A steel rod is held tightly between two rigid walls and heated through $50^\circ$C. For steel, $\alpha = 1.2 \times 10^{-5}\ \text{K}^{-1}$ and $Y = 2 \times 10^{11}$ N m$^{-2}$. The stress developed in the rod is
Answer: (C) $1.2 \times 10^8$ N m$^{-2}$
Concept. The walls stop the rod expanding, so they compress it by the strain it would have expanded, $\alpha\Delta T$. Stress $= Y \times$ strain.
Step 1. Prevented strain. $\alpha\Delta T = 1.2 \times 10^{-5} \times 50 = 6 \times 10^{-4}$.
Step 2. Stress.
$$\text{stress} = Y\alpha\Delta T = 2 \times 10^{11} \times 6 \times 10^{-4} = 1.2 \times 10^8\ \text{N m}^{-2}$$
It does not depend on the rod's length or area.
Answer. $1.2 \times 10^8$ N m$^{-2}$ (Option C).
Solution by Sreeraj P, M.Sc Physics