A rod of length l and cross-section area A has a variable thermal conductivity given by K = `alpha` T, where `alpha` is a positive constant and T is temperature in kelvin. Two ends of the rod are maintained at temperature `T_(1)` and `T_(2)` `(T_(1)gtT_(2))`. Heat current flowing through the rod will be

Options

A. `(Aalpha(T_(1)^(2)-T_(2)^(2)))/(1)`
B. `(Aalpha(T_(1)^(2)+T_(2)^(2)))/(1)`
C. `(Aalpha(T_(1)^(2)+T_(2)^(2)))/(31)`
D. `(Aalpha(T_(1)^(2)-T_(2)^(2)))/(21)`

Explanation

To find the heat current flowing through a rod with variable thermal conductivity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a rod of length \( L \) and cross-sectional area \( A \) with a thermal conductivity that varies with temperature, given by \( K = \alpha T \), where \( \alpha \) is a constant and \( T \) is the temperature in Kelvin. The temperatures at the two ends of the rod are \( T_1 \) and \( T_2 \) (with \( T_1 > T_2 \)).

2. **Heat Current Formula**: The heat current \( \frac{dQ}{dt} \) through the rod can be expressed using Fourier's law of heat conduction:
\[ \frac{dQ}{dt} = -K A \frac{dT}{dx} \]
Here, \( \frac{dT}{dx} \) is the temperature gradient along the rod.

3. **Express \( K \) in Terms of Temperature**: Since \( K \) is a function of temperature, we can substitute \( K = \alpha T \) into the heat current formula:
\[ \frac{dQ}{dt} = -(\alpha T) A \frac{dT}{dx} \]

4. **Integrate Over the Length of the Rod**: We need to integrate this expression from \( T_1 \) to \( T_2 \). The temperature \( T \) varies along the length of the rod, so we can express \( \frac{dT}{dx} \) as:
\[ \frac{dT}{dx} = \frac{T_2 - T_1}{L} \]

5. **Substituting into the Heat Current Equation**: We can substitute \( \frac{dT}{dx} \) into our equation:
\[ \frac{dQ}{dt} = -\alpha A \left( \frac{T_2 + T_1}{2} \right) \frac{(T_2 - T_1)}{L} \]
Here, we take the average temperature \( \frac{T_2 + T_1}{2} \) because \( K \) varies linearly with \( T \).

6. **Integrate the Heat Current**: We can also directly integrate the expression for heat current over the temperature range from \( T_1 \) to \( T_2 \):
\[ \frac{dQ}{dt} = \frac{\alpha A}{L} \int_{T_1}^{T_2} T \, dT \]
The integral of \( T \) gives:
\[ \int T \, dT = \frac{T^2}{2} \]
Thus,
\[ \frac{dQ}{dt} = \frac{\alpha A}{L} \left[ \frac{T^2}{2} \right]_{T_1}^{T_2} = \frac{\alpha A}{L} \left( \frac{T_2^2}{2} - \frac{T_1^2}{2} \right) \]

7. **Final Expression for Heat Current**: Therefore, the heat current flowing through the rod is:
\[ \frac{dQ}{dt} = \frac{\alpha A}{2L} (T_2^2 - T_1^2) \]
Since \( T_1 > T_2 \), we can express this as:
\[ \frac{dQ}{dt} = -\frac{\alpha A}{2L} (T_1^2 - T_2^2) \]

### Final Answer:
\[ \frac{dQ}{dt} = -\frac{\alpha A}{2L} (T_1^2 - T_2^2) \]