If the resistance of a conductor is `5 Omega` at `50^(@)C` and `7Omega` at `100^(@)C` then the mean temperature coefficient of resistance of the material is

Options

A. `0.001//^(@)"C"`
B. `0.004//^(@)"C"`
C. `0.006//^(@)"C"`
D. `0.008//^(@)"C"`

Answer

To find the mean temperature coefficient of resistance (α) of the material, we can use the formula that relates resistance at two different temperatures:

\[ R_2 = R_1 \left( 1 + \alpha (T_2 - T_1) \right) \]

Where:
- \( R_1 \) is the resistance at temperature \( T_1 \)
- \( R_2 \) is the resistance at temperature \( T_2 \)
- \( \alpha \) is the temperature coefficient of resistance
- \( T_1 \) and \( T_2 \) are the initial and final temperatures respectively

### Step-by-step Solution:

1. **Identify the given values:**
- \( R_1 = 5 \, \Omega \) at \( T_1 = 50 \, °C \)
- \( R_2 = 7 \, \Omega \) at \( T_2 = 100 \, °C \)

2. **Substitute the values into the formula:**
\[ 7 = 5 \left( 1 + \alpha (100 - 50) \right) \]

3. **Simplify the equation:**
\[ 7 = 5 \left( 1 + \alpha \cdot 50 \right) \]
\[ 7 = 5 + 5 \alpha \cdot 50 \]

4. **Rearrange the equation to isolate \( \alpha \):**
\[ 7 - 5 = 5 \alpha \cdot 50 \]
\[ 2 = 5 \alpha \cdot 50 \]

5. **Solve for \( \alpha \):**
\[ \alpha = \frac{2}{5 \cdot 50} \]
\[ \alpha = \frac{2}{250} \]
\[ \alpha = 0.008 \]

6. **Convert to appropriate units:**
\[ \alpha = 0.008 \, °C^{-1} \]

### Final Answer:
The mean temperature coefficient of resistance \( \alpha \) is \( 0.008 \, °C^{-1} \).