Question

A thin copper wire of length L increases in length by 1%, when heated from \(T_1\) to \(T_2\). What is the percentage change in area when a thin copper plate having dimensions \((10L \times 2L)\) is heated from \(T_1\) to \(T_2\)?

• A. 2%
• B. 20%
• C. 10%
• D. 40%

Hint:

Area expansion is approximately twice the linear expansion for small changes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The coefficient of superficial (area) expansion (\( \beta \)) is twice the coefficient of linear expansion (\( \alpha \)).
For small percentage changes, the percentage change in area is twice the percentage change in length.
: Key Formula or Approach:
1. \( \frac{\Delta L}{L} \times 100 = 1% \).
2. \( \frac{\Delta A}{A} = 2 \frac{\Delta L}{L} \).
Step 2: Detailed Explanation:
- Given that the percentage increase in length is 1%. This corresponds to linear expansion.
- The area \( A \) of the plate is \( (10L) \times (2L) = 20L^2 \).
- When heated, every linear dimension (length and width) increases by 1%.
- Percentage change in area \( = 2 \times (\text{Percentage change in length}) \).
\[ % \Delta A = 2 \times 1% = 2% \]
Step 3: Final Answer:
The percentage change in area is 2%.