There is a definite potential difference between the two ends of a potentiometer. Two cells are connected in such a way that first time help each other, and second time they oppose each other. They are balanced on the potentiometer wire at 120cm and 60cm length respectively. compare the electromotive force of the cells. potentiometer wire at 120cm and 60cm length respectively. Compare the electromotive force of the cells.

Options

A. `53^(@)`
B. `37^(@)`
C. `45^(@)`
D. `30^(@)`

Explanation

To solve the problem of comparing the electromotive forces (emf) of two cells connected in a potentiometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a potentiometer wire with a potential difference across it.
- Two cells are connected in two different configurations:
- In the first configuration, they help each other (E1 + E2).
- In the second configuration, they oppose each other (E1 - E2).

2. **Establish the Relationships**:
- The potential gradient (k) of the potentiometer wire relates the emf (E) and the length (l) of the wire:
\[ E = k \cdot l \]

3. **Write the Equations**:
- For the first configuration (helping each other), the length of the wire is 120 cm:
\[ E1 + E2 = k \cdot 120 \]
- For the second configuration (opposing each other), the length of the wire is 60 cm:
\[ E1 - E2 = k \cdot 60 \]

4. **Label the Equations**:
- Let’s label the first equation as (1) and the second equation as (2):
- (1) \( E1 + E2 = k \cdot 120 \)
- (2) \( E1 - E2 = k \cdot 60 \)

5. **Divide the Equations**:
- Divide equation (1) by equation (2):
\[ \frac{E1 + E2}{E1 - E2} = \frac{k \cdot 120}{k \cdot 60} \]
- The k cancels out:
\[ \frac{E1 + E2}{E1 - E2} = \frac{120}{60} = 2 \]

6. **Cross-Multiply**:
- Cross-multiplying gives:
\[ E1 + E2 = 2(E1 - E2) \]

7. **Expand and Rearrange**:
- Expanding the right side:
\[ E1 + E2 = 2E1 - 2E2 \]
- Rearranging gives:
\[ E1 + E2 + 2E2 = 2E1 \]
\[ 3E2 = E1 \]

8. **Final Comparison**:
- From the equation \( 3E2 = E1 \), we can express the ratio of the emfs:
\[ \frac{E1}{E2} = \frac{3}{1} \]

### Conclusion:
The ratio of the electromotive forces of the two cells is \( E1 : E2 = 3 : 1 \).

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