Two identical conducting rods are first connected independently to two vessels, one containing water at `100^(@)C` and the other containing ice at `0^(@)C`. In the second case, the rods are joined end to end connected to the same vessels. Let `m_(1) and m_(2)` g/s be the rate of melting of ice in the two cases respectively, the ratio `(m_(1))/(m_(2))` is

Options

A. `1:2`
B. `2:1`
C. `4:1`
D. `1:4`

Answer

To solve the problem, we will analyze the heat transfer through the rods in both scenarios and derive the ratio of the rates of melting of ice, \( \frac{m_1}{m_2} \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical conducting rods.
- In the first case, one rod is connected to water at \( 100^\circ C \) and the other to ice at \( 0^\circ C \).
- In the second case, the rods are connected end to end between the same two vessels.

2. **Heat Transfer Formula**:
- The heat transfer rate (heat current) \( H \) can be expressed as:
\[ H = \frac{\Delta T}{R} \]
where \( \Delta T \) is the temperature difference and \( R \) is the thermal resistance of the rods.

3. **Case 1: Independent Connection**:
- For the first case, the effective resistance \( R_{eq} \) when the rods are connected in parallel is:
\[ R_{eq} = \frac{R \cdot R}{R + R} = \frac{R^2}{2R} = \frac{R}{2} \]
- The temperature difference \( \Delta T \) is:
\[ \Delta T = 100^\circ C - 0^\circ C = 100^\circ C \]
- Thus, the heat current \( H_1 \) for the first case is:
\[ H_1 = \frac{100}{R/2} = \frac{200}{R} \]

4. **Case 2: End-to-End Connection**:
- For the second case, the effective resistance \( R_{eq} \) when the rods are connected end to end is:
\[ R_{eq} = R + R = 2R \]
- The temperature difference remains the same:
\[ \Delta T = 100^\circ C - 0^\circ C = 100^\circ C \]
- Thus, the heat current \( H_2 \) for the second case is:
\[ H_2 = \frac{100}{2R} = \frac{50}{R} \]

5. **Relating Heat Transfer to Melting of Ice**:
- The heat transferred is related to the rate of melting of ice:
\[ H_1 = L \cdot m_1 \quad \text{and} \quad H_2 = L \cdot m_2 \]
where \( L \) is the latent heat of fusion of ice.

6. **Setting Up the Equations**:
- From the heat currents, we have:
\[ L \cdot m_1 = \frac{200}{R} \quad \text{(1)} \]
\[ L \cdot m_2 = \frac{50}{R} \quad \text{(2)} \]

7. **Finding the Ratio**:
- Dividing equation (1) by equation (2):
\[ \frac{L \cdot m_1}{L \cdot m_2} = \frac{\frac{200}{R}}{\frac{50}{R}} \]
- Simplifying gives:
\[ \frac{m_1}{m_2} = \frac{200}{50} = 4 \]

8. **Final Result**:
- Thus, the ratio of the rates of melting of ice is:
\[ \frac{m_1}{m_2} = 4 \]

### Conclusion:
The final answer is \( \frac{m_1}{m_2} = 4 \).