Resistance of a metal wire of length 1 m is `26 Omega` at `20^@ C`. If the diameter of the wire is `0.3 mm`, what will be the resistivity of the metal at that temperature ? Using Table `1.1`, predict the material of the wire.

Explanation

To find the resistivity of the metal wire, we can use the formula that relates resistance (R), resistivity (ρ), length (L), and cross-sectional area (A):

\[ R = \frac{\rho L}{A} \]

### Step-by-Step Solution:

**Step 1: Write down the given information.**
- Length of the wire (L) = 1 m
- Resistance (R) = 26 Ω
- Diameter of the wire = 0.3 mm = 0.3 × 10^-3 m = 3.0 × 10^-4 m
- Temperature = 20 °C

**Step 2: Calculate the radius of the wire.**
- Radius (r) = Diameter / 2 = (0.3 mm) / 2 = 0.15 mm = 0.15 × 10^-3 m = 1.5 × 10^-4 m

**Step 3: Calculate the cross-sectional area (A) of the wire.**
- The area of a circle is given by the formula:
\[ A = \pi r^2 \]
- Substituting the radius:
\[ A = \pi (1.5 × 10^{-4})^2 \]
\[ A = \pi (2.25 × 10^{-8}) \]
\[ A ≈ 3.14 × 2.25 × 10^{-8} \]
\[ A ≈ 7.07 × 10^{-8} \, \text{m}^2 \]

**Step 4: Rearrange the resistance formula to solve for resistivity (ρ).**
\[ \rho = \frac{R \cdot A}{L} \]

**Step 5: Substitute the values into the formula.**
\[ \rho = \frac{26 \, \Omega \cdot 7.07 × 10^{-8} \, \text{m}^2}{1 \, \text{m}} \]
\[ \rho = 26 \cdot 7.07 × 10^{-8} \]
\[ \rho ≈ 1.8382 × 10^{-6} \, \Omega \cdot \text{m} \]

**Step 6: Round the result.**
\[ \rho ≈ 1.84 × 10^{-6} \, \Omega \cdot \text{m} \]

**Step 7: Identify the material using Table 1.1.**
- By comparing the calculated resistivity value with standard values in Table 1.1, we find that the resistivity of approximately \( 1.84 × 10^{-6} \, \Omega \cdot \text{m} \) corresponds to manganese.

### Final Answer:
- The resistivity of the metal at 20 °C is \( 1.84 × 10^{-6} \, \Omega \cdot \text{m} \).
- The material of the wire is manganese.