It is observed in a potentiometer experiment that no current passes through the galvanometer, when the terminals of the potentiometer, wire. On shunting the cell by a `2Omega` resistance, the balancing length is reduced to half. The internal resistance of the cell is:-

Options

A. `4Omega`
B. `2Omega`
C. `9Omega`
D. `18Omega`

Answer

To solve the problem, we need to determine the internal resistance of the cell based on the information provided in the potentiometer experiment.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- In the potentiometer experiment, the initial balancing length is denoted as \( L_1 \).
- When a \( 2 \, \Omega \) resistance is shunted across the cell, the new balancing length becomes \( L_2 = \frac{L_1}{2} \).

2. **Using the Potentiometer Formula**:
- The formula relating the internal resistance \( r \) of the cell, the external resistance \( R \), and the lengths \( L_1 \) and \( L_2 \) is given by:
\[ r = \left( \frac{L_1}{L_2} - 1 \right) R \]

3. **Substituting Known Values**:
- From the problem, we know:
- \( R = 2 \, \Omega \)
- \( L_2 = \frac{L_1}{2} \)
- Plugging \( L_2 \) into the formula:
\[ r = \left( \frac{L_1}{\frac{L_1}{2}} - 1 \right) \cdot 2 \]

4. **Simplifying the Expression**:
- Simplifying \( \frac{L_1}{\frac{L_1}{2}} \) gives:
\[ \frac{L_1}{\frac{L_1}{2}} = 2 \]
- Thus, the equation for \( r \) becomes:
\[ r = (2 - 1) \cdot 2 \]

5. **Calculating the Internal Resistance**:
- Therefore:
\[ r = 1 \cdot 2 = 2 \, \Omega \]

6. **Final Answer**:
- The internal resistance of the cell is \( 2 \, \Omega \).