A copper strip is introduced in the left gap and a resistance `R-0.4 Omega` is placed in the right gap of a meter bridge experiment. The balance points before and after interchanging the copper strip and the resistance R are 30 cm and 60 cm respectively. The resistance per unit length of the bridge wire is

Options

A. `4 Omega.m^(-1)`
B. `3 Omega m^(-1)`
C. `(3)/(4) Omega m^(-1)`
D. `(4)/(3) Omega m^(-1)`

Answer

To solve the problem regarding the meter bridge experiment with a copper strip and a resistance \( R = 0.4 \, \Omega \), we need to analyze the balance points before and after interchanging the copper strip and the resistance.

### Step-by-Step Solution:

1. **Understanding the Meter Bridge Setup**:
In a meter bridge, the balance point \( L \) is given by the formula:
\[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \]
where \( R_1 \) is the resistance in one gap and \( R_2 \) is the resistance in the other gap, and \( L_1 \) and \( L_2 \) are the lengths from the ends of the bridge wire to the balance point.

2. **Identifying the Given Values**:
- When the copper strip is in the left gap, the balance point is at \( 30 \, \text{cm} \).
- When the copper strip is interchanged with the resistance \( R \), the balance point shifts to \( 60 \, \text{cm} \).
- The resistance \( R = 0.4 \, \Omega \).

3. **Setting Up the Equations**:
- Let the resistance of the copper strip be \( R_c \).
- For the first balance point (copper strip in the left gap):
\[ \frac{R_c}{0.4} = \frac{30}{70} \quad \text{(since total length is 100 cm)} \]
- Simplifying this gives:
\[ R_c = 0.4 \times \frac{30}{70} = 0.4 \times \frac{3}{7} = \frac{1.2}{7} \approx 0.1714 \, \Omega \]

4. **For the Second Balance Point**:
- Now, when the copper strip is in the right gap:
\[ \frac{0.4}{R_c} = \frac{60}{40} \]
- Simplifying this gives:
\[ 0.4 \times \frac{40}{60} = R_c \quad \Rightarrow \quad R_c = 0.4 \times \frac{2}{3} = \frac{0.8}{3} \approx 0.2667 \, \Omega \]

5. **Calculating the Resistance per Unit Length**:
- The average resistance of the copper strip from both calculations can be used to find the resistance per unit length of the bridge wire.
- Let \( r \) be the resistance per unit length of the wire.
- The total resistance of the bridge wire is \( 100r \).
- The balance point equations can be used to express \( r \):
\[ R_c = r \times L_c \quad \text{(where \( L_c \) is the length of the copper strip)} \]
- Using the average \( R_c \) value, we can solve for \( r \).

6. **Final Calculation**:
- The average \( R_c \) from both balance points is:
\[ R_c = \frac{0.1714 + 0.2667}{2} \approx 0.21905 \, \Omega \]
- Assuming the length of the copper strip is \( L_c = 30 \, \text{cm} \):
\[ r = \frac{R_c}{L_c} = \frac{0.21905}{0.30} \approx 0.7302 \, \Omega/m \]

### Conclusion:
The resistance per unit length of the bridge wire is approximately \( 0.7302 \, \Omega/m \).