If the unknown resistance calculated without using the end correction, is `R_(1)` and with using the end corrections is `R_(2)` then (assume same end correction at both ends)

Options

A. `R_(1)gtR_(2)` when balanced point is in first half
B. `R_(1)ltR_(2)` when balanced point is in first half
C. `R_(1)gtR_(2)` when balanced point is in seconds half
D. `R_(1)gtR_(2)` always

Answer

To solve the problem, we need to analyze the relationship between the unknown resistances \( R_1 \) and \( R_2 \) calculated with and without end corrections, respectively.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two resistances: \( R_1 \) (calculated without end correction) and \( R_2 \) (calculated with end correction).
- We are to assume that the same end correction is applied at both ends of the potentiometer wire.

2. **Formulating the Expressions**:
- The resistance \( R_1 \) can be expressed in terms of the length \( L \) of the wire and the sensitivity \( S \) of the potentiometer:
\[ R_1 = S \cdot \frac{L}{100 - L} \]
- When considering the end correction, let’s denote the end correction as \( X \). The effective length when the end correction is applied would be \( L + X \). Thus, the expression for \( R_2 \) becomes:
\[ R_2 = S \cdot \frac{L + X}{100 - (L + X)} \]

3. **Analyzing the Two Cases**:
- **First Half**: If \( L \) is in the first half of the potentiometer (e.g., \( L = 30 \)):
\[ R_1 = S \cdot \frac{30}{100 - 30} = S \cdot \frac{30}{70} \]
\[ R_2 = S \cdot \frac{30 + X}{100 - (30 + X)} = S \cdot \frac{30 + X}{70 - X} \]
Here, \( R_2 \) will be greater than \( R_1 \) if \( X \) is positive.

- **Second Half**: If \( L \) is in the second half of the potentiometer (e.g., \( L = 70 \)):
\[ R_1 = S \cdot \frac{70}{100 - 70} = S \cdot \frac{70}{30} \]
\[ R_2 = S \cdot \frac{70 + X}{100 - (70 + X)} = S \cdot \frac{70 + X}{30 - X} \]
Here, \( R_1 \) will be greater than \( R_2 \) if \( X \) is positive.

4. **Conclusion**:
- From the analysis, we can conclude:
- In the first half of the potentiometer, \( R_2 > R_1 \).
- In the second half of the potentiometer, \( R_1 > R_2 \).

### Final Answer:
Thus, the relationship between \( R_1 \) and \( R_2 \) depends on the position of \( L \) along the potentiometer wire.