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. `44.0 J//s`
B. `16.8 J//s`
C. `8.0 J//s`
D. `4.0 J//s`

Explanation

To solve the problem, we will use the principle of heat conduction, which is described by Fourier's law of heat conduction. The rate of heat flow (Q/t) through a material is proportional to the temperature difference (ΔT) across it, given by the formula:

\[ \frac{Q}{t} = k \cdot A \cdot \frac{\Delta T}{x} \]

Where:
- \(Q/t\) is the rate of heat flow (in Joules per second),
- \(k\) is the thermal conductivity of the material,
- \(A\) is the cross-sectional area of the rod,
- \(\Delta T\) is the temperature difference across the rod,
- \(x\) is the length of the rod.

### Step-by-step Solution:

1. **Identify the initial conditions**:
- For the first scenario, the temperatures at the ends of the rod are \(T_1 = 100^\circ C\) and \(T_2 = 110^\circ C\).
- The temperature difference \(\Delta T_1 = T_2 - T_1 = 110^\circ C - 100^\circ C = 10^\circ C\).
- The rate of heat flow is given as \(Q/t = 4.0 \, J/s\).

2. **Identify the new conditions**:
- For the second scenario, the temperatures at the ends of the rod are \(T_1' = 200^\circ C\) and \(T_2' = 210^\circ C\).
- The temperature difference \(\Delta T_2 = T_2' - T_1' = 210^\circ C - 200^\circ C = 10^\circ C\).

3. **Compare the temperature differences**:
- We observe that \(\Delta T_1 = 10^\circ C\) and \(\Delta T_2 = 10^\circ C\).

4. **Apply Fourier's law**:
- Since the rod's material (thermal conductivity \(k\)), cross-sectional area \(A\), and length \(x\) remain constant, we can conclude that the rate of heat flow is directly proportional to the temperature difference.
- Therefore, if \(\Delta T_1 = \Delta T_2\), then the rate of heat flow in both cases will be the same.

5. **Conclude the result**:
- Since the temperature differences are equal, the rate of heat flow in the second case will also be \(4.0 \, J/s\).

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