91Ó°ÊÓ

Temperature of steel rod,$T_i={25}^{\text{0}}\text{C}$.

The Young's modulus of the substance defines its property of getting stretched and then deforming into its original shape after the pressure is released. The higher the value of Young's modulus of the substance, the stiffer is the substance. It is the ratio of the stress applied to the body and the strain it has due to the linear expansion resulting from increasing temperature. We can find the temperature at which the steel rod will rupture by using the equation of Young’s modulus in the equation of thermal expansion.

Formula:

Linear expansion of a body due to thermal radiation,$\mathrm{Δ}L=\mathrm{Î\pm }L\mathrm{Δ}T$ …(¾±)

where$\mathrm{Î\pm }$ is the coefficient of thermal linear expansion of the substance,$\mathrm{Δ}T$is the temperature difference at both the ends of the body,and $L$is the original length.

Relation of stress and strain of a body in terms of Young’s modulus, E:

$\frac{F}{A}=E\left(\frac{\mathrm{Δ}L}{L}\right)$ …(¾±¾±)

where $F$is the force applied on the body,$A$ is the area on which the force is applied,$\mathrm{Δ}L$ is the change in length, and $L$is the original length of th body

From equation (i), we get the ratio of change in length to the original length as

$\frac{\mathrm{Δ}L}{L}=\mathrm{Î\pm }\mathrm{Δ}T$ …(¾±¾±¾±)

Substituting the above value of equation (iii) in equation (ii), we get the change in temperature as

$\mathrm{Δ}T=\frac{\left(\frac{F}{A}\right)}{E\mathrm{Î\pm }}$ …(¾±±¹)

From table:

For steel, the value of stress applied on it is given by

$\left(\frac{F}{A}\right)=400×{10}^6{\text{N/m}}^{\text{2}}$

For steel, the value of Young’s modulus applied on it is given by

$E=200×{10}^9{\text{N/m}}^{\text{2}}$

For steel, the value of coefficient of linear expansion is given by

$\mathrm{Î\pm }=11×{10}^{−6}{/}^0\text{C}$

Substitutingthe above values in equation (iv), the change in temperature is given by

$\begin{array}{c}\mathrm{Δ}T=\frac{(400×{10}^6\frac{\text{N}}{{\text{m}}^{\text{2}}})}{(200×{10}^9\frac{\text{N}}{{\text{m}}^{\text{2}}})(11×{10}^{−6}\frac{\text{1}}{{}^{\text{0}}\text{C}})}\\ ={182}^0\text{C}\end{array}$

Therefore, the final temperature at which the steel rod will rupture is given by

$\begin{array}{c}{T}_i−T_f={182}^0\text{C}\\ {25}^0\text{C}−T_f={182}^0\text{C}\\ T_f=−{157}^0\text{C}\end{array}$

Therefore, the temperature at which the steel rod will rupture is$−{157}^0\text{C}$ .