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

Explanation

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

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

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 in degrees Celsius.

### Step 1: Identify the given values
- \( R_1 = 5 \, \Omega \) at \( T_1 = 50^\circ C \)
- \( R_2 = 7 \, \Omega \) at \( T_2 = 100^\circ C \)

### Step 2: Substitute the values into the formula
Using the formula, we substitute the known values:

\[ 7 = 5 \times (1 + \alpha (100 - 50)) \]

### Step 3: Simplify the equation
First, calculate \( T_2 - T_1 \):

\[ 100 - 50 = 50 \]

Now, substitute this back into the equation:

\[ 7 = 5 \times (1 + 50\alpha) \]

### Step 4: Divide both sides by 5
To isolate the term with α, divide both sides by 5:

\[ \frac{7}{5} = 1 + 50\alpha \]

### Step 5: Rearrange to solve for α
Subtract 1 from both sides:

\[ \frac{7}{5} - 1 = 50\alpha \]

Convert 1 to a fraction with a denominator of 5:

\[ \frac{7}{5} - \frac{5}{5} = 50\alpha \]

This simplifies to:

\[ \frac{2}{5} = 50\alpha \]

### Step 6: Solve for α
Now, divide both sides by 50:

\[ \alpha = \frac{2}{5 \times 50} \]

Calculating the right side gives:

\[ \alpha = \frac{2}{250} = 0.008 \, \text{per degree Celsius} \]

### Final Answer
The mean temperature coefficient of resistance of the material is:

\[ \alpha = 0.008 \, \text{per degree Celsius} \]

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