To determine the specific resistance (ρ) of a conducting wire using the meter bridge experiment, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Setup**:
- We have a meter bridge where a standard 6 Ω resistance is connected on the left gap and a conducting wire of length 10 cm and diameter 1 mm is connected on the right gap.
- The balance point is obtained at 60 cm on the meter bridge.
2. **Using the Meter Bridge Formula**:
- The balance condition of the meter bridge is given by the formula:
\[
\frac{R_1}{R} = \frac{L_1}{L}
\]
- Where:
- \( R_1 \) is the resistance on the left (6 Ω),
- \( R \) is the resistance of the wire,
- \( L_1 \) is the length of the bridge wire on the left (60 cm),
- \( L \) is the total length of the bridge wire (100 cm).
3. **Calculating the Resistance of the Wire**:
- From the balance condition, we can rearrange the formula to find \( R \):
\[
R = \frac{L}{L_1} \times R_1
\]
- Substituting the values:
\[
R = \frac{100 \, \text{cm}}{60 \, \text{cm}} \times 6 \, \Omega = \frac{100}{60} \times 6 = \frac{10}{6} \times 6 = 10 \, \Omega
\]
4. **Finding the Specific Resistance**:
- The resistance of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
- Where:
- \( \rho \) is the specific resistance,
- \( L \) is the length of the wire (10 cm = 0.1 m),
- \( A \) is the cross-sectional area of the wire.
- The cross-sectional area \( A \) for a wire with diameter \( d \) is calculated as:
\[
A = \frac{\pi d^2}{4}
\]
- Given that the diameter \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \):
\[
A = \frac{\pi (1 \times 10^{-3})^2}{4} = \frac{\pi \times 10^{-6}}{4} = \frac{3.14 \times 10^{-6}}{4} \approx 7.85 \times 10^{-7} \, \text{m}^2
\]
5. **Substituting Values to Find \( \rho \)**:
- Rearranging the resistance formula to find \( \rho \):
\[
\rho = R \frac{A}{L}
\]
- Substituting the known values:
\[
\rho = 10 \, \Omega \times \frac{7.85 \times 10^{-7} \, \text{m}^2}{0.1 \, \text{m}} = 10 \times 7.85 \times 10^{-6} \, \Omega \cdot \text{m} = 7.85 \times 10^{-5} \, \Omega \cdot \text{m}
\]
6. **Final Result**:
- The specific resistance of the material of the given wire is approximately:
\[
\rho \approx 7.85 \times 10^{-5} \, \Omega \cdot \text{m}
\]
