Question

A rod of iron of Young’s modulus $Y = 2.0 \times 10^{11}N/m^{2}$ just fits the gap between two rigid supports 1m apart. If the rod is heated through $100^{o}C$ the strain energy of the rod is ($\alpha = 18 \times 1{0^{- 6}}^{o}C^{- 1}$ and area of cross-section $A = 1cm^{2}$)

• A. 32.4 J
• B. 32.4 Mj
• C. 26.4 J
• D. 26.4 Mj

Answer 1

Answer: (A)

32.4 J

Answer 2

$U = \frac{1}{2} \times Y \times (\text{strain})^{2} \times \text{volume} = \frac{1}{2} \times Y(\alpha\Delta\theta)^{2} \times A \times L = \frac{1}{2} \times (2 \times 10^{11}) \times (18 \times 10^{- 6} \times 100)^{2} \times 1 \times 10^{- 4} \times 1 = 324 \times 10^{- 1}$= 32.4 J.