One end of thermally insulated rod is kept at a temperature `T_(1)` and the other at `T_(2)`. The rod is composed of two section of length `l_(1)` and `l_(2)` thermal conductivities `k_(1)` and `k_(2)` respectively. The temerature at the interface of two section is

Options

A. (a)`(K_(2)l_(2)T_(1)+K_(1)l_(1)T_(2))//(K_(1)l_(1)+K_(2)l_(2))`
B. (b)`(K_(2)l_(1)T_(1)+K_(1)l_(2)T_(2))//(K_(2)l_(1)+K_(1)l_(2))`
C. (c)`(K_(1)l_(2)T_(1)+K_(2)l_(2)T_(2))//(K_(1)l_(2)+k_(2)l_(1))`
D. (d)`(K_(1)l_(1)T_(1)+K_(2)l_(2)T_(2))//(K_(1)l_(1)+K_(2)l_(2))`

Explanation

To find the temperature at the interface of the two sections of a thermally insulated rod, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
We have a rod divided into two sections. The left section has a length \( l_1 \) and thermal conductivity \( k_1 \), while the right section has a length \( l_2 \) and thermal conductivity \( k_2 \). The left end of the rod is at temperature \( T_1 \) and the right end is at temperature \( T_2 \).

2. **Apply the Concept of Steady-State Heat Flow**:
In a steady-state condition, the rate of heat flow through both sections of the rod must be equal. This can be expressed mathematically as:
\[ \frac{dQ_1}{dt} = \frac{dQ_2}{dt} \]
where \( dQ_1 \) and \( dQ_2 \) are the heat flows through the first and second sections, respectively.

3. **Write the Heat Flow Equations**:
The heat flow through the first section can be given by:
\[ \frac{dQ_1}{dt} = k_1 A \frac{T_1 - T}{l_1} \]
The heat flow through the second section can be expressed as:
\[ \frac{dQ_2}{dt} = k_2 A \frac{T - T_2}{l_2} \]
Here, \( A \) is the cross-sectional area of the rod.

4. **Set the Heat Flow Equations Equal**:
Since \( \frac{dQ_1}{dt} = \frac{dQ_2}{dt} \), we can equate the two equations:
\[ k_1 A \frac{T_1 - T}{l_1} = k_2 A \frac{T - T_2}{l_2} \]
The area \( A \) cancels out from both sides:
\[ k_1 \frac{T_1 - T}{l_1} = k_2 \frac{T - T_2}{l_2} \]

5. **Rearrange the Equation**:
Rearranging gives:
\[ k_1 (T_1 - T) l_2 = k_2 (T - T_2) l_1 \]

6. **Expand and Collect Terms**:
Expanding both sides:
\[ k_1 l_2 T_1 - k_1 l_2 T = k_2 l_1 T - k_2 l_1 T_2 \]
Rearranging to isolate terms involving \( T \):
\[ k_1 l_2 T_1 + k_2 l_1 T_2 = k_1 l_2 T + k_2 l_1 T \]

7. **Factor Out \( T \)**:
Factoring \( T \) from the right side:
\[ k_1 l_2 T_1 + k_2 l_1 T_2 = T (k_1 l_2 + k_2 l_1) \]

8. **Solve for \( T \)**:
Finally, we can solve for \( T \):
\[ T = \frac{k_1 l_2 T_1 + k_2 l_1 T_2}{k_1 l_2 + k_2 l_1} \]

### Final Answer:
The temperature at the interface of the two sections is:
\[ T = \frac{k_1 l_2 T_1 + k_2 l_1 T_2}{k_1 l_2 + k_2 l_1} \]