Two rods of length l and `2l` thermal conductivities `2K` and `K` are connected end to end. If cross sectional areas of two rods are eual, then equivalent thermal conductivity of the system is .

Options

A. `(5//6)K`
B. `1.5K`
C. `1.2K`
D. `(8//9)K`

Explanation

To find the equivalent thermal conductivity of the system formed by two rods connected end to end, we can follow these steps:

### Step 1: Identify the parameters of the rods
- Rod 1:
- Length (L1) = l
- Thermal conductivity (K1) = 2K
- Cross-sectional area (A) = A (same for both rods)

- Rod 2:
- Length (L2) = 2l
- Thermal conductivity (K2) = K
- Cross-sectional area (A) = A (same for both rods)

### Step 2: Calculate the thermal resistance of each rod
The thermal resistance (R) of a rod is given by the formula:
\[ R = \frac{L}{K \cdot A} \]

- For Rod 1 (R1):
\[ R_1 = \frac{L_1}{K_1 \cdot A} = \frac{l}{2K \cdot A} \]

- For Rod 2 (R2):
\[ R_2 = \frac{L_2}{K_2 \cdot A} = \frac{2l}{K \cdot A} \]

### Step 3: Combine the thermal resistances
Since the rods are connected end to end, the total thermal resistance (R_total) is the sum of the individual resistances:
\[ R_{\text{total}} = R_1 + R_2 \]
\[ R_{\text{total}} = \frac{l}{2K \cdot A} + \frac{2l}{K \cdot A} \]

### Step 4: Simplify the expression for total resistance
To combine the fractions:
\[ R_{\text{total}} = \frac{l}{2K \cdot A} + \frac{4l}{2K \cdot A} = \frac{l + 4l}{2K \cdot A} = \frac{5l}{2K \cdot A} \]

### Step 5: Relate total resistance to equivalent thermal conductivity
The equivalent thermal resistance (R_eff) can also be expressed in terms of the equivalent thermal conductivity (K_eff):
\[ R_{\text{eff}} = \frac{L_{\text{total}}}{K_{\text{eff}} \cdot A} \]
Where \( L_{\text{total}} = l + 2l = 3l \).

Setting the two expressions for resistance equal:
\[ \frac{5l}{2K \cdot A} = \frac{3l}{K_{\text{eff}} \cdot A} \]

### Step 6: Solve for equivalent thermal conductivity (K_eff)
Cancel out common terms (l and A):
\[ \frac{5}{2K} = \frac{3}{K_{\text{eff}}} \]

Cross-multiply to solve for K_eff:
\[ 5K_{\text{eff}} = 6K \]
\[ K_{\text{eff}} = \frac{6K}{5} \]

### Final Answer
The equivalent thermal conductivity of the system is:
\[ K_{\text{eff}} = 1.2K \]

---