In an experiment with potentiometer to measure the internal resistance of a cell, when the cell is shunted by `5 Omega`, the null point is obtained at 2m. when cell is shunted by `20 Omega` the null point is obtained at 3m.The internal resistance of cell is

Options

A. `2 Omega `
B. `4 Omega `
C. `6 Omega`
D. `8 Omega `

Answer

To solve the problem of finding the internal resistance of the cell using the potentiometer method, we will follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( E \) be the emf of the cell.
- Let \( r \) be the internal resistance of the cell.
- Let \( k \) be the potential gradient of the potentiometer wire.

2. **Set Up the First Condition (5 Ω Shunt)**:
- When the cell is shunted by \( 5 \, \Omega \), the null point is obtained at \( 2 \, m \).
- According to the potentiometer principle, we can write:
\[ \frac{E}{r + 5} \times 5 = k \times 2 \]
- Rearranging gives:
\[ \frac{E}{r + 5} = \frac{2k}{5} \quad \text{(Equation 1)} \]

3. **Set Up the Second Condition (20 Ω Shunt)**:
- When the cell is shunted by \( 20 \, \Omega \), the null point is obtained at \( 3 \, m \).
- Similarly, we can write:
\[ \frac{E}{r + 20} \times 20 = k \times 3 \]
- Rearranging gives:
\[ \frac{E}{r + 20} = \frac{3k}{20} \quad \text{(Equation 2)} \]

4. **Equate the Two Conditions**:
- From Equation 1 and Equation 2, we can equate:
\[ \frac{E}{r + 5} \cdot \frac{20}{3} = \frac{E}{r + 20} \cdot \frac{5}{2} \]
- This simplifies to:
\[ \frac{r + 20}{r + 5} = \frac{2}{3} \cdot \frac{20}{5} \]
- Simplifying further gives:
\[ \frac{r + 20}{r + 5} = \frac{8}{3} \]

5. **Cross-Multiply and Solve for \( r \)**:
- Cross-multiplying gives:
\[ 3(r + 20) = 8(r + 5) \]
- Expanding both sides:
\[ 3r + 60 = 8r + 40 \]
- Rearranging gives:
\[ 60 - 40 = 8r - 3r \]
- Thus:
\[ 5r = 20 \quad \Rightarrow \quad r = 4 \, \Omega \]

6. **Conclusion**:
- The internal resistance of the cell is \( r = 4 \, \Omega \).