Physics - Elasticity Question with Solution | TestHub

Increase in length of the iron wire$=\text{L}\alpha \Delta \text{t}=\text{L}\times \text{12}\times {\text{10}}^{-6}\times \text{40}=\text{48L}\times {\text{10}}^{-5}m$

Increase in length of the brass tube$=\text{L}\alpha \Delta \text{t}=\text{L}\times \text{18}\times {\text{10}}^{-6}\times \text{40}=\text{72L}\times {\text{10}}^{-5}m$

Since the increase in length of the brass tube is greater than that of the iron wire, the latter will be tightened. Let T be the tension in the wire. Then

$\text{Stress}=\frac{T}{\pi r^2}=\frac{T}{\pi {\left(3\times 10^{-4}\right)}^2}=\frac{T}{9\pi \times 10^{-8}}$

$\text{Strain}=\frac{\left(72L-48L\right)\times 10^{-5}}{L}=24\times 10^{-5}$

$\text{Since stress}=Y\times \text{strain}\text{,}\frac{T}{9\pi \times 10^{-8}}=21\times 10^{10}\times 24\times 10^{-5}$

$\text{T}=21\times 10^{10}\times 24\times 10^{-5}\times 9\pi \times 10^{-8}=\text{14.2 N}$