A thin copper wire of length l increases in length by 1% when heated from `0^(@)C` to `100^(@)C`. If a then copper plate of area `2lxxl` is heated from `0^(@)C` to `100^(@)C`, the percentage increases in its area will be

Options

A. `1%`
B. `2%`
C. `3%`
D. `4%`

Explanation

To solve the problem of finding the percentage increase in the area of a copper plate when heated from \(0^\circ C\) to \(100^\circ C\), we can follow these steps:

### Step 1: Understand the linear expansion of materials
When a material is heated, it expands. For a linear dimension (like length), the change in length (\(\Delta L\)) is given by:
\[ \frac{\Delta L}{L} = \alpha \Delta T \]
where \(\alpha\) is the coefficient of linear expansion and \(\Delta T\) is the change in temperature.

### Step 2: Given information
From the problem, we know that a thin copper wire of length \(L\) increases in length by 1% when heated from \(0^\circ C\) to \(100^\circ C\). This means:
\[ \frac{\Delta L}{L} = 0.01 \quad (\text{which is } 1\%) \]
Thus, we can conclude that:
\[ \alpha \Delta T = 0.01 \]
Since \(\Delta T = 100^\circ C\), we can find \(\alpha\):
\[ \alpha = \frac{0.01}{100} = 0.0001 \, \text{per degree Celsius} \]

### Step 3: Relate area expansion to linear expansion
For an area, the change in area (\(\Delta A\)) can be related to the change in length as follows:
\[ \frac{\Delta A}{A} = 2\alpha \Delta T \]
This is because the area is a function of two linear dimensions (length and width).

### Step 4: Calculate the percentage increase in area
Now, substituting the values we have:
\[ \frac{\Delta A}{A} = 2 \times 0.0001 \times 100 = 0.02 \]
To convert this to a percentage, we multiply by 100:
\[ \text{Percentage increase in area} = 0.02 \times 100 = 2\% \]

### Conclusion
Thus, the percentage increase in the area of the copper plate when heated from \(0^\circ C\) to \(100^\circ C\) is \(2\%\).

---