The ends of a thin bar are maintained at different temperatures. The temperature of the cooler end is `11^(@)C` , while the temperature at a point 0.13m from the cooler end is `23^(@)C` and the temperature of the warmer end is `48^(@)C`. Assuming that heat flows only along the length of the bar ( the sides are insulated), find the length of the bar.

Options

A. 0.50m
B. 0.40m
C. 0.23m
D. 0.64m

Answer

To solve the problem, we will use the concept of heat conduction along the bar and apply the principle that the rate of heat flow is the same throughout the bar.

### Step-by-Step Solution:

1. **Identify the given temperatures and distances:**
- Cooler end temperature, \( T_1 = 11^\circ C \)
- Temperature at a point \( 0.13 \, m \) from the cooler end, \( T_2 = 23^\circ C \)
- Warmer end temperature, \( T_3 = 48^\circ C \)
- Distance from the cooler end to the point where temperature is \( 23^\circ C \), \( d_1 = 0.13 \, m \)

2. **Define the length of the bar:**
- Let the total length of the bar be \( L \).
- The length from the cooler end to the point where the temperature is \( 23^\circ C \) is \( d_1 = 0.13 \, m \).
- The remaining length from this point to the warmer end is \( d_2 = L - 0.13 \).

3. **Set up the heat conduction equations:**
- According to Fourier's law of heat conduction, the heat flow rate \( Q \) is given by:
\[ Q = \frac{K \cdot A \cdot (T_2 - T_1)}{d_1} = \frac{K \cdot A \cdot (T_3 - T_2)}{d_2} \]
- Here, \( K \) is the thermal conductivity, and \( A \) is the cross-sectional area of the bar.

4. **Substitute the known values:**
- For the first segment (from cooler end to point at \( 23^\circ C \)):
\[ Q = \frac{K \cdot A \cdot (23 - 11)}{0.13} = \frac{K \cdot A \cdot 12}{0.13} \]
- For the second segment (from point at \( 23^\circ C \) to warmer end):
\[ Q = \frac{K \cdot A \cdot (48 - 23)}{d_2} = \frac{K \cdot A \cdot 25}{d_2} \]

5. **Equate the two expressions for \( Q \):**
\[ \frac{K \cdot A \cdot 12}{0.13} = \frac{K \cdot A \cdot 25}{d_2} \]
- Cancel \( K \) and \( A \) from both sides:
\[ \frac{12}{0.13} = \frac{25}{d_2} \]

6. **Solve for \( d_2 \):**
\[ d_2 = \frac{25 \cdot 0.13}{12} \]
\[ d_2 = \frac{3.25}{12} \approx 0.2708 \, m \]

7. **Calculate the total length \( L \):**
\[ L = d_1 + d_2 = 0.13 + 0.2708 \approx 0.4008 \, m \]
- Rounding to two decimal places, we find:
\[ L \approx 0.40 \, m \]

### Final Answer:
The total length of the bar is approximately \( 0.40 \, m \).