Consider two rods of same length and different specific heats (`S_1` and `S_2`), conductivities `K_1` and `K_2` and area of cross section (`A_1` and `A_2`) and both having temperature `T_1` and `T_2` at their ends. If the rate of heat loss due to conduction is equal then

Options

A. `K_1A_1=K_2A_2`
B. `K_2A_1=K_1A_2`
C. `(K_1A_1)/(S_1)=(K_2A_2)/(S_2)`
D. `(K_2A_1)/(S_2)=(K_1A_2)/(S_1)`

Explanation

To solve the problem, we need to analyze the heat conduction through two rods with different properties but equal rates of heat loss. The key concept here is to use Fourier's law of heat conduction, which states that the rate of heat transfer (Q/t) through a material is proportional to the temperature difference across the material and the area, while inversely proportional to the length of the material.

### Step-by-Step Solution:

1. **Understand the Formula for Heat Conduction**:
The rate of heat conduction (Q/t) through a rod can be expressed using Fourier's law:
\[ \frac{Q}{t} = \frac{K \cdot A \cdot (T_1 - T_2)}{L} \]
where:
- \( K \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area,
- \( T_1 \) and \( T_2 \) are the temperatures at the ends of the rod,
- \( L \) is the length of the rod.

2. **Set Up the Equations for Both Rods**:
For the first rod (Rod 1):
\[ \frac{Q}{t} = \frac{K_1 \cdot A_1 \cdot (T_1 - T_2)}{L} \]
For the second rod (Rod 2):
\[ \frac{Q}{t} = \frac{K_2 \cdot A_2 \cdot (T_1 - T_2)}{L} \]

3. **Equate the Heat Loss Rates**:
Since the rate of heat loss is equal for both rods, we can set the two equations equal to each other:
\[ \frac{K_1 \cdot A_1 \cdot (T_1 - T_2)}{L} = \frac{K_2 \cdot A_2 \cdot (T_1 - T_2)}{L} \]

4. **Cancel Out Common Terms**:
The length \( L \) and the temperature difference \( (T_1 - T_2) \) are common to both sides of the equation and can be canceled out (assuming \( T_1 \neq T_2 \)):
\[ K_1 \cdot A_1 = K_2 \cdot A_2 \]

5. **Solve for the Relationship Between Areas and Conductivities**:
Rearranging gives us the relationship between the conductivities and the areas:
\[ \frac{K_1}{K_2} = \frac{A_2}{A_1} \]

### Final Result:
The relationship between the thermal conductivities and cross-sectional areas of the two rods is given by:
\[ K_1 \cdot A_1 = K_2 \cdot A_2 \]