The end of two rods of different materials with their thermal conductivities, area of cross-section and lengths all in the ratio 1:2 are maintained at the same temperature difference. If the rate of flow of heat in the first rod is `4 cal//s`. Then, in the second rod rate of heat flow in `cal//s ` will be

Options

A. 1
B. 2
C. 8
D. 16

Explanation

To solve the problem, we need to analyze the heat transfer through two rods with different materials but with specific ratios in their thermal conductivities, areas of cross-section, and lengths. Let's break it down step by step.

### Step 1: Understand the given ratios
We are given that the thermal conductivities (K), areas of cross-section (A), and lengths (L) of the two rods are in the ratio 1:2. This means:
- \( K_1 : K_2 = 1 : 2 \)
- \( A_1 : A_2 = 1 : 2 \)
- \( L_1 : L_2 = 1 : 2 \)

### Step 2: Write the formula for thermal resistance
The thermal resistance (R) for heat transfer through a rod is given by:
\[ R = \frac{L}{KA} \]
Where:
- \( L \) is the length of the rod,
- \( K \) is the thermal conductivity,
- \( A \) is the area of cross-section.

### Step 3: Calculate the resistance for both rods
Using the ratios provided:
1. For rod 1:
\[ R_1 = \frac{L_1}{K_1 A_1} \]
2. For rod 2:
\[ R_2 = \frac{L_2}{K_2 A_2} \]

Substituting the ratios:
- \( L_1 = L \), \( L_2 = 2L \)
- \( K_1 = K \), \( K_2 = 2K \)
- \( A_1 = A \), \( A_2 = 2A \)

Thus,
\[ R_1 = \frac{L}{K \cdot A} \]
\[ R_2 = \frac{2L}{2K \cdot 2A} = \frac{2L}{4KA} = \frac{L}{2KA} \]

### Step 4: Find the ratio of resistances
Now, we can find the ratio of the resistances:
\[ \frac{R_1}{R_2} = \frac{\frac{L}{KA}}{\frac{L}{2KA}} = 2 \]

### Step 5: Relate heat flow to resistance
The rate of heat flow (H) is inversely proportional to the thermal resistance (R):
\[ H \propto \frac{1}{R} \]
Thus, we can write:
\[ \frac{H_1}{H_2} = \frac{R_2}{R_1} \]

### Step 6: Substitute known values
Given that \( H_1 = 4 \, \text{cal/s} \):
\[ \frac{4}{H_2} = \frac{1}{2} \]

### Step 7: Solve for \( H_2 \)
Cross-multiplying gives:
\[ 4 = \frac{H_2}{2} \]
Thus,
\[ H_2 = 4 \times 2 = 8 \, \text{cal/s} \]

### Conclusion
The rate of heat flow in the second rod is \( 8 \, \text{cal/s} \).