A railway track (made of iron) is laid in winter when the average temperature is $18^{\circ} C$ . The track consists of sections of $12.0 m$ placed one after the other. How much gap should be left between two such section so that there is no compression during summer when the maximum temperature goes to $48^{\circ} C$ ? Coefficient of linear expansion of iron $= {11 \times 10^{- 6}}^{\circ} C^{- 1}$ .

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Solution

$L 0 = 12 c m$

$α = 11 \times 10^{- 5} /^{\circ} C$

$t w = 18^{\circ} C$

$t s = 48^{\circ} C$

$L w = L 0 (1 + α t w) = 12 (1 + 11 \times 10^{- 5} \times 18) = 12.002 m$

$L s = L 0 (1 + α t s) = 12 (1 + 11 \times 10^{- 5} \times 48) = 12.006 m$

$Δ L = 12.006 - 12.002 = 0.004 m ≅ 0.4 c m$

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L0=12cmα=11×10–5/∘Ctw=18∘Cts=48∘CLw=L0(1+αtw)=12(1+11×10–5×18)=12.002mLs=L0(1+αts)=12(1+11×10–5×48)=12.006mΔL=12.006–12.002=0.004m≅0.4cm

$L 0 = 12 c m$

$α = 11 \times 10^{- 5} /^{\circ} C$

$t w = 18^{\circ} C$

$t s = 48^{\circ} C$

$L w = L 0 (1 + α t w) = 12 (1 + 11 \times 10^{- 5} \times 18) = 12.002 m$

$L s = L 0 (1 + α t s) = 12 (1 + 11 \times 10^{- 5} \times 48) = 12.006 m$

$Δ L = 12.006 - 12.002 = 0.004 m ≅ 0.4 c m$