Two cells of E.M.F. `E_(1)` and `E_(2)(E_(2)gtE_(1))` are connected in series in the secondary circuit of a potentiometer experiment for determination of E.M.F. The balancing length is found to be 825 cm. Now when the terminals to cell `E_(1)` are reversed, then the balancing length is found to be 225 cm. The ratio of `E_(1)` and `E_(2)` is

Options

A. `2:3`
B. `4:7`
C. `7:4`
D. none of these

Answer

To solve the problem, we need to analyze the two scenarios given in the potentiometer experiment involving two cells of EMF \( E_1 \) and \( E_2 \).

### Step-by-Step Solution:

1. **Understanding the First Scenario:**
In the first case, both cells \( E_1 \) and \( E_2 \) are connected in series. The total EMF in this case is:
\[ E_{\text{total}} = E_1 + E_2 \]
The balancing length \( L_1 \) is given as 825 cm. According to the potentiometer principle, the balancing length is proportional to the EMF. Therefore, we can write:
\[ k (E_1 + E_2) = 825 \quad \text{(1)} \]
where \( k \) is a constant of proportionality.

2. **Understanding the Second Scenario:**
In the second case, the terminals of cell \( E_1 \) are reversed. This means that the effective EMF in the circuit becomes:
\[ E_{\text{effective}} = E_2 - E_1 \]
The balancing length \( L_2 \) is given as 225 cm. Thus, we can write:
\[ k (E_2 - E_1) = 225 \quad \text{(2)} \]

3. **Setting Up the Equations:**
From equations (1) and (2), we have:
\[ E_1 + E_2 = \frac{825}{k} \quad \text{(3)} \]
\[ E_2 - E_1 = \frac{225}{k} \quad \text{(4)} \]

4. **Adding and Subtracting the Equations:**
Now, we can add equations (3) and (4) to eliminate \( E_1 \):
\[ (E_1 + E_2) + (E_2 - E_1) = \frac{825}{k} + \frac{225}{k} \]
Simplifying this gives:
\[ 2E_2 = \frac{1050}{k} \quad \Rightarrow \quad E_2 = \frac{525}{k} \quad \text{(5)} \]

Next, we subtract equation (4) from equation (3):
\[ (E_1 + E_2) - (E_2 - E_1) = \frac{825}{k} - \frac{225}{k} \]
Simplifying this gives:
\[ 2E_1 = \frac{600}{k} \quad \Rightarrow \quad E_1 = \frac{300}{k} \quad \text{(6)} \]

5. **Finding the Ratio \( \frac{E_1}{E_2} \):**
Now we can find the ratio of \( E_1 \) to \( E_2 \) using equations (5) and (6):
\[ \frac{E_1}{E_2} = \frac{\frac{300}{k}}{\frac{525}{k}} = \frac{300}{525} = \frac{4}{7} \]

### Final Result:
Thus, the ratio of \( E_1 \) to \( E_2 \) is:
\[ \frac{E_1}{E_2} = \frac{4}{7} \]