The two ends of a metal rod are maintained at temperature `100^(@)C` and `110^(@)C`. The rate of heat flow in the rod is found to be `4.0 J//s`. If the ends are maintained at temperature s `200^(@)C` and `210^(@)C`. The rate of heat flow will be

Options

A. 44 J`s^(-1)`
B. 16.8 J`s^(-1)`
C. 8 J`s^(-1)`
D. 4 J`s^(-1)`

Explanation

To solve the problem, we will use the principle of heat conduction, which states that the rate of heat flow (Q/t) through a material is directly proportional to the temperature difference (ΔT) across the material.

### Step-by-Step Solution:

1. **Identify the Temperature Differences**:
- For the first scenario, the temperatures are \(100^\circ C\) and \(110^\circ C\).
- The temperature difference (ΔT1) is:
\[ ΔT_1 = 110^\circ C - 100^\circ C = 10^\circ C \]

2. **Given Rate of Heat Flow**:
- The rate of heat flow (Q/t) for the first scenario is given as \(4.0 \, J/s\).

3. **Identify the Second Scenario**:
- For the second scenario, the temperatures are \(200^\circ C\) and \(210^\circ C\).
- The temperature difference (ΔT2) is:
\[ ΔT_2 = 210^\circ C - 200^\circ C = 10^\circ C \]

4. **Apply the Principle of Heat Conduction**:
- According to Fourier's law of heat conduction, the rate of heat flow is directly proportional to the temperature difference:
\[ \frac{Q_1}{t} \propto ΔT_1 \quad \text{and} \quad \frac{Q_2}{t} \propto ΔT_2 \]
- Since both ΔT1 and ΔT2 are equal (both are \(10^\circ C\)), the rate of heat flow in both scenarios will be the same.

5. **Conclusion**:
- Therefore, the rate of heat flow in the second scenario will also be:
\[ \frac{Q_2}{t} = 4.0 \, J/s \]

### Final Answer:
The rate of heat flow when the ends are maintained at \(200^\circ C\) and \(210^\circ C\) is \(4.0 \, J/s\).

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