Four cylindrical rods of same material with length and radius (l,r),(2l,r),(l/2,r) and (l,2r) are connected between two reservoirs at `0^@C` and `100^@C`. Find the ratio of the maximum to minimum rate of cunduction in them.

Explanation

To solve the problem, we need to find the ratio of the maximum to minimum rate of conduction through four cylindrical rods of the same material. The formula for the rate of heat conduction \( \frac{dq}{dt} \) through a rod is given by:

\[ \frac{dq}{dt} = k \cdot A \cdot \frac{\Delta T}{L} \]

Where:
- \( k \) is the thermal conductivity (same for all rods),
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference (constant for all rods),
- \( L \) is the length of the rod.

Since \( k \) and \( \Delta T \) are constant for all rods, we can express the rate of conduction as proportional to the ratio of the cross-sectional area \( A \) to the length \( L \):

\[ \frac{dq}{dt} \propto \frac{A}{L} \]

### Step 1: Calculate the cross-sectional area for each rod
The cross-sectional area \( A \) of a cylindrical rod is given by:

\[ A = \pi r^2 \]

### Step 2: Analyze each rod
1. **Rod 1**: Length \( L \), Radius \( r \)
\[ A_1 = \pi r^2, \quad L_1 = L \quad \Rightarrow \quad \frac{dq}{dt} \propto \frac{\pi r^2}{L} \]

2. **Rod 2**: Length \( 2L \), Radius \( r \)
\[ A_2 = \pi r^2, \quad L_2 = 2L \quad \Rightarrow \quad \frac{dq}{dt} \propto \frac{\pi r^2}{2L} \]

3. **Rod 3**: Length \( \frac{L}{2} \), Radius \( r \)
\[ A_3 = \pi r^2, \quad L_3 = \frac{L}{2} \quad \Rightarrow \quad \frac{dq}{dt} \propto \frac{\pi r^2}{\frac{L}{2}} = \frac{2\pi r^2}{L} \]

4. **Rod 4**: Length \( L \), Radius \( 2r \)
\[ A_4 = \pi (2r)^2 = 4\pi r^2, \quad L_4 = L \quad \Rightarrow \quad \frac{dq}{dt} \propto \frac{4\pi r^2}{L} \]

### Step 3: Express the rates of conduction
Now we can summarize the proportional rates of conduction for each rod:
- For Rod 1: \( \frac{dq}{dt} \propto \frac{r^2}{L} \)
- For Rod 2: \( \frac{dq}{dt} \propto \frac{r^2}{2L} \)
- For Rod 3: \( \frac{dq}{dt} \propto \frac{2r^2}{L} \)
- For Rod 4: \( \frac{dq}{dt} \propto \frac{4r^2}{L} \)

### Step 4: Identify maximum and minimum rates
From the expressions above, we can see:
- Maximum rate of conduction is from Rod 4: \( \frac{dq}{dt} \propto \frac{4r^2}{L} \)
- Minimum rate of conduction is from Rod 2: \( \frac{dq}{dt} \propto \frac{r^2}{2L} \)

### Step 5: Calculate the ratio of maximum to minimum rates
Now we can find the ratio of maximum to minimum rates of conduction:

\[ \frac{\frac{dq}{dt}_{\text{max}}}{\frac{dq}{dt}_{\text{min}}} = \frac{4r^2/L}{r^2/(2L)} = \frac{4r^2 \cdot 2L}{r^2 \cdot L} = \frac{8}{1} \]

Thus, the ratio of the maximum to minimum rate of conduction is:

\[ \text{Ratio} = 8:1 \]