A 10 m long potentiometer wire is connected to a battery having a steady voltage . A leclanche cell is balanced at 4 m length of the wire . If the length is kept the same, but its cross - section is doubled the null point will be obtained at .

Options

A. 8 m
B. 4 m
C. 2m
D. None of these

Answer

To solve the problem, we need to analyze how the change in the cross-section of the potentiometer wire affects the balance length when measuring the EMF of a Leclanché cell.

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- We have a potentiometer wire of length \( L = 10 \, \text{m} \).
- A Leclanché cell is balanced at a length \( L_B = 4 \, \text{m} \).

2. **Potentiometer Principle**:
- The balance length \( L_B \) is related to the EMF of the cell (\( \epsilon_1 \)) and the total EMF of the potentiometer circuit (\( E \)) by the formula:
\[ \epsilon_1 = \frac{L_B}{L} \cdot E \]
- Here, \( L \) is the total length of the wire.

3. **Effect of Doubling the Cross-Section**:
- When the cross-section of the wire is doubled, the resistance of the wire changes. The resistance \( R \) of a wire is given by:
\[ R = \frac{\rho L}{A} \]
- If the cross-section \( A \) is doubled, the new resistance \( R' \) becomes:
\[ R' = \frac{\rho L}{2A} \]
- This means the resistance is halved.

4. **Current in the Circuit**:
- The current \( I \) through the wire can be expressed as:
\[ I = \frac{E}{R} \]
- With the new resistance, the current becomes:
\[ I' = \frac{E}{R'} = \frac{E}{\frac{\rho L}{2A}} = \frac{2EA}{\rho L} \]

5. **Potential Drop Across the Balance Length**:
- The potential difference across the balance length \( L_B \) is given by:
\[ V_B = I \cdot R_B \]
- The resistance of the balance length \( R_B \) is:
\[ R_B = \frac{\rho L_B}{A} \]
- Thus, the potential drop across \( L_B \) is:
\[ V_B = I \cdot R_B = \left(\frac{E}{R}\right) \cdot \left(\frac{\rho L_B}{A}\right) \]

6. **Finding the New Balance Length**:
- Since the EMF of the Leclanché cell remains unchanged, we can set up the equation for the new balance length \( L_B' \):
\[ \epsilon_1 = \frac{L_B'}{L} \cdot E \]
- Since \( \epsilon_1 \) and \( E \) remain constant, and the total length \( L \) is unchanged, we conclude that the new balance length \( L_B' \) will still equal the original balance length \( L_B \):
\[ L_B' = L_B = 4 \, \text{m} \]

### Conclusion:
The null point will still be obtained at **4 m** even after doubling the cross-section of the wire.