In a potentiometer a cell of emf 1.5 V gives a balanced point at 32 cm length of the wire. If the cell is replaced by another cell then the balance point shifts to 65 cm. Then the emf of second cell is

Options

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

Answer

To solve the problem step by step, we will use the concept of a potentiometer and the relationship between the electromotive force (emf) of the cells and the lengths at which they balance.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a potentiometer wire where a cell of emf 1.5 V gives a balanced point at 32 cm. When this cell is replaced with another cell, the balance point shifts to 65 cm. We need to find the emf of the second cell.

2. **Define Variables**:
- Let \( E_1 = 1.5 \, \text{V} \) (emf of the first cell)
- Let \( L_1 = 32 \, \text{cm} \) (length at which the first cell balances)
- Let \( E_2 \) be the emf of the second cell (unknown)
- Let \( L_2 = 65 \, \text{cm} \) (length at which the second cell balances)

3. **Calculate the Potential Gradient**:
The potential gradient (k) can be defined as the ratio of the emf to the length of the wire:
\[ k = \frac{E_1}{L_1} = \frac{1.5 \, \text{V}}{32 \, \text{cm}} \]

4. **Set Up the Equation for the Second Cell**:
For the second cell, we can express the potential gradient as:
\[ k = \frac{E_2}{L_2} = \frac{E_2}{65 \, \text{cm}} \]

5. **Equate the Two Expressions for Potential Gradient**:
Since both expressions represent the same potential gradient, we can set them equal to each other:
\[ \frac{1.5 \, \text{V}}{32 \, \text{cm}} = \frac{E_2}{65 \, \text{cm}} \]

6. **Cross-Multiply to Solve for \( E_2 \)**:
Cross-multiplying gives us:
\[ 1.5 \times 65 = E_2 \times 32 \]
\[ E_2 = \frac{1.5 \times 65}{32} \]

7. **Calculate \( E_2 \)**:
Now, calculate \( E_2 \):
\[ E_2 = \frac{97.5}{32} \approx 3.05 \, \text{V} \]

### Final Answer:
The emf of the second cell \( E_2 \) is approximately **3.05 V**.