One end of conducting rod is maintained at temperature `50^(@)C` and at the other end ice is melting at `0^(@)C`. The rate of melting of ice is doubled if:

Options

A. the temperature is made `200^(@)C` and the area of cross-section of the rod is doubled
B. the temperature is made `100^(@)C` and length of the rod is made four times
C. area of cross-section of rod is halved and length doubled
D. the temperature is made `100^(@)C` and area cross-section of rod and length both are doubled

Answer

To solve the problem, we need to analyze the heat transfer through the conducting rod and how changes in temperature, area of cross-section, and length affect the rate of heat transfer.

### Step-by-Step Solution:

1. **Understand the Heat Transfer Formula:**
The rate of heat transfer (Q) through a conducting rod is given by the formula:
\[ Q = \frac{K \cdot A \cdot (T_1 - T_2)}{L} \]
where:
- \( K \) = thermal conductivity of the material
- \( A \) = area of cross-section of the rod
- \( T_1 \) = temperature at one end of the rod
- \( T_2 \) = temperature at the other end of the rod
- \( L \) = length of the rod

2. **Initial Conditions:**
In our case:
- \( T_1 = 50^\circ C \)
- \( T_2 = 0^\circ C \)
- The initial heat transfer rate \( Q \) can be expressed as:
\[ Q = \frac{K \cdot A \cdot (50 - 0)}{L} = \frac{K \cdot A \cdot 50}{L} \]

3. **Evaluate Each Option:**
We will evaluate each option to see if it results in doubling the rate of melting of ice (i.e., \( Q = 2Q \)).

**Option 1:** Temperature is made \( 200^\circ C \) and area is doubled.
- New conditions:
\[ Q_1 = \frac{K \cdot (2A) \cdot (200 - 0)}{L} = \frac{K \cdot 2A \cdot 200}{L} = \frac{K \cdot 400A}{L} \]
- Comparing with initial \( Q \):
\[ Q_1 = 8Q \quad (\text{since } 400A/L = 8 \cdot 50A/L) \]
- **Conclusion:** This option is incorrect (rate is 8 times).

**Option 2:** Temperature is made \( 100^\circ C \) and length is made 4 times.
- New conditions:
\[ Q_2 = \frac{K \cdot A \cdot (100 - 0)}{4L} = \frac{K \cdot A \cdot 100}{4L} = \frac{K \cdot 25A}{L} \]
- Comparing with initial \( Q \):
\[ Q_2 = \frac{1}{2}Q \quad (\text{since } 25A/L = \frac{1}{2} \cdot 50A/L) \]
- **Conclusion:** This option is incorrect (rate is half).

**Option 3:** Area of cross-section is halved and length is doubled.
- New conditions:
\[ Q_3 = \frac{K \cdot \left(\frac{A}{2}\right) \cdot (50 - 0)}{2L} = \frac{K \cdot 25A}{2L} \]
- Comparing with initial \( Q \):
\[ Q_3 = \frac{1}{4}Q \quad (\text{since } 25A/L = \frac{1}{4} \cdot 50A/L) \]
- **Conclusion:** This option is incorrect (rate is one-fourth).

**Option 4:** Temperature is made \( 100^\circ C \) and both area and length are doubled.
- New conditions:
\[ Q_4 = \frac{K \cdot (2A) \cdot (100 - 0)}{2L} = \frac{K \cdot 2A \cdot 100}{2L} = \frac{K \cdot 100A}{L} \]
- Comparing with initial \( Q \):
\[ Q_4 = 2Q \quad (\text{since } 100A/L = 2 \cdot 50A/L) \]
- **Conclusion:** This option is correct (rate is doubled).

### Final Answer:
The rate of melting of ice is doubled if the temperature is made \( 100^\circ C \) and both the area of cross-section and length of the rod are doubled (Option 4).