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

Options

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

Answer

To find the neutral temperature for the thermocouple given the relation \( E = 30\theta - \frac{\theta^2}{15} \), we need to determine the temperature difference \( \theta \) at which the electromotive force (emf) \( E \) is maximized. Here’s a step-by-step solution:

### Step 1: Understand the relationship
The emf \( E \) is given by the equation:
\[ E = 30\theta - \frac{\theta^2}{15} \]
where \( \theta \) is the temperature difference in degrees Celsius between the two junctions.

### Step 2: Differentiate the equation
To find the maximum emf, we need to differentiate \( E \) with respect to \( \theta \) and set the derivative equal to zero:
\[ \frac{dE}{d\theta} = 30 - \frac{2\theta}{15} \]

### Step 3: Set the derivative to zero
Setting the derivative equal to zero to find the critical points:
\[ 30 - \frac{2\theta}{15} = 0 \]

### Step 4: Solve for \( \theta \)
Rearranging the equation gives:
\[ \frac{2\theta}{15} = 30 \]
Multiplying both sides by 15:
\[ 2\theta = 450 \]
Dividing by 2:
\[ \theta = 225 \]

### Step 5: Confirm it's a maximum
To confirm that this value of \( \theta \) corresponds to a maximum, we need to check the second derivative:
\[ \frac{d^2E}{d\theta^2} = -\frac{2}{15} \]
Since the second derivative is negative, this indicates that \( E \) is at a maximum when \( \theta = 225 \).

### Step 6: Conclusion
The neutral temperature \( T \) for the thermocouple, which corresponds to the temperature difference \( \theta \) when one junction is at 0°C, is:
\[ T = \theta = 225^\circ C \]

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

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