A tungsten wire, 0.5 mm in diameter, is just stretched between two fixed points at a temperature of 40⁰C. Determine the tension in the wire when the temperature falls to 20⁰C. (coefficient of linear expansion of tungsten = 4.5 × 10^–6/⁰C; Young’s modulus of tungsten = 3.45 × 10¹¹Nm^–2)

Given,

$\mathrm{Y}=3.45\times {10}^{10}{\mathrm{Nm}}^{-2}; \mathrm{\alpha }=4.5\times {10}^{-6}/°\mathrm{C}; \mathrm{D}=0.5\mathrm{mm}; {\mathrm{T}}_1={40}^{°}\mathrm{C}; {\mathrm{T}}_2={20}^{°}\mathrm{C}; ∆\mathrm{T}={20}^{°}\mathrm{C}$

We know that,

$\mathrm{Young} \mathrm{modulus}, \mathrm{Y}=\frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{∆l}{l}}}\Rightarrow \frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y}\frac{∆l}{l}\mathrm{\_\_\_\_\_\_\_}\left(1\right)$

thermal expansion in length,

$\mathrm{\Delta }l=l\mathrm{\alpha \Delta T}\Rightarrow \frac{∆l}{l}=\mathrm{\alpha \Delta T\_\_\_\_\_\_\_}\left(2\right)$

$\mathrm{Area} \mathrm{A}=\frac{{\mathrm{\pi D}}^2}{4}\mathrm{\_\_\_\_\_\_\_\_}\left(3\right)$

Substituting 2 and 3 in equation 1,

$\mathrm{F}=\frac{{\mathrm{\pi D}}^2}{4}\mathrm{Y\alpha \Delta T}=\frac{3.14\times 0.5\times 0.5\times {10}^{-6}}{4}\times 3.45\times {10}^{10}\times 4.5\times {10}^{-6}\times 20=6.097\mathrm{N}$

Hence the correct answer is 6.097N.