Question

If the gap between steel sails on the railway track of 66 m long is 3.63 cm at 10°C. Then at what value of temperature will just touch of steel is 11 × 10^-6 °C.

Answer 1

L₀ = 66 m = 6600 cm 

α = 11 × 10^-6 °C. 

∆L = L_t – L₀ = 3.63

t₁ = 10°C 

t₂ = ?

\(a = \frac{Δ L}{L_0Δ T}\)

ΔT = \(\frac{Δ L}{L_0 \times a}\)

Δ T = \(\frac{3.63}{6600 \times 11 \times 10^{-6}}\)

Δ T = t₂ - t₁ = 50

⇒ t₂ – 10 = 50

⇒ t₂ = 50 + 10 = 60°C

so final temperature t₂ = 60°C