A steel rod of length $1m$ is heated from ${25}^{\circ }C$ to ${75}^{\circ }C$ keeping its length constant. The longitudinal strain developed in the rod is (Given: Coefficient of Linear expansion of steel $=12\times {10}^{-6}{/}^{\circ }C)$

The correct option is B $-6\times {10}^{-4}$
We know that: $\epsilon =\alpha \Delta T$
Given: ${\alpha }_s=12\times {10}^{-6}{/}^{\circ }C$,
$\Delta T=75-25={50}^{\circ }C$
Strain developed will be given by
$\epsilon =\alpha \Delta T=(12\times {10}^{-6})\times 50=6\times {10}^{-4}$
Strain will be negative, as the rod is in a compressed state.
Final answer (c)