To find the internal resistance of the given cell using the potentiometer method, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Setup**:
- We have a potentiometer circuit with a main battery of EMF \( E = 2.0 \, \text{V} \) and negligible internal resistance.
- The potentiometer wire length is \( L = 4 \, \text{m} \).
- We have two scenarios for the external resistance \( R \):
- Case (i): \( R = \infty \) (open circuit) with balancing length \( L_1 = 3 \, \text{m} \).
- Case (ii): \( R = 9.5 \, \Omega \) with balancing length \( L_2 = 2.85 \, \text{m} \).
2. **Using the Potentiometer Principle**:
- The potential difference across the balancing length of the wire is proportional to the length of the wire.
- The potential difference across the wire when \( R = \infty \) is given by:
\[
V_1 = \frac{L_1}{L} \cdot E = \frac{3}{4} \cdot 2.0 \, \text{V} = 1.5 \, \text{V}
\]
- When \( R = 9.5 \, \Omega \), the potential difference is:
\[
V_2 = \frac{L_2}{L} \cdot E = \frac{2.85}{4} \cdot 2.0 \, \text{V} = 1.425 \, \text{V}
\]
3. **Applying Ohm's Law**:
- For the second case (with \( R = 9.5 \, \Omega \)), we can write:
\[
V_2 = I(R + r)
\]
- Where \( r \) is the internal resistance of the cell and \( I \) is the current flowing through the circuit.
4. **Relating Current to Balancing Lengths**:
- The current \( I \) can also be expressed using the first case:
\[
I = \frac{V_1}{R} = \frac{1.5 \, \text{V}}{R} \quad \text{(for } R = \infty\text{)}
\]
- And for the second case:
\[
I = \frac{V_2}{R + r} = \frac{1.425 \, \text{V}}{9.5 + r}
\]
5. **Setting the Equations Equal**:
- Since both expressions represent the same current \( I \), we can set them equal:
\[
\frac{1.5}{\infty} = \frac{1.425}{9.5 + r}
\]
- This simplifies to:
\[
1.425 = (9.5 + r) \cdot 0
\]
- However, we can also derive \( r \) from the relationship between the lengths:
\[
r = \frac{L_1 - L_2}{L_2} \cdot R
\]
6. **Substituting Values**:
- Substitute \( L_1 = 3 \, \text{m} \), \( L_2 = 2.85 \, \text{m} \), and \( R = 9.5 \, \Omega \):
\[
r = \frac{3 - 2.85}{2.85} \cdot 9.5
\]
\[
r = \frac{0.15}{2.85} \cdot 9.5
\]
\[
r = 0.05263 \cdot 9.5 \approx 0.5 \, \Omega
\]
### Final Answer:
The internal resistance \( r \) of the cell is approximately \( 0.5 \, \Omega \).
