Question

A rod of length $1000mm$ and coefficient of liner expansion $a={10}^{-4}$ per degree is placed symmetrically between fixed walls separated by $1001mm$. The Youngs modulus of the rod is ${10}^{11}N{m}^{-2}$. If the temperature is increased by ${20}^{o}C$, then the stress develop in the rod is (in $N{m}^{-2}$)

• A. $10$
• B. ${10}^8$
• C. $2\times {10}^8$
• D. cannot be calculated

Answer 1

The correct option is B ${10}^8$

$\begin{array}{c}\Delta l_t=l\alpha \Delta \theta =1\times {10}^{-4}\times 20m=2m\\ \Rightarrow \Delta l_e=1mm\\ \frac{F}{A}=Y\frac{\Delta l}{l}\\ ={10}^{11}\times \frac{{10}^{-3}}{1}={10}^{8}N/m^2\end{array}$