In a potentiometer experiment, two cells connected in series get balanced at 9 m length of the wire. Now, if the connections of terminals of cell of lower emf are reversed, then the balancing length is obtained at 3 m. The ratio of emf's of two cells will be

Options

A. `1:2`
B. `2:1`
C. `1:4`
D. `4:1`

Answer

To solve the problem, we need to analyze the potentiometer experiment and the information given about the balancing lengths when two cells are connected in series and when the terminals of one cell are reversed.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the EMFs of the two cells be \( E_1 \) and \( E_2 \).
- When the cells are connected in series, the total EMF \( E \) is given by:
\[ E = E_1 + E_2 \]
- The balancing length for this configuration is given as \( L_1 = 9 \, \text{m} \).

2. **Balancing Condition**:
- The potential difference across the potentiometer wire is proportional to the length of the wire. Hence, we can write:
\[ E \propto L_1 \implies E = k \cdot L_1 \]
- Where \( k \) is a constant of proportionality.

3. **Reversing the Terminal of the Lower EMF Cell**:
- When the terminals of the cell with the lower EMF (\( E_2 \)) are reversed, the effective EMF becomes:
\[ E' = E_1 - E_2 \]
- The new balancing length is given as \( L_2 = 3 \, \text{m} \).

4. **Balancing Condition for the Reversed Configuration**:
- Similarly, we can write:
\[ E' \propto L_2 \implies E' = k \cdot L_2 \]

5. **Setting Up the Equations**:
- From the first balancing condition:
\[ E = k \cdot 9 \]
- From the second balancing condition:
\[ E' = k \cdot 3 \]

6. **Relating the EMFs**:
- We can set up the ratio of the two EMFs:
\[ \frac{E}{E'} = \frac{k \cdot 9}{k \cdot 3} = \frac{9}{3} = 3 \]
- Therefore, we have:
\[ E = 3E' \]

7. **Substituting for \( E' \)**:
- Substitute \( E' = E_1 - E_2 \) into the equation:
\[ E = 3(E_1 - E_2) \]
- Now substituting \( E = E_1 + E_2 \):
\[ E_1 + E_2 = 3(E_1 - E_2) \]

8. **Expanding and Rearranging**:
- Expanding the right-hand side:
\[ E_1 + E_2 = 3E_1 - 3E_2 \]
- Rearranging gives:
\[ E_1 + E_2 + 3E_2 = 3E_1 \]
\[ E_1 + 4E_2 = 3E_1 \]
\[ 4E_2 = 3E_1 - E_1 \]
\[ 4E_2 = 2E_1 \]
- Dividing both sides by 2:
\[ 2E_2 = E_1 \]

9. **Finding the Ratio**:
- Thus, the ratio of the EMFs is:
\[ \frac{E_1}{E_2} = \frac{2}{1} \]

### Final Answer:
The ratio of the EMFs of the two cells is \( 2:1 \).