The resistance of a resistance thermometer has values `2.71` and `3.70` ohm at `10^(@) C` and `100^(@) C` . The temperature at which the resistance is `3.26` ohm is

Options

A. `40^(@)C`
B. `50^(@)C`
C. `60^(@)C`
D. `70^(@)C`

Answer

To solve the problem, we will use the concept of linear thermal expansion in relation to resistance. The resistance of a conductor changes with temperature, and we can express this relationship mathematically.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Resistance at \( T_1 = 10^\circ C \): \( R_1 = 2.71 \, \Omega \)
- Resistance at \( T_2 = 100^\circ C \): \( R_2 = 3.70 \, \Omega \)
- Resistance at \( T_3 \): \( R_3 = 3.26 \, \Omega \)
- We need to find the temperature \( T_3 \) at which the resistance is \( 3.26 \, \Omega \).

2. **Use the Resistance-Temperature Relation**:
The relationship between resistance and temperature can be expressed as:
\[ R = R_1 \left(1 + \alpha (T - T_1)\right) \]
where \( \alpha \) is the temperature coefficient of resistance.

3. **Calculate the Coefficient of Resistance (\( \alpha \))**:
We can rearrange the formula to find \( \alpha \) using the known resistances and temperatures:
\[ R_2 = R_1 \left(1 + \alpha (T_2 - T_1)\right) \]
Plugging in the values:
\[ 3.70 = 2.71 \left(1 + \alpha (100 - 10)\right) \]
Simplifying:
\[ 3.70 = 2.71 \left(1 + 90\alpha\right) \]
Dividing both sides by \( 2.71 \):
\[ \frac{3.70}{2.71} = 1 + 90\alpha \]
\[ 1.365 = 1 + 90\alpha \]
\[ 90\alpha = 0.365 \]
\[ \alpha = \frac{0.365}{90} \approx 0.0040556 \, \text{per degree Celsius} \]

4. **Set Up the Equation for \( R_3 \)**:
Now, we can use the same formula to find \( T_3 \):
\[ R_3 = R_1 \left(1 + \alpha (T_3 - T_1)\right) \]
Substituting the known values:
\[ 3.26 = 2.71 \left(1 + 0.0040556 (T_3 - 10)\right) \]
Dividing both sides by \( 2.71 \):
\[ \frac{3.26}{2.71} = 1 + 0.0040556 (T_3 - 10) \]
\[ 1.202 = 1 + 0.0040556 (T_3 - 10) \]
\[ 0.202 = 0.0040556 (T_3 - 10) \]
Dividing both sides by \( 0.0040556 \):
\[ T_3 - 10 = \frac{0.202}{0.0040556} \approx 49.8 \]
Thus,
\[ T_3 \approx 49.8 + 10 \approx 59.8 \, \text{or} \, 60^\circ C \]

5. **Final Answer**:
The temperature at which the resistance is \( 3.26 \, \Omega \) is approximately \( 60^\circ C \).