Question

A rod of length 1000 mm and coefficient of linear expansion $\alpha ={10}^{-4}$ per degree is placed symmetrically
between fixed walls separated by 1001 mm. The Young’s modulus of the rod is ${10}^{11}\frac{N}{m^2}$ If the temperature
is increased by 20⁰C, then the stress developed in the rod is due to a

• A. compression of 2 mm
• B. compression of 1 mm
• C. stretch of 2 mm
• D. stretch of 1 mm

Answer 1

The correct option is B compression of 1 mm

The change in length of rod due to increase in temperature in absence of walls is
$\Delta l=l\alpha \Delta T=1000\times {10}^{-4}\times 20mm=2mm.$
But the rod can expand upto 1001 mm only.
At that temperature its natural length = 1002 mm.
$\therefore$ Compression = 1 mm
$\therefore$ Machanical stress $=Y\frac{\Delta l}{l}={10}^{1}1\times \frac{1}{1000}={10}^8\frac{N}{m^2}$