A rod AB is 1m long. The temperature of its one end A is maintained at `100^(@)`C and other end B at `10^(@)`C, the temperature at a distance of 60 cm from point B is

Options

A. `64^(@)`C
B. `36^(@)`C
C. `46^(@)`C
D. `72^(@)`C

Explanation

To solve the problem of finding the temperature at a distance of 60 cm from point B on a rod AB that is 1 meter long, we can use the concept of steady-state heat conduction. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Understand the Setup**:
- The rod AB is 1 meter long.
- Temperature at end A (0 cm) = 100°C.
- Temperature at end B (100 cm) = 10°C.
- We need to find the temperature (θ) at a point 60 cm from B (which is 40 cm from A).

2. **Define the Lengths**:
- Distance from B to the point where we want to find the temperature = 60 cm = 0.6 m.
- Distance from A to the point = 1 m - 0.6 m = 0.4 m.

3. **Set Up the Heat Transfer Equation**:
- The heat transfer rate (Q) through the rod can be expressed using Fourier's law of heat conduction:
\[ Q = \frac{kA(T_1 - T_2)}{L} \]
- For the section from A to the point (0.4 m), we have:
\[ Q = \frac{kA(100 - \theta)}{0.4} \]
- For the section from the point to B (0.6 m), we have:
\[ Q = \frac{kA(\theta - 10)}{0.6} \]

4. **Equate the Heat Transfer Rates**:
- Since the heat transfer rate must be the same at steady state, we can set the two equations equal to each other:
\[ \frac{kA(100 - \theta)}{0.4} = \frac{kA(\theta - 10)}{0.6} \]
- The \(kA\) terms cancel out:
\[ \frac{100 - \theta}{0.4} = \frac{\theta - 10}{0.6} \]

5. **Cross-Multiply to Solve for θ**:
- Cross-multiplying gives:
\[ 0.6(100 - \theta) = 0.4(\theta - 10) \]
- Expanding both sides:
\[ 60 - 0.6\theta = 0.4\theta - 4 \]

6. **Combine Like Terms**:
- Rearranging the equation:
\[ 60 + 4 = 0.6\theta + 0.4\theta \]
\[ 64 = θ \]

7. **Final Result**:
- The temperature at a distance of 60 cm from point B is:
\[ θ = 64°C \]