A potentiometer circuit has been setup for finding. The internal resistance of a given cell. The main battery used a negligible internal resistance. The potentiometer wire itsefl is `4 m` long. When the resistance, `R`, connected across the given cell, has value of
(i) Infinity `(ii)9.5 Omega`,
the 'balancing length' , on the potentiometer wire are found to be `3 m` and `2.85 m`, respectively.
The value of internal resistance of the cell is

Options

A. `0.25 Omega`
B. `9.5 Omega`
C. `0.5 Omega`
D. `0.75 Omega`

Answer

To find the internal resistance of the given cell using the potentiometer, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Length of the potentiometer wire, \( L_{\text{total}} = 4 \, \text{m} \)
- Balancing length for \( R = \infty \), \( L_1 = 3 \, \text{m} \)
- Balancing length for \( R = 9.5 \, \Omega \), \( L_2 = 2.85 \, \text{m} \)
- External resistance, \( R = 9.5 \, \Omega \)

2. **Use the Potentiometer Formula**:
The formula to find the internal resistance \( r \) of the cell is given by:
\[ r = \left( \frac{L_1}{L_2} - 1 \right) R \]

3. **Substitute the Values**:
Substitute the known values into the formula:
\[ r = \left( \frac{3}{2.85} - 1 \right) \times 9.5 \]

4. **Calculate \( \frac{L_1}{L_2} \)**:
First, calculate \( \frac{3}{2.85} \):
\[ \frac{3}{2.85} \approx 1.0526 \]

5. **Subtract 1**:
Now, subtract 1 from the result:
\[ 1.0526 - 1 = 0.0526 \]

6. **Multiply by the External Resistance**:
Multiply the result by the external resistance \( R \):
\[ r = 0.0526 \times 9.5 \approx 0.5 \, \Omega \]

7. **Final Result**:
The internal resistance of the cell is approximately \( 0.5 \, \Omega \).

### Final Answer:
The value of the internal resistance of the cell is \( 0.5 \, \Omega \).

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