The ratio of the resistances of a conductor at a temperature of `15^(@)C` to its resistance at a temperature of `37.5^(@)C` is `4:5`. The temperature coefficient of resistance of the conductor is

Options

A. `(1)/(25).^(@)C^(-1)`
B. `(1)/(50).^(@)C^(-1)`
C. `(1)/(80).^(@)C^(-1)`
D. `(1)/(75).^(@)C^(-1)`

Answer

To find the temperature coefficient of resistance of the conductor given the ratio of its resistances at two different temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables:**
Let \( R_1 \) be the resistance at \( 15^\circ C \) and \( R_2 \) be the resistance at \( 37.5^\circ C \). According to the problem, the ratio of the resistances is given as:
\[ \frac{R_1}{R_2} = \frac{4}{5} \]

2. **Use the Formula for Resistance:**
The resistance of a conductor at a temperature \( T \) can be expressed using the formula:
\[ R = R_0 \left(1 + \alpha T\right) \]
where \( R_0 \) is the resistance at \( 0^\circ C \), \( \alpha \) is the temperature coefficient of resistance, and \( T \) is the temperature in degrees Celsius.

3. **Express \( R_1 \) and \( R_2 \):**
For \( R_1 \) at \( 15^\circ C \):
\[ R_1 = R_0 \left(1 + \alpha \cdot 15\right) \]
For \( R_2 \) at \( 37.5^\circ C \):
\[ R_2 = R_0 \left(1 + \alpha \cdot 37.5\right) \]

4. **Set up the Ratio:**
Substitute \( R_1 \) and \( R_2 \) into the ratio:
\[ \frac{R_0 \left(1 + \alpha \cdot 15\right)}{R_0 \left(1 + \alpha \cdot 37.5\right)} = \frac{4}{5} \]
The \( R_0 \) cancels out:
\[ \frac{1 + 15\alpha}{1 + 37.5\alpha} = \frac{4}{5} \]

5. **Cross Multiply:**
Cross multiplying gives:
\[ 5(1 + 15\alpha) = 4(1 + 37.5\alpha) \]

6. **Expand and Rearrange:**
Expanding both sides:
\[ 5 + 75\alpha = 4 + 150\alpha \]
Rearranging gives:
\[ 75\alpha - 150\alpha = 4 - 5 \]
\[ -75\alpha = -1 \]

7. **Solve for \( \alpha \):**
Dividing both sides by -75:
\[ \alpha = \frac{1}{75} \]

### Final Answer:
The temperature coefficient of resistance \( \alpha \) is:
\[ \alpha = \frac{1}{75} \, \text{per degree Celsius} \]