A wire has a resistance of `2.5Omega at 28^(@)C` and a resistance of `2.9Omega at 100^(@)C`. The temperature coefficient of resistivity of material of the wire is

Options

A. `1.06xx10^(-3).^(@)C^(-1)`
B. `3.5xx10^(-2).^(@)C^(-1)`
C. `2.22xx10^(-3).^(@)C^(-1)`
D. `9.95xx10^(-2).^(@)C^(-1)`

Explanation

To find the temperature coefficient of resistivity (α) of the material of the wire, we can use the formula that relates the resistance of a conductor to its temperature:

\[ R = R_0 (1 + \alpha (T - T_0)) \]

Where:
- \( R \) is the resistance at temperature \( T \)
- \( R_0 \) is the resistance at a reference temperature \( T_0 \)
- \( \alpha \) is the temperature coefficient of resistivity
- \( T \) is the final temperature
- \( T_0 \) is the initial temperature

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance at \( T_1 = 28^\circ C \): \( R_1 = 2.5 \, \Omega \)
- Resistance at \( T_2 = 100^\circ C \): \( R_2 = 2.9 \, \Omega \)

2. **Set Up the Equation:**
Using the formula for resistance, we can write:
\[ R_2 = R_1 (1 + \alpha (T_2 - T_1)) \]
Substituting the known values:
\[ 2.9 = 2.5 (1 + \alpha (100 - 28)) \]

3. **Simplify the Equation:**
Calculate \( T_2 - T_1 \):
\[ T_2 - T_1 = 100 - 28 = 72 \, \text{degrees Celsius} \]
Now substitute this back into the equation:
\[ 2.9 = 2.5 (1 + 72\alpha) \]

4. **Divide Both Sides by 2.5:**
\[ \frac{2.9}{2.5} = 1 + 72\alpha \]
Calculate \( \frac{2.9}{2.5} \):
\[ 1.16 = 1 + 72\alpha \]

5. **Isolate \( \alpha \):**
Subtract 1 from both sides:
\[ 1.16 - 1 = 72\alpha \]
\[ 0.16 = 72\alpha \]

6. **Solve for \( \alpha \):**
\[ \alpha = \frac{0.16}{72} \]
Calculate \( \alpha \):
\[ \alpha \approx 0.002222 \, \text{per degree Celsius} \]

7. **Express in Scientific Notation:**
\[ \alpha \approx 2.22 \times 10^{-3} \, \text{per degree Celsius} \]

### Final Answer:
The temperature coefficient of resistivity of the material of the wire is approximately:
\[ \alpha \approx 2.22 \times 10^{-3} \, \text{per degree Celsius} \]