A potentiometer with a cell of EMF `2 V` and internal resistance `0.4 Omega` is used across the wire `AB`. A standard cadmium cell of EMF `1.02 V` gives a balance point at `66 cm` length of wire. The standard cell is then replaced by a cell of unknows EMF `e` (internal resistance `r`), and the balance. Point found similarly turns out to be `88 cm` length of the wire. The length of potentiometer wire `AB` is `1 m`.
The value of `e` is

Options

A. `1.36 V`
B. `2.63 V`
C. `1.83 V`
D. none

Explanation

To find the unknown EMF \( e \) of the cell using the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Known Values:**
- EMF of the standard cadmium cell, \( e_1 = 1.02 \, \text{V} \)
- Balance point for the standard cell, \( L_1 = 66 \, \text{cm} = 0.66 \, \text{m} \)
- Balance point for the unknown cell, \( L_2 = 88 \, \text{cm} = 0.88 \, \text{m} \)

2. **Use the Potentiometer Principle:**
The relationship between the EMF of the cells and the lengths of the wire at which they balance is given by:
\[ \frac{e_1}{L_1} = \frac{e}{L_2} \]
Rearranging this equation gives:
\[ e = e_1 \cdot \frac{L_2}{L_1} \]

3. **Substitute the Known Values:**
Substitute \( e_1 \), \( L_1 \), and \( L_2 \) into the equation:
\[ e = 1.02 \, \text{V} \cdot \frac{0.88 \, \text{m}}{0.66 \, \text{m}} \]

4. **Calculate the Unknown EMF:**
First, calculate the fraction:
\[ \frac{0.88}{0.66} = 1.3333 \]
Now, multiply by \( 1.02 \, \text{V} \):
\[ e = 1.02 \, \text{V} \cdot 1.3333 \approx 1.36 \, \text{V} \]

5. **Final Result:**
Thus, the value of the unknown EMF \( e \) is approximately:
\[ e \approx 1.36 \, \text{V} \]

### Final Answer:
The value of \( e \) is \( 1.36 \, \text{V} \).

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