Question

A rod of length $1000mm$ and coefficient of linear expansion $\alpha ={10}^{-4}{/}^{\circ }C$ is placed symmetrically
between fixed walls separated by $1001mm$. The Young’s modulus of the rod is ${10}^{11}\frac{N}{m^2}$ If the temperature is increased by ${20}^{\circ }C$, then the stress developed in the rod (in $MPa$) is

Answer 1

The change in length of rod due to increase in temperature in absence of walls is
$\Delta l=l\alpha \Delta T=1000\times {10}^{-4}\times 20mm=2mm.$
At this temperature natural length of rod $=1002mm$.
But the rod can expand upto $1001mm$ only.

$\therefore$ Compression = $1mm$
This compression causes mechanical stress in the rod.
$\therefore$ Mechanical stress $=Y\frac{\Delta l}{l}={10}^{11}\times \frac{1}{1000}={10}^8\frac{N}{m^2}=100MPa$