The change in area of a copper plate on heating obeys $\dfrac{\Delta A}{A_o} = \alpha_A \Delta T$. A plate of area $2\ \text{m}^2$ with $\alpha_A = 3 \times 10^{-5}\ \text{K}^{-1}$ is heated through $50\ \text{K}$. What is the increase in area?

1. Rearrange the formula: $\Delta A = A_o\, \alpha_A\, \Delta T$. 2. Substitute: $\Delta A = 2 \times (3 \times 10^{-5}) \times 50$. 3. Multiply: $2 \times 50 = 100$, then $100 \times 3 \times 10^{-5} = 3 \times 10^{-3}\ \text{m}^2$. 4. So the area increases by $3 \times 10^{-3}\ \text{m}^2$. 5. Forgetting to multiply by $\Delta T$ gives $6 \times 10^{-5}$-type errors, and halving the area term gives $1.5 \times 10^{-3}$, both traps. _Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 3 Thermal Physics "Superficial Expansion", p.42_