Question

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^–1; Young’s modulus of brass = 0.91 × 10^11 Pa.

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Answer 1

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Given :

Initial Temperature = 27° C

Final Temperature = -39°C

Length of the wire = 1.8m

Diameter of the wire = 2mm

Young's Modulus = 0.91 × 10¹¹ Pa

$\sf Young's \ Modulus = \dfrac{Stress}{Strain} \\ \bf Now \\ \begin{gathered}\begin{gathered}\sf Y = \dfrac{\dfrac{F}{A}}{\dfrac{e}{L}} \\ \\ \implies \sf e = \dfrac{FL}{AY}---------(1)\end{gathered} \end{gathered} \\ \bf Also, \\ \sf \: e = \alpha L \Delta{T} \\ \\ \sf \dfrac{F \times L}{A \times Y} = \alpha L \Delta{T} \\ \bf \: Substituting \: values, \: we \: get : \\ \boxed{ \boxed{ \sf F= - 3.8 \times {10}^{2} \: N}}$

(The negative sign indicates that the tension is directed inward.)

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Answer 2

Answer:

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