The temperature co-efficient of resistance of a wire at `0^(@)C` is `0.00125^(@)C^(-1)` . At `25^(@)C` its resistance is one ohm. The resistance of the wire will ne 1.2 ohm at .

Options

A. 225K
B. `190^(@)`
C. `260^(@)C`
D. `185K`

Answer

To solve the problem, we need to determine the temperature at which the resistance of a wire becomes 1.2 ohms, given that the resistance at 25°C is 1 ohm and the temperature coefficient of resistance is 0.00125 °C⁻¹.

### Step-by-Step Solution:

1. **Identify the known values:**
- Temperature coefficient of resistance, \( \alpha = 0.00125 \, °C^{-1} \)
- Resistance at \( 25°C, R(25) = 1 \, \Omega \)
- Resistance at temperature \( T, R(T) = 1.2 \, \Omega \)

2. **Use the formula for resistance change with temperature:**
The formula for the resistance of a conductor as a function of temperature is given by:
\[ R(T) = R_0 (1 + \alpha \Delta T) \]
where \( R_0 \) is the resistance at \( 0°C \) and \( \Delta T \) is the change in temperature.

3. **Set up the equations:**
- For the resistance at \( 25°C \):
\[ 1 = R_0 (1 + \alpha \cdot 25) \]
- For the resistance at temperature \( T \):
\[ 1.2 = R_0 (1 + \alpha T) \]

4. **Solve for \( R_0 \) from the first equation:**
Rearranging the first equation gives:
\[ R_0 = \frac{1}{1 + \alpha \cdot 25} \]

5. **Substitute \( R_0 \) into the second equation:**
Substitute \( R_0 \) into the equation for \( R(T) \):
\[ 1.2 = \frac{1}{1 + \alpha \cdot 25} (1 + \alpha T) \]

6. **Cross-multiply to eliminate the fraction:**
\[ 1.2 (1 + \alpha \cdot 25) = 1 + \alpha T \]

7. **Expand and rearrange the equation:**
\[ 1.2 + 1.2 \alpha \cdot 25 = 1 + \alpha T \]
\[ 1.2 - 1 + 1.2 \alpha \cdot 25 = \alpha T \]
\[ 0.2 + 30 \alpha = \alpha T \]

8. **Solve for \( T \):**
\[ T = \frac{0.2}{\alpha} + 30 \]

9. **Substitute the value of \( \alpha \):**
Substitute \( \alpha = 0.00125 \):
\[ T = \frac{0.2}{0.00125} + 30 \]
\[ T = 160 + 30 = 190°C \]

### Final Answer:
The temperature at which the resistance of the wire will be 1.2 ohms is **190°C**.