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A Steel Wire of Cross-sectional Area 0.5 Mm2 is Held Between Two Fixed Supports.

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प्रश्न

A steel wire of cross-sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10–5 °C–1 and its Young's modulus is 2.0 × 10–11 Nm–2.

थोडक्यात उत्तर
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उत्तर

Given:
Cross-sectional area of the steel wire, A = 0.5 mm2 = 0.5 × 10–6 m2
The wire is taut at a temperature, T1 = 20 °C,
After this, the temperature is reduced to T2 = 0 °C
​So, the change in temperature, Δθ = T1-T2 = 20 °C
Coefficient of linear expansion of steel, α = 1.2 ×10–5 °C​-1
Young's modulus, γ = 2 ×1011 Nm​-2
Let L be the initial length of the steel wire and L' be the length of the steel wire when temperature is reduced to 0°C.

Decrease in length due to compression, ΔL =L' - L = LαΔθ ...(1)

Let the tension applied be F.

`γ = "stress"/"strain" =("F"/"A")/((triangle"L")/"L")` 

`=> γ = "F"/"A" xx "L"/(triangle"L")`

`=> triangle"L" = "FL"/("A"gamma)`     ..(2)

Change in length due to tension produced is given by (1) and (2).
So, on equating (1) and (2), we get:

`"L"αΔθ ="FL"/("A"gamma)`

⇒ F = αΔθAγ

= 1.2 × 10-5 × (20-0) × 0.5 ×10-6 × 2 ×1011

= 1.2 × 20

⇒ F = 24 N

Therefore, the tension produced when the temperature falls to 0°C

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पाठ 23: Heat and Temperature - Exercises [पृष्ठ १३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 23 Heat and Temperature
Exercises | Q 28 | पृष्ठ १३

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