Ans ITT t fds srightt t

nitial temperature, T ₁ = 27 ° C

Length of the brass wire at T ₁ , l = 2 m

Final temperature, T ₂ = –43 ° C

Diameter of the wire, d = 2.0 mm = 2 × 10 ^–3 m

Tension developed in the wire = F

Coefficient of linear expansion of brass, α = 11 × 10 ^–6 K ^–1

Young’s modulus of brass, Y = 2 × 10 ¹¹ Pa

Young’s modulus is given by the relation:

$$\begin{gathered} \mathrm{Y}=\frac{\mathrm{stress}}{\mathrm{strain}} \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{F}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$∆\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ ∆\mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}} .................. \left(1\right) \end{gathered}$$

Where,

F = Tension developed in the wire

A = Area of cross-section of the wire.

Δ L = Change in the length, given by the relation:

Δ L = α L ( T ₂ – T ₁ ) … ( ii )

Equating equations ( i ) and ( ii ), we get:
$$\begin{gathered} \alpha L\left(T_2-T_1\right)=\frac{FL}{\mathrm{\pi }{\left({\displaystyle \raisebox{1ex}{$\mathrm{d}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2\mathrm{Y}} \\ \mathrm{F}=\mathrm{\alpha L}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\mathrm{\pi }{\left(\raisebox{1ex}{$\mathrm{d}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2 \\ \mathrm{F}=11\times {10}^{-6}\left(-43-27\right)\times 3.14\times 2\times {10}^{11}{\left(\frac{2\times {10}^{-3}}{2}\right)}^2 \\ \mathrm{F}=-483.56 \mathrm{N} \end{gathered}$$

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

Regards