One ideal cell is used to drive current in a metre bridge circuit. In an experiment x is the balancing length obtained. If we replace the wire of the metre bridge with another wire of material having more rsistivity and lesser diameter, the new balancing length will be

Options

A. less than x
B. more than x
C. equal to x
D. more or less that depends on the ratio of resistances used in the left and right gap of metre bridge.

Explanation

To solve the problem, we need to analyze how the balancing length in a meter bridge circuit is affected when we replace the wire with another wire that has higher resistivity and a smaller diameter.

### Step-by-Step Solution:

1. **Understanding the Meter Bridge**:
The meter bridge consists of a wire of uniform cross-section and length 1 meter. When a current is passed through it, a galvanometer is used to find the balancing point (null point) where the potential difference across the two segments of the wire is equal.

2. **Initial Balancing Condition**:
Let the initial balancing length be \( x \). According to the principle of the meter bridge, we have:
\[ \frac{x}{L - x} = \frac{R_1}{R_2} \]
where \( L = 100 \, \text{cm} \) (the total length of the wire), \( R_1 \) is the resistance in one arm, and \( R_2 \) is the resistance in the other arm.

3. **Replacing the Wire**:
When we replace the wire with another wire that has higher resistivity (\( \rho' > \rho \)) and a smaller diameter, the resistance of the new wire will increase. The resistance of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where \( A \) is the cross-sectional area of the wire. A smaller diameter means a smaller area, which increases the resistance.

4. **Effect on Balancing Length**:
The new resistance \( R' \) can be expressed as:
\[ R' = \rho' \frac{L}{A'} \]
Since \( R_1 \) and \( R_2 \) are constant (not changing), the new balancing length \( x' \) will still satisfy the meter bridge equation:
\[ \frac{x'}{L - x'} = \frac{R_1}{R_2} \]
This implies that the ratio \( \frac{x'}{L - x'} \) remains the same as \( \frac{x}{L - x} \).

5. **Conclusion**:
Since the ratio of the resistances \( R_1 \) and \( R_2 \) does not change, the new balancing length \( x' \) will remain equal to the original balancing length \( x \). Thus, we conclude:
\[ x' = x \]

### Final Answer:
The new balancing length will be equal to \( x \).

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