Question 9 of 20intermediate🔧 ApplyNumerical3 marks

A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0×10−5 K−12.0 \times 10^{-5} \text{ K}^{-1}; Young’s modulus of brass = 0.91×1011 Pa0.91 \times 10^{11} \text{ Pa}.

Correct Answer

The tension developed in the wire is 377 N (approximately).

Exercise: EXERCISES | Q: 10.9 | (Page 23)
For More Understanding

Explanation

When the brass wire is cooled, it tends to contract. However, since it is held taut between two rigid supports, this contraction is prevented, resulting in thermal stress and tension in the wire. The thermal strain developed equals α × ΔT, and the corresponding stress is Y × strain. The tension is then calculated as stress × cross-sectional area.

Solution Steps

  1. Step 1: Calculate the temperature change. ΔT = T_final - T_initial = -39 - 27 = -66 °C Magnitude of temperature change = 66 °C

  2. Step 2: Calculate the thermal strain. Thermal strain = α × ΔT = 2.0×102.0 \times 10⁻⁵ × 66 = 1.32×101.32 \times 10⁻³

  3. Step 3: Calculate the thermal stress developed. Stress = Y × strain = 0.91×100.91 \times 10¹¹ × 1.32×101.32 \times 10⁻³ = 1.20×101.20 \times 10⁸ Pa

  4. Step 4: Calculate the cross-sectional area. Area = π × (d/2)² = π × (2.0×102.0 \times 10⁻³/2)² = π × 10⁻⁶ m²

  5. Step 5: Calculate the tension. Tension = Stress × Area = 1.20×101.20 \times 10⁸ × π × 10⁻⁶ = 120π ≈\approx 377 N

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