A thin copper wire of length `l` metre increases in length by `2%` when heated through `10^(@)C`. What is the percentage increase in area when a square copper sheet of length `l` metre is heated through `10^(@)C`

Options

A. (a)`4%`
B. (b)`8%`
C. (c)`16%`
D. (d)None of the above

Answer

To solve the problem, we need to determine the percentage increase in the area of a square copper sheet when it is heated through \(10^\circ C\), given that a thin copper wire of length \(l\) metre increases in length by \(2\%\) under the same conditions.

### Step-by-Step Solution:

1. **Understand the Change in Length**:
- The problem states that the length of the copper wire increases by \(2\%\) when heated. This can be expressed mathematically as:
\[ \frac{dL}{L} = 0.02 \]
where \(dL\) is the change in length and \(L\) is the original length.

2. **Relate Length to Area**:
- For a square copper sheet with side length \(l\), the area \(A\) is given by:
\[ A = l^2 \]

3. **Differentiate the Area**:
- To find the change in area \(dA\) when the length changes, we differentiate the area with respect to length:
\[ dA = 2l \cdot dL \]

4. **Express the Change in Area as a Fraction of Original Area**:
- We can express the fractional change in area as:
\[ \frac{dA}{A} = \frac{2l \cdot dL}{l^2} = \frac{2 \cdot dL}{l} \]
- Substituting \(dL = 0.02L\) into the equation gives:
\[ \frac{dA}{A} = 2 \cdot 0.02 = 0.04 \]

5. **Convert to Percentage**:
- To find the percentage increase in area, we multiply the fractional change by \(100\):
\[ \text{Percentage Increase} = \frac{dA}{A} \times 100 = 0.04 \times 100 = 4\% \]

### Final Answer:
The percentage increase in area when the square copper sheet is heated through \(10^\circ C\) is **4%**.

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