The ends of the two rods of different materials with their lengths, diameters of cross-section and thermal conductivities all in the ratio 1 : 2 are maintained at the same temperature difference. The rate of flow of heat in the shorter rod is `1cals^(-1)`. What is the rate of flow of heat in the larger rod:

Options

A. `1 cal s^(-1)`
B. `4 cal s^(-1)`
C. `8 cal s^(-1)`
D. `16 cal s^(-1)`

Explanation

To solve the problem, we will apply the formula for the rate of heat transfer through a rod, which is given by Fourier's law of heat conduction:

\[ \frac{Q}{t} = \frac{k \cdot A \cdot \Delta T}{L} \]

Where:
- \( Q/t \) is the rate of heat transfer (in calories per second),
- \( k \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the length of the rod.

### Step 1: Define the parameters for both rods
Let:
- For the shorter rod (Rod 1):
- Length \( L_1 = L \)
- Diameter \( d_1 = d \)
- Area \( A_1 = \frac{\pi d_1^2}{4} = A \)
- Thermal conductivity \( k_1 = k \)

- For the longer rod (Rod 2):
- Length \( L_2 = 2L \)
- Diameter \( d_2 = 2d \) (since the diameter is in the ratio 1:2)
- Area \( A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (2d)^2}{4} = 4A \)
- Thermal conductivity \( k_2 = 2k \) (since thermal conductivity is also in the ratio 1:2)

### Step 2: Calculate the rate of heat transfer for both rods
Using the formula for heat transfer, we can calculate the rate for both rods.

For Rod 1:
\[ \frac{Q_1}{t} = \frac{k_1 \cdot A_1 \cdot \Delta T}{L_1} = \frac{k \cdot A \cdot \Delta T}{L} \]

For Rod 2:
\[ \frac{Q_2}{t} = \frac{k_2 \cdot A_2 \cdot \Delta T}{L_2} = \frac{(2k) \cdot (4A) \cdot \Delta T}{2L} \]

### Step 3: Simplify the expression for Rod 2
Substituting the values into the equation for Rod 2:
\[ \frac{Q_2}{t} = \frac{(2k) \cdot (4A) \cdot \Delta T}{2L} = \frac{8k \cdot A \cdot \Delta T}{2L} = \frac{4k \cdot A \cdot \Delta T}{L} \]

### Step 4: Relate the heat flow rates
From the calculations, we have:
\[ \frac{Q_2}{t} = 4 \cdot \frac{Q_1}{t} \]

Given that the rate of flow of heat in the shorter rod (Rod 1) is \( 1 \, \text{cal/s} \):
\[ \frac{Q_1}{t} = 1 \, \text{cal/s} \]

Thus, for Rod 2:
\[ \frac{Q_2}{t} = 4 \cdot 1 \, \text{cal/s} = 4 \, \text{cal/s} \]

### Final Answer
The rate of flow of heat in the larger rod is \( 4 \, \text{cal/s} \).

---