Choose the correct option:
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:

Options

A. (a) `1/2`
B. (b)`2/1`
C. (c)`4/1`
D. (d)`1/4`

Answer

To solve the problem, we need to analyze the heat transfer through the two identical conducting rods in two different configurations: independent connections and a series connection.

### 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 joined end to end (in series) and connected to the same vessels.

2. **Heat Transfer and Melting of Ice**:
- The rate of melting of ice, denoted as \(q\), is proportional to the amount of heat transferred per second.
- According to Fourier's law of heat conduction, the rate of heat transfer \(Q\) through a conductor is given by:
\[ Q = \frac{\Delta T}{R} \]
where \(\Delta T\) is the temperature difference and \(R\) is the thermal resistance.

3. **Thermal Resistance**:
- For two identical rods, the thermal resistance \(R\) can be expressed as:
\[ R = \frac{L}{kA} \]
where \(L\) is the length of the rod, \(k\) is the thermal conductivity, and \(A\) is the cross-sectional area.
- Since the rods are identical, we can denote the resistance of one rod as \(R = r\).

4. **Case 1: Independent Connection**:
- In this case, the rods are connected independently to the heat sources. The thermal resistance for each rod remains \(r\).
- The effective resistance for the melting of ice (which is at \(0^\circ C\)) is:
\[ R_1 = r \quad \text{(for the rod connected to ice)} \]
- The temperature difference for the rod connected to water is:
\[ \Delta T_1 = 100^\circ C - 0^\circ C = 100^\circ C \]
- The rate of melting of ice in this case, \(q_1\), can be expressed as:
\[ q_1 \propto \frac{100}{r} \]

5. **Case 2: Series Connection**:
- When the rods are joined end to end, the total resistance becomes:
\[ R_2 = r + r = 2r \]
- The temperature difference remains the same:
\[ \Delta T_2 = 100^\circ C - 0^\circ C = 100^\circ C \]
- The rate of melting of ice in this case, \(q_2\), can be expressed as:
\[ q_2 \propto \frac{100}{2r} = \frac{50}{r} \]

6. **Finding the Ratio**:
- Now, we can find the ratio of the rates of melting of ice in the two cases:
\[ \frac{q_1}{q_2} = \frac{\frac{100}{r}}{\frac{50}{r}} = \frac{100}{50} = 2 \]

7. **Final Ratio**:
- Therefore, the ratio \(q_1/q_2\) is:
\[ \frac{q_1}{q_2} = 2 \quad \text{or} \quad 2:1 \]

### Conclusion:
The correct option is **B: 2 by 1**.