A negligbly small current is passed through a wire of length 15 cm and uniform cross-section `6.0xx10^(-7) m^(2)` and its resistance is measured to be `5.0 Omega`. What is the resistivity of the material at the temperature of the experiment ?

Answer

To find the resistivity of the material at the temperature of the experiment, we can use the formula for resistance in terms of resistivity:

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

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

### Step 1: Identify the given values
- Length of the wire, \( L = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} \)
- Cross-sectional area, \( A = 6.0 \times 10^{-7} \, \text{m}^2 \)
- Resistance, \( R = 5.0 \, \Omega \)

### Step 2: Rearrange the formula to solve for resistivity \( \rho \)
We rearrange the formula to isolate \( \rho \):

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

### Step 3: Substitute the known values into the formula
Now we substitute the known values into the rearranged formula:

\[ \rho = 5.0 \, \Omega \times \frac{6.0 \times 10^{-7} \, \text{m}^2}{15 \times 10^{-2} \, \text{m}} \]

### Step 4: Calculate the resistivity
Now we perform the calculation step-by-step:

1. Calculate the fraction \( \frac{6.0 \times 10^{-7}}{15 \times 10^{-2}} \):

\[ \frac{6.0 \times 10^{-7}}{15 \times 10^{-2}} = \frac{6.0}{15} \times 10^{-7 + 2} = 0.4 \times 10^{-5} = 4.0 \times 10^{-6} \]

2. Now multiply by the resistance:

\[ \rho = 5.0 \times 4.0 \times 10^{-6} = 20.0 \times 10^{-6} \, \Omega \cdot \text{m} \]

3. Finally, express it in standard form:

\[ \rho = 2.0 \times 10^{-5} \, \Omega \cdot \text{m} \]

### Final Answer
The resistivity of the material at the temperature of the experiment is:

\[ \rho = 2.0 \times 10^{-5} \, \Omega \cdot \text{m} \]