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 rod with half the length and double the radius of the first and if the thermal conductivity of material of the second rod is 0.25 times that of first, the rate at which ice melts in `gs^(-1)` will be:

Options

A. 0.1
B. 0.2
C. 1.6
D. 3.2

Explanation

To solve the problem, we will use the principle of heat transfer through a cylindrical rod and the relationship between the heat transferred and the amount of ice melted.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have a cylindrical rod with one end in a steam chamber (at 100°C) and the other end in ice (at 0°C).
- The rate of melting of ice with the first rod is given as 0.1 g/s.

2. **Heat Transfer Equation**:
- The heat transfer through the rod can be described by Fourier's law of heat conduction:
\[ \frac{dq}{dt} = \frac{K \cdot A \cdot (T_1 - T_2)}{L} \]
- Where:
- \( \frac{dq}{dt} \) = rate of heat transfer (J/s)
- \( K \) = thermal conductivity of the material of the rod
- \( A \) = cross-sectional area of the rod
- \( T_1 - T_2 \) = temperature difference (100°C - 0°C = 100°C)
- \( L \) = length of the rod

3. **Calculating the Heat Required to Melt Ice**:
- The heat required to melt ice can be expressed as:
\[ \frac{dq}{dt} = m \cdot L_f \]
- Where:
- \( m \) = mass of ice melted per second (0.1 g = 0.1 × 10^-3 kg)
- \( L_f \) = latent heat of fusion of ice (approximately 334,000 J/kg)

4. **Setting Up the First Rod**:
- For the first rod:
\[ \frac{dq}{dt} = K \cdot A_1 \cdot \frac{100}{L} \]
- And from the melting ice:
\[ \frac{dq}{dt} = 0.1 \times 10^{-3} \cdot 334,000 \]

5. **Setting Up the Second Rod**:
- The second rod has half the length (\( \frac{L}{2} \)) and double the radius (\( 2R \)).
- The cross-sectional area \( A_2 \) of the second rod is:
\[ A_2 = \pi (2R)^2 = 4\pi R^2 = 4A_1 \]
- The thermal conductivity of the second rod is \( K_2 = 0.25K_1 \).
- The heat transfer for the second rod becomes:
\[ \frac{dq}{dt} = \frac{0.25K_1 \cdot 4A_1 \cdot 100}{\frac{L}{2}} = \frac{0.25K_1 \cdot 4A_1 \cdot 100 \cdot 2}{L} = \frac{2K_1 \cdot A_1 \cdot 100}{L} \]

6. **Comparing Heat Transfer Rates**:
- Now we can compare the heat transfer rates of both rods:
\[ \frac{dq}{dt}_{\text{second rod}} = 2 \cdot \frac{dq}{dt}_{\text{first rod}} \]
- Since the first rod melts 0.1 g of ice per second, the second rod will melt:
\[ m_2 = 2 \cdot 0.1 \text{ g/s} = 0.2 \text{ g/s} \]

7. **Final Answer**:
- Therefore, the rate at which ice melts with the second rod is:
\[ \boxed{0.2 \text{ g/s}} \]