To determine which cylindrical rod will conduct the most heat when their ends are maintained at the same steady temperature, we can use the formula for heat conduction through a cylindrical rod:
\[ Q = -K \cdot A \cdot \frac{\Delta T}{L} \]
Where:
- \( Q \) is the amount of heat conducted,
- \( K \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the length of the rod.
Since the ends of the rods are maintained at the same steady temperature, we can assume that \( \Delta T \) is constant for all rods.
1. **Identify the Cross-Sectional Area**:
The cross-sectional area \( A \) of a cylindrical rod is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the rod.
2. **Substitute into the Heat Conduction Formula**:
The heat conducted can be expressed as:
\[ Q = -K \cdot (\pi r^2) \cdot \frac{\Delta T}{L} \]
This can be simplified to:
\[ Q = C \cdot \frac{r^2}{L} \]
where \( C = -K \cdot \pi \cdot \Delta T \) is a constant for the same material and temperature difference.
3. **Calculate \( \frac{r^2}{L} \) for Each Rod**:
For each rod, we will calculate the ratio \( \frac{r^2}{L} \):
- **Rod A**: \( r = 0.2 \, \text{m}, L = 1 \, \text{m} \)
\[ \frac{r^2}{L} = \frac{(0.2)^2}{1} = \frac{0.04}{1} = 0.04 \]
- **Rod B**: \( r = 0.1 \, \text{m}, L = 1 \, \text{m} \)
\[ \frac{r^2}{L} = \frac{(0.1)^2}{1} = \frac{0.01}{1} = 0.01 \]
- **Rod C**: \( r = 0.1 \, \text{m}, L = 10 \, \text{m} \)
\[ \frac{r^2}{L} = \frac{(0.1)^2}{10} = \frac{0.01}{10} = 0.001 \]
- **Rod D**: \( r = 0.3 \, \text{m}, L = 0.1 \, \text{m} \)
\[ \frac{r^2}{L} = \frac{(0.3)^2}{0.1} = \frac{0.09}{0.1} = 0.9 \]
4. **Compare the Values**:
Now we compare the calculated values of \( \frac{r^2}{L} \):
- Rod A: \( 0.04 \)
- Rod B: \( 0.01 \)
- Rod C: \( 0.001 \)
- Rod D: \( 0.9 \)
The maximum value is for Rod D, which indicates that it will conduct the most heat.
5. **Conclusion**:
Therefore, the rod that will conduct the most heat is **Rod D**.
