Two identical rods are connected between two containers. One of them is at `100^(@)C` containing water and another is at `0^(@)C` containing ice. If rods are connected in parallel then the rate of melting of ice is `q_(1)g//s` . If they are connected in series then teh rate is `q_(2)g//s` . The ratio `q_(2)//q_(1)` is

Options

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

Explanation

To solve the problem, we need to analyze the heat transfer through the rods in both parallel and series configurations and find the ratio of the rates of melting of ice, \( \frac{q_2}{q_1} \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical rods connecting two containers: one with water at \( 100^\circ C \) and the other with ice at \( 0^\circ C \).
- The rods are made of the same material, so they have the same thermal conductivity \( k \), cross-sectional area \( A \), and length \( L \).

2. **Parallel Configuration**:
- In the parallel configuration, both rods conduct heat simultaneously.
- The thermal resistance \( R \) for each rod is given by:
\[ R = \frac{L}{kA} \]
- For two rods in parallel, the equivalent thermal resistance \( R_{eq} \) is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{\frac{L}{kA}} + \frac{1}{\frac{L}{kA}} = \frac{2kA}{L} \]
- Therefore, the equivalent resistance is:
\[ R_{eq} = \frac{L}{2kA} \]
- The rate of heat transfer \( \frac{dq}{dt} \) through the rods is given by:
\[ \frac{dq}{dt} = \frac{\Delta T}{R_{eq}} = \frac{100 - 0}{\frac{L}{2kA}} = \frac{200kA}{L} \]
- This heat is used to melt the ice, so we can relate it to the rate of melting:
\[ \frac{dm}{dt} = \frac{dq}{dt} \cdot \frac{1}{L_f} = \frac{200kA}{L} \cdot \frac{1}{L_f} \]
- Thus, we have:
\[ q_1 = \frac{200kA}{L} \cdot \frac{1}{L_f} \]

3. **Series Configuration**:
- In the series configuration, the heat must pass through both rods sequentially.
- The total thermal resistance is:
\[ R_{eq} = R_1 + R_2 = \frac{L}{kA} + \frac{L}{kA} = \frac{2L}{kA} \]
- The rate of heat transfer \( \frac{dq}{dt} \) is:
\[ \frac{dq}{dt} = \frac{100 - 0}{\frac{2L}{kA}} = \frac{100kA}{L} \]
- This heat is also used to melt the ice:
\[ \frac{dm}{dt} = \frac{dq}{dt} \cdot \frac{1}{L_f} = \frac{100kA}{L} \cdot \frac{1}{L_f} \]
- Thus, we have:
\[ q_2 = \frac{100kA}{L} \cdot \frac{1}{L_f} \]

4. **Finding the Ratio**:
- Now we can find the ratio \( \frac{q_2}{q_1} \):
\[ \frac{q_2}{q_1} = \frac{\frac{100kA}{L} \cdot \frac{1}{L_f}}{\frac{200kA}{L} \cdot \frac{1}{L_f}} = \frac{100}{200} = \frac{1}{2} \]

### Conclusion:
The ratio \( \frac{q_2}{q_1} \) is \( \frac{1}{2} \).