In a potentiometer a cell of emf 1.5V gives a balanced point 32cm length of the cell is replaced by another cell then the balance point shifts to 65.0cm the emf of second cell is

Options

A. 3.05V
B. 2.05V
C. 4.05V
D. 6.05V

Explanation

To find the EMF of the second cell in the potentiometer setup, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: In a potentiometer, the EMF (E) of a cell is directly proportional to the balancing length (L) on the wire. This can be expressed as:
\[ E_1 \propto L_1 \quad \text{and} \quad E_2 \propto L_2 \]

2. **Set Up the Ratio**: Since the EMFs are proportional to their respective lengths, we can set up a ratio:
\[ \frac{E_1}{E_2} = \frac{L_1}{L_2} \]

3. **Substitute Known Values**: We know the following values:
- \( E_1 = 1.5 \, \text{V} \) (EMF of the first cell)
- \( L_1 = 32 \, \text{cm} \) (balancing length for the first cell)
- \( L_2 = 65 \, \text{cm} \) (balancing length for the second cell)

Substituting these values into the ratio gives:
\[ \frac{1.5}{E_2} = \frac{32}{65} \]

4. **Cross-Multiply to Solve for \( E_2 \)**: Rearranging the equation to find \( E_2 \):
\[ 1.5 \times 65 = E_2 \times 32 \]
\[ E_2 = \frac{1.5 \times 65}{32} \]

5. **Calculate \( E_2 \)**: Now, perform the calculation:
\[ E_2 = \frac{97.5}{32} \approx 3.04 \, \text{V} \]

6. **Conclusion**: Therefore, the EMF of the second cell is:
\[ E_2 \approx 3.04 \, \text{V} \]

### Final Answer:
The EMF of the second cell is **3.04 V**.