Two rods made of same material having same length and diameter are joined in series. The thermal power dissipated through then is 2W.If they are joined in parallel, the thermal power dissipated under the same conditions on the two ends of the rods, will be :

Options

A. 16 W
B. 8 W
C. 4 W
D. 2 W

Explanation

To solve the problem step by step, we will analyze the thermal power dissipation in both series and parallel configurations of the two rods.

### Step 1: Understand the Series Configuration
When the two rods are connected in series, the total thermal resistance (R_total) is the sum of the individual resistances of the rods.

For each rod:
- Let \( R_1 = \frac{L}{kA} \) (where \( L \) is the length, \( k \) is the thermal conductivity, and \( A \) is the cross-sectional area).
- Since both rods are identical, \( R_2 = R_1 \).

Thus, the total resistance in series is:
\[ R_{total} = R_1 + R_2 = R_1 + R_1 = 2R_1 = 2 \left(\frac{L}{kA}\right) \]

### Step 2: Calculate Thermal Power in Series
The thermal power (P) dissipated through the rods in series can be expressed as:
\[ P = \frac{\Delta T}{R_{total}} \]
Given that the power is 2 W:
\[ 2 = \frac{\Delta T}{2 \left(\frac{L}{kA}\right)} \]
From this equation, we can express \( \Delta T \):
\[ \Delta T = 2 \cdot 2 \left(\frac{L}{kA}\right) = \frac{4L}{kA} \]

### Step 3: Understand the Parallel Configuration
When the two rods are connected in parallel, the total resistance (R_parallel) can be calculated using the formula for resistors in parallel:
\[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Since \( R_1 = R_2 \):
\[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_1} = \frac{2}{R_1} \]
Thus,
\[ R_{parallel} = \frac{R_1}{2} = \frac{L}{2kA} \]

### Step 4: Calculate Thermal Power in Parallel
The thermal power in the parallel configuration can be expressed as:
\[ P_{parallel} = \frac{\Delta T}{R_{parallel}} \]
Substituting \( R_{parallel} \):
\[ P_{parallel} = \frac{\Delta T}{\frac{L}{2kA}} = \Delta T \cdot \frac{2kA}{L} \]

### Step 5: Relate Thermal Power in Parallel to Series
Now we can substitute \( \Delta T \) from the series configuration into the equation for parallel:
\[ P_{parallel} = \left(\frac{4L}{kA}\right) \cdot \frac{2kA}{L} \]
This simplifies to:
\[ P_{parallel} = 4 \cdot 2 = 8 \, \text{W} \]

### Final Answer
Thus, the thermal power dissipated when the rods are joined in parallel is:
\[ \boxed{8 \, \text{W}} \]