The temperature coefficient of resistance of platinum is `alpha=3.92xx10^(-3)K^(-1)` at `20^(@)C`. Find the temperature at which the increase in the resistance of platinum wire is `10%` of its value at `20^(@)C`

Options

A. `40.5^(@)C`
B. `45.5^(@)C`
C. `48.5^(@)C`
D. `43.5^(@)C`

Explanation

To solve the problem, we need to determine the temperature at which the resistance of a platinum wire increases by 10% from its value at 20°C.

### Step-by-step Solution:

1. **Understand the given data**:
- Temperature coefficient of resistance, \( \alpha = 3.92 \times 10^{-3} \, \text{K}^{-1} \)
- Initial temperature, \( T_1 = 20^\circ C \)
- Increase in resistance = 10% of the resistance at \( T_1 \)

2. **Define the initial resistance**:
- Let the resistance at \( 20^\circ C \) be \( R_1 \).

3. **Calculate the new resistance**:
- The new resistance \( R_2 \) after a 10% increase is given by:
\[ R_2 = R_1 + 0.1 R_1 = 1.1 R_1 \]

4. **Use the formula for resistance change with temperature**:
- The resistance at a new temperature \( T_2 \) can be expressed as:
\[ R_2 = R_1 (1 + \alpha (T_2 - T_1)) \]
- Substituting \( T_1 = 20^\circ C \):
\[ R_2 = R_1 (1 + \alpha (T_2 - 20)) \]

5. **Set the two expressions for \( R_2 \) equal**:
- From the two expressions for \( R_2 \):
\[ 1.1 R_1 = R_1 (1 + \alpha (T_2 - 20)) \]
- Dividing both sides by \( R_1 \) (assuming \( R_1 \neq 0 \)):
\[ 1.1 = 1 + \alpha (T_2 - 20) \]

6. **Rearranging the equation**:
- Subtract 1 from both sides:
\[ 0.1 = \alpha (T_2 - 20) \]

7. **Solve for \( T_2 \)**:
- Substitute \( \alpha \):
\[ 0.1 = 3.92 \times 10^{-3} (T_2 - 20) \]
- Rearranging gives:
\[ T_2 - 20 = \frac{0.1}{3.92 \times 10^{-3}} \]
- Calculate the right side:
\[ T_2 - 20 = \frac{0.1}{3.92 \times 10^{-3}} \approx 25.51 \]
- Therefore:
\[ T_2 = 25.51 + 20 = 45.51^\circ C \]

8. **Final Answer**:
- The temperature at which the resistance increases by 10% is approximately \( T_2 = 45.51^\circ C \).