The length of each rail is $$10m$$. The linear expansion of steel is $$0.000012/0C$$ and the range of variation of temperature at the given place is $$150C$$. So the gap to be provided between the rail is

Question

The length of each rail is $$10m$$. The linear expansion of steel is  $$0.000012/0C$$ and the range of variation of temperature at the given place is  $$150C$$. So the gap to be provided between the rail is

Infinity Learn · Accepted Answer

change in length due to variation in temperature is Δ$$l=l$$αΔt here l is initial $$length=10m$$;α is coefficient of linear $$expansion=12$$×$$10$$−$$6/C$$  $$0$$ Δt is change in $$temperature=150C$$ Δ$$l=10$$×$$12$$×$$10$$−$$6$$×$$15$$ Δ$$l=180$$×$$10$$−$$5=18$$×$$10$$−$$4$$ Δ$$l=0.0018m=gap$$ between $$2$$ rails to be maintained

The length of each rail is 10m. The linear expansion of steel is $0.000012 /^{0} C$ and the range of variation of temperature at the given place is $15^{0} C$ . So the gap to be provided between the rail is

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The length of each rail is 10m. The linear expansion of steel is 0.000012/0C and the range of variation of temperature at the given place is 150C. So the gap to be provided between the rail is

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The length of each rail is 10m. The linear expansion of steel is $0.000012 /^{0} C$ and the range of variation of temperature at the given place is $15^{0} C$ . So the gap to be provided between the rail is