In Q.No.4-49, the lowest rate of flow of heat is for:

Options

A. l = 2m, r=1 cm
B. l=4m,r=2cm
C. l = 2 m, r = 2 cm
D. l = 2 m, r = 4 cm

Explanation

To solve the problem of determining which cylinder rod has the highest rate of flow of heat, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Heat Transfer**:
The rate of heat transfer (dq/dt) through a rod can be expressed using the formula:
\[ \frac{dq}{dt} = \frac{k \cdot A \cdot \Delta T}{L} \]
where:
- \( k \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference across the rod,
- \( L \) is the length of the rod.

2. **Identify Constants**:
Since all rods are made of the same material and have the same temperature difference, \( k \) and \( \Delta T \) are constant for all rods. Thus, the rate of heat transfer simplifies to:
\[ \frac{dq}{dt} \propto \frac{A}{L} \]

3. **Calculate the Area and Length for Each Rod**:
- For Rod 1 (R1):
- Area \( A_1 = 1^2 = 1 \)
- Length \( L_1 = 2 \)
- Rate of heat transfer \( \propto \frac{1}{2} = 0.5 \)

- For Rod 2 (R2):
- Area \( A_2 = 2^2 = 4 \)
- Length \( L_2 = 4 \)
- Rate of heat transfer \( \propto \frac{4}{4} = 1 \)

- For Rod 3 (R3):
- Area \( A_3 = 2^2 = 4 \)
- Length \( L_3 = 2 \)
- Rate of heat transfer \( \propto \frac{4}{2} = 2 \)

- For Rod 4 (R4):
- Area \( A_4 = 4^2 = 16 \)
- Length \( L_4 = 2 \)
- Rate of heat transfer \( \propto \frac{16}{2} = 8 \)

4. **Compare the Rates of Heat Transfer**:
- R1: \( 0.5 \)
- R2: \( 1 \)
- R3: \( 2 \)
- R4: \( 8 \)

5. **Determine the Highest Rate of Flow**:
From the calculations, it is clear that Rod 4 (R4) has the highest rate of heat transfer.

### Conclusion:
The rod with the highest rate of flow of heat is **Rod 4 (R4)**.

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