The balancing lengths of potentiometer wire are 800 cm and 600cm when two cells of emf's `E_(1) and E_(2)` are connected in the secondary circuit first in series and then terminals of one cell is reversed,`E_(1)/E_(2)`is equal to

Options

A. `1//11`
B. `14//11`
C. `7//1`
D. `4//3`

Answer

To solve the problem, we need to find the ratio of the EMFs \( \frac{E_1}{E_2} \) based on the balancing lengths of a potentiometer wire when two cells are connected in different configurations.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two cells with EMFs \( E_1 \) and \( E_2 \). The balancing lengths of the potentiometer wire are given as 800 cm when the cells are connected in series and 600 cm when the terminals of one cell are reversed.

2. **Setting Up the Equations**:
- When the cells are connected in series, the total EMF is \( E_1 + E_2 \). The balancing length is proportional to the EMF, so we can write:
\[ E_1 + E_2 = k \cdot 800 \quad \text{(1)} \]
where \( k \) is a constant of proportionality.

- When one terminal of \( E_2 \) is reversed, the effective EMF becomes \( E_1 - E_2 \). The balancing length in this case is:
\[ E_1 - E_2 = k \cdot 600 \quad \text{(2)} \]

3. **Dividing the Equations**:
We can divide equation (1) by equation (2):
\[ \frac{E_1 + E_2}{E_1 - E_2} = \frac{800}{600} = \frac{4}{3} \]

4. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ 3(E_1 + E_2) = 4(E_1 - E_2) \]

5. **Expanding and Rearranging**:
Expanding both sides:
\[ 3E_1 + 3E_2 = 4E_1 - 4E_2 \]
Rearranging gives:
\[ 3E_2 + 4E_2 = 4E_1 - 3E_1 \]
Simplifying this results in:
\[ 7E_2 = E_1 \]

6. **Finding the Ratio**:
Thus, we can express the ratio of \( E_1 \) to \( E_2 \):
\[ \frac{E_1}{E_2} = \frac{7}{1} \]

### Final Answer:
The ratio \( \frac{E_1}{E_2} \) is \( 7:1 \).