Two identical rods are connected in parallel. One end of the rods is maintained at `100^(@)C` and other end is kept in `0^(@)C` ice . The rate at which the ice melts is `Q_(1)` gm/sec . Now the rods are connected in series . The ends are again maintained at `100^(@)`C and kept in `0^(@)C` ice . Now the rate at which ice melts is `Q_(2)` gm/sec . Then `Q_(2) //Q_(1)` is

Options

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

Answer

To solve the problem, we need to analyze the heat transfer through the rods in both parallel and series configurations. Let's break it down step by step.

### Step 1: Understand the Parallel Configuration
When the two identical rods are connected in parallel, the heat transfer rate \( Q_1 \) can be calculated using the formula:

\[ Q_1 = \frac{(T_2 - T_1)}{R_{\text{eq}}} \]

Where:
- \( T_2 = 100^\circ C \)
- \( T_1 = 0^\circ C \)
- \( R_{\text{eq}} \) is the equivalent thermal resistance for the parallel configuration.

### Step 2: Calculate the Equivalent Resistance in Parallel
For two identical rods in parallel, the equivalent thermal resistance \( R_{\text{eq}} \) is given by:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Since \( R_1 = R_2 = \frac{L}{kA} \) (where \( L \) is the length, \( k \) is the thermal conductivity, and \( A \) is the cross-sectional area), we have:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{\frac{L}{kA}} + \frac{1}{\frac{L}{kA}} = \frac{2kA}{L} \]

Thus,

\[ R_{\text{eq}} = \frac{L}{2kA} \]

### Step 3: Substitute into the Heat Transfer Equation for Parallel
Now substituting \( R_{\text{eq}} \) into the equation for \( Q_1 \):

\[ Q_1 = \frac{100 - 0}{\frac{L}{2kA}} = \frac{100 \cdot 2kA}{L} = \frac{200kA}{L} \]

### Step 4: Understand the Series Configuration
When the rods are connected in series, the equivalent thermal resistance \( R_{\text{eq}} \) is:

\[ R_{\text{eq}} = R_1 + R_2 = \frac{L}{kA} + \frac{L}{kA} = \frac{2L}{kA} \]

### Step 5: Calculate the Heat Transfer Rate in Series
Now, for the series configuration, the heat transfer rate \( Q_2 \) is given by:

\[ Q_2 = \frac{(T_2 - T_1)}{R_{\text{eq}}} = \frac{100 - 0}{\frac{2L}{kA}} = \frac{100 \cdot kA}{2L} = \frac{100kA}{2L} \]

### Step 6: Find the Ratio \( \frac{Q_2}{Q_1} \)
Now we can find the ratio of the heat transfer rates:

\[ \frac{Q_2}{Q_1} = \frac{\frac{100kA}{2L}}{\frac{200kA}{L}} = \frac{100kA}{2L} \cdot \frac{L}{200kA} = \frac{100}{2 \cdot 200} = \frac{1}{4} \]

### Final Answer
Thus, the ratio \( \frac{Q_2}{Q_1} \) is:

\[ \frac{Q_2}{Q_1} = \frac{1}{4} \]