The thermo e.m.f. E in volts of a certain thermocouple is found to vary with temperature difference q in `.^(@)C` between the two junctions according to the relation `E=30 theta-(theta^(2))/(15)`
The neutral temperature for the thermo-couple will be:-

Options

A. `400^(@)C`
B. `225^(@)C`
C. `30^(@)C`
D. `450^(@)C`

Answer

To find the neutral temperature of the thermocouple, we need to determine the temperature at which the thermo e.m.f. \( E \) is maximized. The given relation for the thermo e.m.f. is:

\[ E = 30\theta - \frac{\theta^2}{15} \]

### Step-by-Step Solution:

1. **Differentiate the e.m.f. with respect to temperature**:
We need to find the derivative of \( E \) with respect to \( \theta \) and set it to zero to find the maximum point.

\[ \frac{dE}{d\theta} = \frac{d}{d\theta} \left( 30\theta - \frac{\theta^2}{15} \right) \]

2. **Calculate the derivative**:
Using basic differentiation rules, we differentiate each term:

\[ \frac{dE}{d\theta} = 30 - \frac{2\theta}{15} \]

3. **Set the derivative equal to zero**:
To find the neutral temperature, we set the derivative equal to zero:

\[ 30 - \frac{2\theta}{15} = 0 \]

4. **Solve for \( \theta \)**:
Rearranging the equation gives:

\[ \frac{2\theta}{15} = 30 \]

Multiplying both sides by 15:

\[ 2\theta = 30 \times 15 \]

Simplifying gives:

\[ 2\theta = 450 \]

Dividing both sides by 2:

\[ \theta = \frac{450}{2} = 225 \]

5. **Conclusion**:
The neutral temperature \( \theta \) is:

\[ \theta = 225 \, ^\circ C \]

### Final Answer:
The neutral temperature for the thermocouple is \( 225 \, ^\circ C \).

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