If resistance`R_(1)` in resistance box is `300 Omega`, then the balanced length is found to be `75.0 cm` from end `A`.The diameter of known wire is `1 mm` and length of the unknown wire is `31.4 cm`. Find the specific resistance of the unknown wire.

Explanation

To find the specific resistance (resistivity) of the unknown wire, we will follow these steps:

### Step 1: Understand the relationship in the potentiometer
The potentiometer works on the principle of a Wheatstone bridge, which states that the ratio of the resistances is equal to the ratio of the lengths of the wire. This can be expressed as:
\[ \frac{R_1}{R_x} = \frac{L}{100 - L} \]
where \( R_1 \) is the known resistance, \( R_x \) is the unknown resistance, \( L \) is the balanced length, and \( 100 - L \) is the remaining length of the wire.

### Step 2: Substitute the known values
Given:
- \( R_1 = 300 \, \Omega \)
- \( L = 75.0 \, \text{cm} = 0.75 \, \text{m} \)
- Total length of wire \( = 100 \, \text{cm} = 1.0 \, \text{m} \)

We can calculate \( 100 - L \):
\[ 100 - L = 100 \, \text{cm} - 75 \, \text{cm} = 25 \, \text{cm} = 0.25 \, \text{m} \]

### Step 3: Calculate the unknown resistance \( R_x \)
Using the formula:
\[ \frac{300}{R_x} = \frac{0.75}{0.25} \]
Cross-multiplying gives:
\[ 300 \times 0.25 = R_x \times 0.75 \]
\[ 75 = 0.75 R_x \]
Now, solving for \( R_x \):
\[ R_x = \frac{75}{0.75} = 100 \, \Omega \]

### Step 4: Calculate the specific resistance (resistivity) of the unknown wire
The formula for resistance is:
\[ R = \frac{\rho L}{A} \]
Where:
- \( R \) is the resistance,
- \( \rho \) is the specific resistance (resistivity),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

The cross-sectional area \( A \) of the wire can be calculated using the diameter:
\[ A = \frac{\pi D^2}{4} \]
Given:
- Diameter \( D = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Length of the unknown wire \( L = 31.4 \, \text{cm} = 0.314 \, \text{m} \)

Calculating the area:
\[ A = \frac{\pi (1 \times 10^{-3})^2}{4} = \frac{\pi (1 \times 10^{-6})}{4} = \frac{\pi}{4} \times 10^{-6} \, \text{m}^2 \]

### Step 5: Substitute values into the resistivity formula
Now substituting \( R_x = 100 \, \Omega \), \( L = 0.314 \, \text{m} \), and \( A \):
\[ 100 = \frac{\rho (0.314)}{\frac{\pi}{4} \times 10^{-6}} \]
Rearranging gives:
\[ \rho = \frac{100 \times \frac{\pi}{4} \times 10^{-6}}{0.314} \]

### Step 6: Calculate the specific resistance
Calculating the above expression:
\[ \rho = \frac{100 \times 0.7854 \times 10^{-6}}{0.314} \approx 2.5 \times 10^{-4} \, \Omega \cdot \text{m} \]

### Final Answer
The specific resistance (resistivity) of the unknown wire is approximately:
\[ \rho \approx 2.5 \times 10^{-4} \, \Omega \cdot \text{m} \]

---