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NCERT Exemplar · Q25

Q.A straight rail track made of steel, of length L=10L = 10 m, is clamped rigidly to a railway line at both of its ends so that it cannot expand freely. On a summer day the temperature rises by 20 ∘20\,^\circC, so the track's natural expansion has nowhere to go and it buckles: it takes the shape of two equal straight segments meeting at the centre, which is pushed out by a distance xx from the original straight line. The distance between the two clamped ends stays fixed at LL, while each of the two buckled segments has length 12(L+ΔL)\tfrac{1}{2}(L + \Delta L), where ΔL\Delta L is the total increase in the track's length caused by the heating. Taking the coefficient of linear expansion of steel as αsteel=1.2×10−5 /∘C\alpha_{steel} = 1.2 \times 10^{-5}\ /^\circ\mathrm{C}, find xx, the displacement of the centre.

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Because both ends are clamped, the steel cannot lengthen when heated, so its thermal expansion forces it to bow out. The buckled track forms two equal straight arms of length 12(L+ΔL)\tfrac12(L+\Delta L) on a base that is still LL, and the centre lifts by xx. Applying Pythagoras to one half and using ΔL=LαΔT\Delta L = L\alpha\Delta T gives x≈0.11x \approx 0.11 m.

Concept

A freely heated rod would grow by ΔL=L α ΔT\Delta L = L\,\alpha\,\Delta T. Here the ends are fixed, so the extra length has nowhere to go along the line; the track relieves it by bending out of the straight line (buckling). The material length becomes L+ΔLL + \Delta L, split into two equal straight arms meeting at the raised centre, while the straight-line distance between the clamps is still LL.

Geometry (why Pythagoras)

Each arm is the hypotenuse of a right triangle whose base is half the clamp separation, L/2L/2, and whose height is the centre displacement xx:

(L+ΔL2)2=(L2)2+x2.\left(\frac{L+\Delta L}{2}\right)^2 = \left(\frac{L}{2}\right)^2 + x^2.

Solving for xx:

x=(L+ΔL2)2−(L2)2=12(L+ΔL)2−L2=122L ΔL+ΔL2.x = \sqrt{\left(\frac{L+\Delta L}{2}\right)^2 - \left(\frac{L}{2}\right)^2} = \frac{1}{2}\sqrt{(L+\Delta L)^2 - L^2} = \frac{1}{2}\sqrt{2L\,\Delta L + \Delta L^2}.

Steps

  1. Thermal expansion: ΔL=L αsteel ΔT=10×(1.2×10−5)×20=2.4×10−3 m.\Delta L = L\,\alpha_{steel}\,\Delta T = 10 \times (1.2\times10^{-5}) \times 20 = 2.4\times10^{-3}\ \text{m}. …

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