A thin brass sheet at `10^(@)C` and a thin steel sheet at `20^(@)C` have the same surface area. The common temperature at which both would have the same area is (Coefficient of linear expansion for brass and steel are respectively, `19 xx 10^(-6//@)C` are `11 xx 10^(-6//@)C)`

Options

A. `-3.75@C`
B. `-2.75^(@)C`
C. `2.75^(@)C`
D. `3.75^(@)C`

Explanation

To solve the problem, we need to find the common temperature at which the areas of the brass and steel sheets become equal after thermal expansion.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial temperature of the brass sheet, \( T_1 = 10^\circ C \).
- Let the initial temperature of the steel sheet, \( T_2 = 20^\circ C \).
- Both sheets have the same initial area, \( A_1 = A_2 \).

2. **Define the Coefficients of Linear Expansion**:
- Coefficient of linear expansion for brass, \( \alpha_1 = 19 \times 10^{-6} / ^\circ C \).
- Coefficient of linear expansion for steel, \( \alpha_2 = 11 \times 10^{-6} / ^\circ C \).
- The area expansion coefficient \( \beta \) is related to the linear expansion coefficient \( \alpha \) by \( \beta = 2\alpha \).

3. **Write the Area Expansion Formula**:
- The area of the brass sheet at temperature \( T \) will be:
\[ A_1' = A_1 \left(1 + \beta_1 (T - T_1)\right) = A_1 \left(1 + 2\alpha_1 (T - 10)\right) \]
- The area of the steel sheet at temperature \( T \) will be:
\[ A_2' = A_2 \left(1 + \beta_2 (T - T_2)\right) = A_2 \left(1 + 2\alpha_2 (T - 20)\right) \]

4. **Set the Areas Equal**:
- Since \( A_1' = A_2' \), we can set the equations equal to each other:
\[ A_1 \left(1 + 2\alpha_1 (T - 10)\right) = A_2 \left(1 + 2\alpha_2 (T - 20)\right) \]
- Since \( A_1 = A_2 \), we can cancel \( A_1 \) and \( A_2 \):
\[ 1 + 2\alpha_1 (T - 10) = 1 + 2\alpha_2 (T - 20) \]

5. **Simplify the Equation**:
- Cancel the 1s from both sides:
\[ 2\alpha_1 (T - 10) = 2\alpha_2 (T - 20) \]
- Divide by 2:
\[ \alpha_1 (T - 10) = \alpha_2 (T - 20) \]

6. **Substitute the Values of Coefficients**:
- Substitute \( \alpha_1 \) and \( \alpha_2 \):
\[ 19 \times 10^{-6} (T - 10) = 11 \times 10^{-6} (T - 20) \]

7. **Eliminate the Common Factor**:
- Divide both sides by \( 10^{-6} \):
\[ 19 (T - 10) = 11 (T - 20) \]

8. **Expand and Collect Terms**:
- Expand both sides:
\[ 19T - 190 = 11T - 220 \]
- Rearranging gives:
\[ 19T - 11T = -220 + 190 \]
\[ 8T = -30 \]

9. **Solve for T**:
- Divide by 8:
\[ T = -\frac{30}{8} = -3.75^\circ C \]

### Final Answer:
The common temperature at which both sheets would have the same area is \( T = -3.75^\circ C \).