Two identical long, solid cylinders are used to conduct heat from temp `T_(1)` to Temp `T_(2)`. Originally the cylinders are connected in series and the rate of heat transfer is H. If the cylinders are connected in parallel then the rate of heat transer would be:

Options

A. H/4
B. 2H
C. 4H
D. 8H

Explanation

To solve the problem, we need to analyze the heat transfer through the two identical long solid cylinders when they are connected in series and then in parallel.

### Step-by-Step Solution:

1. **Understanding Heat Transfer in Series:**
- When the two cylinders are connected in series, the total thermal resistance \( R_{\text{series}} \) is the sum of the individual resistances:
\[ R_{\text{series}} = R_1 + R_2 = R + R = 2R \]
- The rate of heat transfer \( H \) through the cylinders can be expressed using the formula:
\[ H = \frac{T_1 - T_2}{R_{\text{series}}} \]
- Substituting \( R_{\text{series}} \):
\[ H = \frac{T_1 - T_2}{2R} \]

2. **Understanding Heat Transfer in Parallel:**
- When the two cylinders are connected in parallel, the total thermal resistance \( R_{\text{parallel}} \) can be calculated using the formula for resistances in parallel:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \]
- Therefore, the equivalent resistance in parallel is:
\[ R_{\text{parallel}} = \frac{R}{2} \]
- The rate of heat transfer \( H' \) through the cylinders in parallel is given by:
\[ H' = \frac{T_1 - T_2}{R_{\text{parallel}}} \]
- Substituting \( R_{\text{parallel}} \):
\[ H' = \frac{T_1 - T_2}{\frac{R}{2}} = \frac{2(T_1 - T_2)}{R} \]

3. **Relating \( H' \) to \( H \):**
- From our earlier expression for \( H \):
\[ H = \frac{T_1 - T_2}{2R} \]
- We can express \( H' \) in terms of \( H \):
\[ H' = 4H \]

### Conclusion:
Thus, when the two identical long solid cylinders are connected in parallel, the rate of heat transfer increases to \( 4H \).