A rod of length l (laterally thermally insulated) of the uniform cross sectional area A consists of a material whose thermal conductivity varies with temperature as `K=(K_(0))/(a+bT')` where `K_(0)`, a and b are constants. `T_(1)` and `T_(2)(lt T_(1))` are the temperatures of the ends of the rod. Then, the rate of flow of heat across the rod is

Options

A. `(AK_(0))/(bl)((a+bT_(1))/(a+bT_(2)))`
B. `(AK_(0))/(bl)((a+bT_(2))/(a+bT_(1)))`
C. `(K_(0)A)/(bl)ln[(a+bT_(1))/(a+bT_(2))]`
D. `(AK_(0))/(bl)ln[(a+bT_(2))/(a+bT_(1))]`

Answer

To solve the problem of finding the rate of flow of heat across a rod with temperature-dependent thermal conductivity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a rod of length \( L \) and cross-sectional area \( A \).
- The thermal conductivity \( K \) of the material varies with temperature \( T \) as:
\[ K = \frac{K_0}{a + bT} \]
- The temperatures at the ends of the rod are \( T_1 \) (hot end) and \( T_2 \) (cold end), where \( T_2 < T_1 \).

2. **Setting Up the Heat Flow Equation**:
- The rate of heat flow \( Q \) through a small element of the rod of length \( dL \) is given by Fourier's law:
\[ dQ = -K \cdot A \cdot \frac{dT}{dL} \]
- Here, \( dT \) is the temperature difference across the small element \( dL \).

3. **Expressing the Thermal Resistance**:
- The thermal resistance \( R \) for a small element of length \( dL \) is:
\[ dR = \frac{dL}{K \cdot A} \]
- Substituting \( K \):
\[ dR = \frac{dL}{\frac{K_0}{a + bT} \cdot A} = \frac{(a + bT) \cdot dL}{K_0 \cdot A} \]

4. **Integrating Over the Length of the Rod**:
- To find the total heat flow \( Q \) across the entire rod, we need to integrate from \( T_1 \) to \( T_2 \):
\[ Q = \int_{T_1}^{T_2} \frac{K_0 \cdot A}{(a + bT)} \cdot \frac{1}{L} \, dT \]
- This can be rewritten as:
\[ Q = \frac{K_0 \cdot A}{L} \int_{T_1}^{T_2} \frac{1}{(a + bT)} \, dT \]

5. **Calculating the Integral**:
- The integral \( \int \frac{1}{(a + bT)} \, dT \) can be solved using the natural logarithm:
\[ \int \frac{1}{(a + bT)} \, dT = \frac{1}{b} \ln |a + bT| \]
- Evaluating this from \( T_1 \) to \( T_2 \):
\[ \int_{T_1}^{T_2} \frac{1}{(a + bT)} \, dT = \frac{1}{b} \left( \ln(a + bT_2) - \ln(a + bT_1) \right) = \frac{1}{b} \ln \left( \frac{a + bT_2}{a + bT_1} \right) \]

6. **Final Expression for Heat Flow**:
- Substituting back into the equation for \( Q \):
\[ Q = \frac{K_0 \cdot A}{L} \cdot \frac{1}{b} \ln \left( \frac{a + bT_2}{a + bT_1} \right) \]

### Final Answer:
\[ Q = \frac{K_0 \cdot A}{bL} \ln \left( \frac{a + bT_2}{a + bT_1} \right) \]