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Q11.9:
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 = $2.0 \times 10^{-5} K^{-1}$; Young’s modulus of brass = $0.91 \times 10^{11}$ Pa.

Solution :

Step 1: Given Variables & Initial Setup

To determine the tension ($F$) developed in the brass wire due to thermal contraction, we utilize the relationship between thermal strain, stress, and Young's Modulus ($Y$).

  • Initial length ($L$) = $1.8\text{ m}$
  • Initial temperature ($T_1$) = $27\text{ °C}$
  • Final temperature ($T_2$) = $-39\text{ °C}$
  • Diameter ($d$) = $2.0\text{ mm} = 2.0 \times 10^{-3}\text{ m}$
  • Coefficient of linear expansion ($\alpha$) = $2.0 \times 10^{-5}\text{ K}^{-1}$
  • Young’s modulus ($Y$) = $0.91 \times 10^{11}\text{ Pa}$

Step 2: Calculating Change in Temperature and Strain

The change in temperature ($\Delta T$) is calculated as follows:

$\Delta T = T_2 - T_1 = -39\text{ °C} - 27\text{ °C} = -66\text{ °C} = -66\text{ K}$

The thermal strain ($\frac{\Delta L}{L}$) that the wire would experience if free to contract is given by [Per the definition of linear thermal expansion]:

$\frac{\Delta L}{L} = \alpha |\Delta T|$

$\frac{\Delta L}{L} = (2.0 \times 10^{-5}\text{ K}^{-1}) \times (66\text{ K}) = 1.32 \times 10^{-3}$

Step 3: Relating Stress to Young’s Modulus

Since the wire is held between rigid supports, it cannot contract. This induces a tensile stress ($\sigma$) equivalent to the strain. By the definition of Young's Modulus ($Y = \frac{\text{Stress}}{\text{Strain}}$):

$\sigma = Y \times \left( \frac{\Delta L}{L} \right)$

$\sigma = (0.91 \times 10^{11}\text{ Pa}) \times (1.32 \times 10^{-3}) = 1.2012 \times 10^8\text{ N/m}^2$

Step 4: Calculating the Cross-sectional Area and Tension

The cross-sectional area ($A$) of the wire is calculated as [Geometry of a cylinder]:

$A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{2.0 \times 10^{-3}\text{ m}}{2} \right)^2 = \pi \times (10^{-3}\text{ m})^2 = 3.1416 \times 10^{-6}\text{ m}^2$

The tension ($F$) is defined by $F = \sigma \times A$:

$F = (1.2012 \times 10^8\text{ N/m}^2) \times (3.1416 \times 10^{-6}\text{ m}^2)$

$F \approx 377.38\text{ N}$

Visual Representation of Thermal Constraint

Brass Wire Rigid Supports (Fixed) Tension F = 3.8 x 10² N

Final Solution: The tension developed in the wire is approximately $3.8 \times 10^2\text{ N}$.


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