A thermocouple of negligible resistance produces an emf fo `40 mu V//^(@)C` in the linear range of temperature. A galvanometer of resistance `10 Omega` whose sensitivity is `1 mu A//div` is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will

Options

A. `0.25^(@)C`
B. `0.5^(@)C`
C. `1^(@)C`
D. `0.1^(@)C`

Explanation

To solve the problem, we need to determine the smallest temperature difference that can be detected by the system consisting of a thermocouple and a galvanometer.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- The thermocouple produces an emf of \(40 \, \mu V/^{\circ}C\).
- The galvanometer has a resistance of \(10 \, \Omega\) and a sensitivity of \(1 \, \mu A/div\).

2. **Calculate the Current for 1 Degree Celsius**:
- The emf produced by the thermocouple for a temperature difference of \(1^{\circ}C\) is \(40 \, \mu V\).
- Using Ohm's Law, the current (\(I\)) through the galvanometer can be calculated as:
\[ I = \frac{V}{R} = \frac{40 \, \mu V}{10 \, \Omega} \]
- Converting \(40 \, \mu V\) to volts gives:
\[ I = \frac{40 \times 10^{-6} \, V}{10 \, \Omega} = 4 \times 10^{-6} \, A = 4 \, \mu A \]

3. **Determine the Current Sensitivity**:
- The galvanometer's sensitivity is \(1 \, \mu A/div\). This means that for every \(1 \, \mu A\) of current, the galvanometer will show a deflection of 1 division.

4. **Calculate the Temperature Difference for 1 Microampere**:
- Since \(4 \, \mu A\) corresponds to a temperature difference of \(1^{\circ}C\), we can find the temperature difference corresponding to \(1 \, \mu A\):
\[ \text{Temperature difference} = \frac{1^{\circ}C}{4 \, \mu A} = 0.25^{\circ}C \]

5. **Conclusion**:
- The smallest temperature difference that can be detected by the system is \(0.25^{\circ}C\).

### Final Answer:
The smallest value of temperature difference that can be detected by the system is \(0.25^{\circ}C\).