Two identical rods are connected between two containers on of tehm is at `100^@C` and another is at `0^@C.` If rods are connected in parralel them the rate of melting of ice is `q_1 gm//sec.`
If they are connected in series then the rate is `q_2.` 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 when they are connected in parallel and in series. We will use the formula for heat transfer, which is given by:

\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \]

where:
- \( Q \) is the rate of heat transfer,
- \( K \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the length of the rod.

### Step 1: Calculate \( Q_1 \) for the parallel connection

When the rods are connected in parallel, the effective area for heat transfer doubles. Therefore, the heat transfer rate \( Q_1 \) can be calculated as follows:

\[ Q_1 = \frac{K \cdot (2A) \cdot (100 - 0)}{L} = \frac{K \cdot 2A \cdot 100}{L} = \frac{200KA}{L} \]

### Step 2: Calculate \( Q_2 \) for the series connection

When the rods are connected in series, the total length of the rods becomes \( 2L \). The heat transfer rate \( Q_2 \) can be calculated as follows:

\[ Q_2 = \frac{K \cdot A \cdot (100 - 0)}{2L} = \frac{K \cdot A \cdot 100}{2L} = \frac{100KA}{2L} = \frac{50KA}{L} \]

### Step 3: Calculate the ratio \( \frac{Q_2}{Q_1} \)

Now we can find the ratio of the rates of melting of ice when the rods are connected in series and in parallel:

\[ \frac{Q_2}{Q_1} = \frac{\frac{50KA}{L}}{\frac{200KA}{L}} = \frac{50}{200} = \frac{1}{4} \]

### Conclusion

The ratio \( \frac{Q_2}{Q_1} \) is \( \frac{1}{4} \).

---