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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. |
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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 = t2 - t1 = 50 ⇒ t2 – 10 = 50 ⇒ t2 = 50 + 10 = 60°C so final temperature t2 = 60°C |
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