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  • A rail track made of steel having length 10 m is clamped on a railway line at its two ends. On a summer day due to rise in temperature by 20oC, it is deformed as shown in the figure.

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

Answers (1)

Given data:

\alpha {steel} = \frac{1.2 \times 10^{-5}}{^{\circ}C}

L0 = 10m

& \Delta T = 20^{\circ}C

Now, by using Pythagoras theorem,

x^{2} = \left [(L + \Delta L){\frac{1}{2}} \right ]^{2} - \left (\frac{L}{2} \right )^{2}

            = \frac{1}{4} [L^{2} + \Delta L^{2} + 2L\Delta L] -\frac{ L^{2}}{4}

Thus, x = \frac{L^{2}}{4} + \frac{\Delta L^{2}}{4} + \frac{2L\Delta L}{4} -\frac{ L^{2}}{4}      ……… (neglecting \Delta L^{2} since \Delta L^{2}<<L)

x^{2 }= \frac{2L\Delta L}{4}

Thus, x= \frac{1}{2} \sqrt{2L\Delta L}

\Delta L = L_{c} \alpha .\Delta T

     = 10 \times 1.2 \times 10^{-5} \times 20

    = 240 \times 10^{-5}

Thus, x= \frac{1}{2} \sqrt{2\times 10\times 240 \times 10^{-5}}

             = \frac{40}{2}\times \sqrt{0.3} \times 10^{-2}

           = 11 \times 10^{-2} \ cm 

Thus,  x = 11 \times 10^{-2} \ cm

Posted by

infoexpert24

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