Q. A rod is fixed between two points at $20℃$. The coefficient of linear expansion of material of rod is $1.1 \times 10^{- 5}/℃$ and Young’s modulus is $1.2 \times 10^{11}N/m^{2}$. Find the stress developed in the rod if temperature of rod becomes $10℃$
A) $1.32 \times 10^{7}N/m^{2}$
B) $1.10 \times 10^{15}N/m^{2}$
C) $1.32 \times 10^{8}N/m^{2}$
D) $1.10 \times 10^{6}N/m^{2}$
Explanation
(a) Thermal stress $= Y\alpha\mathrm{\Delta}\theta$
$= 1.2 \times 10^{11} \times 1.1 \times 10^{- 5} \times (20 - 10) = 1.32 \times 10^{7}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.