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 and connected to the same vessels. Let `q_1 and q_2` gram per second be the rate of melting of ice in the two cases respectively. The ratio `q_1/q_2` is
(a) `1/2` (b)`2/1` (c)`4/1` (d)`1/4`

Options

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

Explanation

To solve the problem, we need to analyze the heat transfer through the conducting rods in both scenarios and find the ratio of the rates of melting of ice, \( q_1 \) and \( q_2 \).

### Step-by-Step Solution

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

2. **Case 1: Rods in Parallel**:
- When the rods are connected independently (in parallel), the equivalent thermal resistance \( R_{eq1} \) is given by:
\[ R_{eq1} = \frac{R \times R}{R + R} = \frac{R}{2} \]
- The heat transfer \( q_1 \) (rate of melting of ice) can be calculated using the formula:
\[ q_1 = \frac{T_{hot} - T_{cold}}{R_{eq1}} = \frac{100 - 0}{\frac{R}{2}} = \frac{200}{R} \]

3. **Case 2: Rods in Series**:
- When the rods are connected end-to-end (in series), the equivalent thermal resistance \( R_{eq2} \) is:
\[ R_{eq2} = R + R = 2R \]
- The heat transfer \( q_2 \) in this case is:
\[ q_2 = \frac{T_{hot} - T_{cold}}{R_{eq2}} = \frac{100 - 0}{2R} = \frac{50}{R} \]

4. **Finding the Ratio \( \frac{q_1}{q_2} \)**:
- Now we can find the ratio of the rates of melting of ice:
\[ \frac{q_1}{q_2} = \frac{\frac{200}{R}}{\frac{50}{R}} = \frac{200}{50} = 4 \]
- Therefore, the ratio \( \frac{q_1}{q_2} = 4:1 \).

5. **Conclusion**:
- The correct answer is option (c) \( 4:1 \).