A solid body of constant heat capacity `1J//^@C` is being heated by keeping it contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoir such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature `100^C` to final temperature `200^@C`. Entropy change of the body in the tow cases respectively is :

Options

A. `ln 2, 4ln2`
B. `ln2, ln 2`
C. `ln2, 2ln, 2`
D. `2ln 2n, 8ln2`

Explanation

To solve the problem, we need to calculate the change in entropy for the solid body in both cases. The key point here is that entropy is a state function, which means it depends only on the initial and final states of the system, not on the path taken to get from one state to another.

### Step-by-Step Solution:

**Step 1: Understand the Problem**
- We have a solid body with a constant heat capacity of \( C = 1 \, \text{J/°C} \).
- The body is heated from an initial temperature \( T_i = 100 \, \text{°C} \) to a final temperature \( T_f = 200 \, \text{°C} \).
- We need to find the change in entropy for two different heating methods.

**Step 2: Calculate the Total Heat Supplied**
- The total heat \( Q \) required to raise the temperature of the body can be calculated using the formula:
\[ Q = C \cdot (T_f - T_i) = 1 \, \text{J/°C} \cdot (200 \, \text{°C} - 100 \, \text{°C}) = 1 \, \text{J/°C} \cdot 100 \, \text{°C} = 100 \, \text{J} \]

**Step 3: Calculate Entropy Change**
- The change in entropy \( \Delta S \) for a process is given by the formula:
\[ \Delta S = \int \frac{dQ}{T} \]
- Since we are looking at two different methods of heating, we will consider the average temperature for each method.

**Case (i): Heating with 2 Reservoirs**
- Each reservoir supplies \( Q/2 = 50 \, \text{J} \).
- The average temperature during the heating process can be approximated as the average of the initial and final temperatures:
\[ T_{\text{avg}} = \frac{T_i + T_f}{2} = \frac{100 + 200}{2} = 150 \, \text{°C} = 423.15 \, \text{K} \]
- The entropy change for this case is:
\[ \Delta S_1 = \frac{Q}{T_{\text{avg}}} = \frac{100 \, \text{J}}{423.15 \, \text{K}} \approx 0.236 \, \text{J/K} \]

**Case (ii): Heating with 8 Reservoirs**
- Each reservoir supplies \( Q/8 = 12.5 \, \text{J} \).
- The average temperature during the heating process remains the same as in the first case because the total heat and temperature change are the same.
- The entropy change for this case is:
\[ \Delta S_2 = \frac{Q}{T_{\text{avg}}} = \frac{100 \, \text{J}}{423.15 \, \text{K}} \approx 0.236 \, \text{J/K} \]

**Step 4: Conclusion**
- The change in entropy for both methods is the same:
\[ \Delta S_1 = \Delta S_2 \approx 0.236 \, \text{J/K} \]

### Final Answer:
The entropy change of the body in both cases is the same.