Two cylindrical rods of lengths `l_(1)` and `l_(2)`, radii `r_(1)` and `r_(2)` have thermal conductivities `k_(1)` and `k_(2)` respectively. The ends of the rods are maintained at the same temperature difference. If `l_(1) =2l_(2)` and `r_(1) =r_(2)//2`, the rates of heat flow in them will be the same if `k_(1)//k_(2)` is:

Options

A. 1
B. 2
C. 4
D. 8

Explanation

To solve the problem, we need to find the ratio of thermal conductivities \( \frac{k_1}{k_2} \) for two cylindrical rods under the given conditions. Let's go through the solution step by step.

### Step 1: Understand the Heat Flow Formula
The rate of heat flow \( \frac{dQ}{dt} \) through a cylindrical rod is given by the formula:
\[ \frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L} \]
where:
- \( k \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( T_1 - T_2 \) is the temperature difference across the ends of the rod,
- \( L \) is the length of the rod.

### Step 2: Define the Areas of the Cylindrical Rods
For a cylindrical rod, the cross-sectional area \( A \) is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the rod.

### Step 3: Set Up the Equations for Both Rods
Let’s denote the parameters for the two rods:
- For rod 1: \( l_1 = 2l_2 \), \( r_1 = \frac{r_2}{2} \)
- For rod 2: \( l_2 \), \( r_2 \)

The areas for the rods are:
- \( A_1 = \pi r_1^2 = \pi \left(\frac{r_2}{2}\right)^2 = \frac{\pi r_2^2}{4} \)
- \( A_2 = \pi r_2^2 \)

### Step 4: Write the Heat Flow Equations
For rod 1:
\[ \frac{dQ_1}{dt} = \frac{k_1 A_1 (T_1 - T_2)}{l_1} = \frac{k_1 \left(\frac{\pi r_2^2}{4}\right) (T_1 - T_2)}{2l_2} \]
For rod 2:
\[ \frac{dQ_2}{dt} = \frac{k_2 A_2 (T_1 - T_2)}{l_2} = \frac{k_2 \pi r_2^2 (T_1 - T_2)}{l_2} \]

### Step 5: Set the Heat Flow Rates Equal
Since the rates of heat flow are the same:
\[ \frac{k_1 \left(\frac{\pi r_2^2}{4}\right) (T_1 - T_2)}{2l_2} = \frac{k_2 \pi r_2^2 (T_1 - T_2)}{l_2} \]

### Step 6: Cancel Common Terms
We can cancel \( \pi r_2^2 (T_1 - T_2) \) and \( l_2 \) from both sides:
\[ \frac{k_1}{2} = k_2 \]

### Step 7: Express \( \frac{k_1}{k_2} \)
Rearranging gives:
\[ k_1 = 2k_2 \]
Thus,
\[ \frac{k_1}{k_2} = 2 \]

### Step 8: Substitute the Radius Relation
Now, substituting the radius relation \( r_1 = \frac{r_2}{2} \):
\[ \frac{k_1}{k_2} = \left(\frac{r_2}{\frac{r_2}{2}}\right)^2 \cdot 2 = \left(2\right)^2 \cdot 2 = 4 \cdot 2 = 8 \]

### Final Result
Thus, the ratio of thermal conductivities is:
\[ \frac{k_1}{k_2} = 8 \]