In potentiometer experiment, a cell of emf `1.25 V` gives balancing length of 30 cm. If the cell is replaced by another cell, then balancing length is found to be 40 cm. What is the emf of second cell ?

Options

A. `~=1.5V`
B. `~=1.67 V`
C. `~=1.47 V`
D. `~=1.37 V`

Answer

To find the emf of the second cell in the potentiometer experiment, we can use the relationship between the emf of the cell and the balancing length on the potentiometer. The emf (E) is directly proportional to the balancing length (L).

### Step-by-Step Solution:

1. **Understand the relationship**:
The relationship can be expressed as:
\[ \frac{E_1}{E_2} = \frac{L_1}{L_2} \]
where:
- \(E_1\) = emf of the first cell = 1.25 V
- \(E_2\) = emf of the second cell (unknown)
- \(L_1\) = balancing length for the first cell = 30 cm
- \(L_2\) = balancing length for the second cell = 40 cm

2. **Set up the equation**:
Plugging in the known values into the equation gives:
\[ \frac{1.25}{E_2} = \frac{30}{40} \]

3. **Simplify the ratio**:
The ratio \(\frac{30}{40}\) simplifies to \(\frac{3}{4}\). Therefore, we can rewrite the equation as:
\[ \frac{1.25}{E_2} = \frac{3}{4} \]

4. **Cross-multiply to solve for \(E_2\)**:
Cross-multiplying gives:
\[ 1.25 \cdot 4 = 3 \cdot E_2 \]
This simplifies to:
\[ 5 = 3E_2 \]

5. **Isolate \(E_2\)**:
Dividing both sides by 3 gives:
\[ E_2 = \frac{5}{3} \]

6. **Convert to decimal**:
Calculating \(\frac{5}{3}\) gives approximately:
\[ E_2 \approx 1.67 \text{ V} \]

### Final Answer:
The emf of the second cell is approximately **1.67 V**.