A rod of length l and cross sectional area A has a variable conductivity given by `K=alphaT`, where `alpha` is a positive constant T is temperatures in Kelvin. Two ends of the rod are maintained at temperatures `T_1` and `T_2(T_1gtT_2)`. Heat current flowing through the rod will be

Options

A. `(A alphaT_1^2-T_2^2)/(l)`
B. `(A alpha T_1^2+T_2^2)/(l)`
C. `(A alpha T_1^2+T_2^2)/(3l)`
D. `(A alpha T_1^2-T_2^2)/(2l)`

Explanation

To solve the problem, we need to find the heat current flowing through a rod with variable thermal conductivity. The thermal conductivity \( K \) is given by \( K = \alpha T \), where \( \alpha \) is a positive constant and \( T \) is the temperature in Kelvin. The ends of the rod are maintained at temperatures \( T_1 \) and \( T_2 \) (with \( T_1 > T_2 \)).

### Step-by-Step Solution:

1. **Understand the Heat Current Formula**:
The heat current \( I \) through a rod can be expressed using Fourier's law of heat conduction:
\[ I = K \cdot A \cdot \frac{dT}{dx} \]
where \( K \) is the thermal conductivity, \( A \) is the cross-sectional area, and \( \frac{dT}{dx} \) is the temperature gradient.

2. **Substitute the Variable Conductivity**:
Given that \( K = \alpha T \), we can substitute this into the heat current formula:
\[ I = \alpha T \cdot A \cdot \frac{dT}{dx} \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ I \, dx = \alpha T \, A \, dT \]

4. **Integrate Both Sides**:
Now, we will integrate both sides. The left side will be integrated from \( 0 \) to \( L \) (length of the rod), and the right side will be integrated from \( T_2 \) to \( T_1 \) (the temperatures at the ends of the rod):
\[ \int I \, dx = \int \alpha A T \, dT \]

5. **Calculate the Integrals**:
The left side becomes:
\[ I \cdot x \Big|_0^L = I \cdot L - I \cdot 0 = I \cdot L \]
The right side becomes:
\[ \alpha A \cdot \frac{T^2}{2} \Big|_{T_2}^{T_1} = \alpha A \left( \frac{T_1^2}{2} - \frac{T_2^2}{2} \right) = \frac{\alpha A}{2} (T_1^2 - T_2^2) \]

6. **Set the Integrals Equal**:
Equating both sides gives:
\[ I \cdot L = \frac{\alpha A}{2} (T_1^2 - T_2^2) \]

7. **Solve for Heat Current \( I \)**:
Rearranging to find \( I \):
\[ I = \frac{\alpha A}{2L} (T_1^2 - T_2^2) \]

8. **Final Expression**:
Thus, the heat current flowing through the rod is:
\[ I = \frac{\alpha A (T_1^2 - T_2^2)}{2L} \]

### Conclusion:
The correct option for the heat current is:
\[ I = \frac{\alpha A (T_1^2 - T_2^2)}{2L} \]