1.

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»

L0 = 66 m = 6600 cm 

α = 11 × 10-6 °C. 

∆L = Lt – 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 = t- t1 = 50

⇒ t2 – 10 = 50

⇒ t2 = 50 + 10 = 60°C

so final temperature t2 = 60°C



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