In a potentiometer arrangement , a cell of EMF 2 V gives a balance point at 40 cm length of the wire . If this cell is replaced by another cell and the balance point shifts to 60 cm , then the EMF of the second cell is

Options

A. 1 V
B. 2 V
C. 3 V
D. 4 V

Answer

To solve the problem, we will use the principle of the potentiometer, which states that the potential difference across a length of wire is directly proportional to that length.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- EMF of the first cell (E1) = 2 V
- Length of the wire for the first cell (L1) = 40 cm
- Length of the wire for the second cell (L2) = 60 cm

2. **Use the Relationship of EMF and Length:**
According to the potentiometer principle:
\[ \frac{E1}{E2} = \frac{L1}{L2} \]
Where E2 is the EMF of the second cell.

3. **Substituting the Known Values:**
Substitute the known values into the equation:
\[ \frac{2 \, \text{V}}{E2} = \frac{40 \, \text{cm}}{60 \, \text{cm}} \]

4. **Simplifying the Ratio:**
Simplifying the right side:
\[ \frac{40}{60} = \frac{2}{3} \]
So, the equation becomes:
\[ \frac{2}{E2} = \frac{2}{3} \]

5. **Cross-Multiplying to Solve for E2:**
Cross-multiplying gives:
\[ 2 \cdot 3 = 2 \cdot E2 \]
\[ 6 = 2 \cdot E2 \]

6. **Dividing Both Sides by 2:**
\[ E2 = \frac{6}{2} = 3 \, \text{V} \]

### Final Answer:
The EMF of the second cell (E2) is **3 V**.