To find the specific resistivity of the unknown wire in the meter bridge experiment, we can follow these steps:
### Step-by-Step Solution:
1. **Identify Given Values:**
- Resistance \( S = 300 \, \Omega \)
- Balanced length from end \( A \) (let's call it \( L_1 \)) = 25.0 cm
- Length of the unknown wire \( L = 31.4 \, \text{cm} \)
- Diameter of the wire \( d = 1 \, \text{mm} = 0.1 \, \text{cm} \)
2. **Calculate the Remaining Length:**
- The total length of the meter bridge wire is 100 cm. Therefore, the length from end \( B \) (let's call it \( L_2 \)) is:
\[
L_2 = 100 \, \text{cm} - L_1 = 100 \, \text{cm} - 25.0 \, \text{cm} = 75.0 \, \text{cm}
\]
3. **Use the Meter Bridge Formula:**
- According to the meter bridge principle:
\[
\frac{S}{R} = \frac{L_2}{L_1}
\]
- Plugging in the values:
\[
\frac{300 \, \Omega}{R} = \frac{75.0 \, \text{cm}}{25.0 \, \text{cm}}
\]
- Simplifying the right side:
\[
\frac{300 \, \Omega}{R} = 3
\]
- Therefore, solving for \( R \):
\[
R = \frac{300 \, \Omega}{3} = 100 \, \Omega
\]
4. **Calculate the Cross-Sectional Area \( A \) of the Wire:**
- The area \( A \) of a circular wire is given by:
\[
A = \pi \left(\frac{d}{2}\right)^2
\]
- Substituting the diameter:
\[
A = \pi \left(\frac{0.1 \, \text{cm}}{2}\right)^2 = \pi \left(0.05 \, \text{cm}\right)^2 = \pi \times 0.0025 \, \text{cm}^2
\]
- Converting to square meters:
\[
A = \pi \times 0.0025 \times 10^{-4} \, \text{m}^2 = \pi \times 2.5 \times 10^{-6} \, \text{m}^2
\]
5. **Calculate the Specific Resistivity \( \rho \):**
- The formula for resistance is:
\[
R = \frac{\rho L}{A}
\]
- Rearranging for \( \rho \):
\[
\rho = \frac{R \cdot A}{L}
\]
- Substituting the known values:
\[
\rho = \frac{100 \, \Omega \cdot \left(\pi \times 2.5 \times 10^{-6} \, \text{m}^2\right)}{31.4 \times 10^{-2} \, \text{m}}
\]
- Simplifying:
\[
\rho = \frac{100 \cdot 3.14 \cdot 2.5 \times 10^{-6}}{0.314} \approx \frac{785 \times 10^{-6}}{0.314} \approx 2.5 \times 10^{-4} \, \Omega \cdot \text{m}
\]
### Final Result:
The specific resistivity of the wire is approximately:
\[
\rho \approx 2.5 \times 10^{-4} \, \Omega \cdot \text{m}
\]
