A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is 1/4 that of first, the rate at which ice melts in `g//s` will be

Options

A. (a) 0.4
B. (b) 0.05
C. (c) 0.2
D. (d) 0.1

Explanation

To solve the problem, we will analyze the heat transfer through the two cylindrical rods and determine the rate of melting of ice based on the heat transfer.

### Step-by-Step Solution:

1. **Understanding the Heat Transfer Formula**:
The rate of heat transfer (dq/dt) through a rod is given by the formula:
\[ \frac{dq}{dt} = \frac{k \cdot A}{L} \cdot \Delta T \]
where:
- \( k \) = thermal conductivity of the material,
- \( A \) = cross-sectional area of the rod,
- \( L \) = length of the rod,
- \( \Delta T \) = temperature difference across the rod.

2. **Identifying Parameters for the First Rod**:
Let the thermal conductivity of the first rod be \( k_1 \), the length be \( L_1 \), and the radius be \( r_1 \).
The cross-sectional area \( A_1 \) of the first rod is given by:
\[ A_1 = \pi r_1^2 \]
Therefore, the rate of heat transfer for the first rod is:
\[ \frac{dq_1}{dt} = \frac{k_1 \cdot A_1}{L_1} \cdot \Delta T \]

3. **Identifying Parameters for the Second Rod**:
The second rod has half the length and double the radius of the first rod:
- Length: \( L_2 = \frac{L_1}{2} \)
- Radius: \( r_2 = 2r_1 \)
The cross-sectional area \( A_2 \) of the second rod is:
\[ A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2 = 4A_1 \]
The thermal conductivity of the second rod is \( k_2 = \frac{1}{4} k_1 \).

4. **Calculating the Rate of Heat Transfer for the Second Rod**:
Now substituting the values for the second rod into the heat transfer formula:
\[ \frac{dq_2}{dt} = \frac{k_2 \cdot A_2}{L_2} \cdot \Delta T \]
Substituting \( k_2 \), \( A_2 \), and \( L_2 \):
\[ \frac{dq_2}{dt} = \frac{\left(\frac{1}{4} k_1\right) \cdot (4A_1)}{\left(\frac{L_1}{2}\right)} \cdot \Delta T \]
Simplifying this:
\[ \frac{dq_2}{dt} = \frac{k_1 \cdot A_1}{L_1} \cdot \Delta T \cdot 2 \]
This shows that:
\[ \frac{dq_2}{dt} = 2 \cdot \frac{dq_1}{dt} \]

5. **Relating Heat Transfer to Ice Melting**:
Since the rate of melting of ice is directly proportional to the rate of heat transfer, we can write:
\[ \text{Rate of melting for second rod} = 2 \cdot \text{Rate of melting for first rod} \]
Given that the rate of melting of ice for the first rod is 0.1 g/s:
\[ \text{Rate of melting for second rod} = 2 \cdot 0.1 \, \text{g/s} = 0.2 \, \text{g/s} \]

### Conclusion:
The rate at which ice melts with the second rod is **0.2 g/s**.