Two rods of same length and areas of cross section `A_1` and `A_2` have their ends at same temperature. If `K_1` and `K_2` are their thermal conductivities, `C_1` and `C_2` their specific heats and `rho_1` and `rho_2` are their densities, then the condition that rate of flow of heat is same in both the rods is

Options

A. `A_(1)//A_(2)=K_(1)//K_(2)`
B. `A_(1)//A_(2)=K_(1)C_(1)rho_(1)//K_(2) C_(2)rho_(2)`
C. `A_(1)//A_(2)=K_(2)C_(1)rho_(1)//K_(1)C_(2)rho_(2)`
D. `A_(1)//A_(2)=K_(2)//K_(1)`

Explanation

To solve the problem, we need to establish the condition under which the rate of heat flow through two rods of different cross-sectional areas and thermal conductivities is the same.

### Step-by-Step Solution:

1. **Understanding Heat Flow**:
The rate of heat flow (dq/dt) through a rod is given by the formula:
\[ \frac{dq}{dt} = K \cdot A \cdot \frac{d\theta}{dx} \]
where \( K \) is the thermal conductivity, \( A \) is the area of cross-section, and \( \frac{d\theta}{dx} \) is the temperature gradient.

2. **Applying the Formula to Both Rods**:
For the first rod with thermal conductivity \( K_1 \) and cross-sectional area \( A_1 \):
\[ \frac{dq}{dt} = K_1 \cdot A_1 \cdot \frac{d\theta}{dx} \]
For the second rod with thermal conductivity \( K_2 \) and cross-sectional area \( A_2 \):
\[ \frac{dq}{dt} = K_2 \cdot A_2 \cdot \frac{d\theta}{dx} \]

3. **Setting the Heat Flow Equal**:
Since the rate of heat flow is the same in both rods, we can equate the two expressions:
\[ K_1 \cdot A_1 \cdot \frac{d\theta}{dx} = K_2 \cdot A_2 \cdot \frac{d\theta}{dx} \]

4. **Canceling the Common Terms**:
Since both rods are at the same temperature difference and have the same length, \( \frac{d\theta}{dx} \) is the same for both rods. We can cancel \( \frac{d\theta}{dx} \) from both sides:
\[ K_1 \cdot A_1 = K_2 \cdot A_2 \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{A_1}{A_2} = \frac{K_2}{K_1} \]

### Conclusion:
The condition that the rate of flow of heat is the same in both rods is:
\[ \frac{A_1}{A_2} = \frac{K_2}{K_1} \]