }∆\mathrm{t}} = \frac{1}{2}\mathrm{L}\sqrt{2\mathrm{\alpha }∆\mathrm{t}}$

    = $\frac{10}{2}\times \sqrt{2\times 1.2\times {10}^{-5}\times 20}$

    = $5\times \sqrt{4\times 1.2\times {10}^{-4}} = 20\sqrt{30} \times {10}^{-3} \mathrm{cm}$