Skip to content
Numericals · Q29

Q.In olden days, while laying the rails for trains, small gaps used to be left between the rail sections to allow for thermal expansion. Suppose the rails are laid at room temperature 27 ºC. If the maximum temperature in the region is 45 ºC and the length of each rail section is 10 m, what should be the gap left, given that α = 1.2 × 10⁻⁵ K⁻¹ for the material of the rail section?

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
76% · 29/38 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Given room temperature 27 °C27\,°\text{C}, maximum temperature 45 °C45\,°\text{C} so ΔT=18 °C\Delta T = 18\,°\text{C}, rail length l=10 ml=10\text{ m}, α=1.2×10−5 K−1\alpha=1.2\times10^{-5}\text{ K}^{-1}. Using Δl=lαΔT\Delta l = l\alpha\Delta T (from Eq. 7.10): Δl=10×1.2×10−5×18=2.16×10−3 m=2.16 mm\Delta l = 10\times1.2\times10^{-5}\times18 = 2.16\times10^{-3}\text{ m} = 2.16\text{ mm}. This is the minimum gap that must be left between consecutive rail sections at the (cooler) laying temperature, so that o …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.