You need produce a set of cylindrical copper wire `3.50 m` long will have a resistance of `0.125Omega` each. What will be the mass of each of these wires? Specific resistance of copper `=1.72xx10^-8Omega-m`,density of copper `=8.9xx10^3kg//m^3`

Explanation

To find the mass of the cylindrical copper wire, we will follow these steps:

### Step 1: Calculate the Cross-Sectional Area of the Wire

The resistance \( R \) of a cylindrical wire can be expressed using the formula:
\[ R = \frac{\rho' \cdot L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho' \) is the specific resistance (resistivity) of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

Rearranging the formula to find the area \( A \):
\[ A = \frac{\rho' \cdot L}{R} \]

Substituting the given values:
- \( \rho' = 1.72 \times 10^{-8} \, \Omega \cdot m \)
- \( L = 3.5 \, m \)
- \( R = 0.125 \, \Omega \)

Calculating:
\[ A = \frac{(1.72 \times 10^{-8}) \cdot (3.5)}{0.125} \]
\[ A = \frac{6.02 \times 10^{-8}}{0.125} = 4.816 \times 10^{-7} \, m^2 \]

### Step 2: Calculate the Volume of the Wire

The volume \( V \) of the cylindrical wire can be calculated using the formula:
\[ V = A \cdot L \]

Substituting the values:
\[ V = (4.816 \times 10^{-7}) \cdot (3.5) \]
\[ V = 1.684 \times 10^{-6} \, m^3 \]

### Step 3: Calculate the Mass of the Wire

Using the density \( \rho \) of copper, we can find the mass \( m \) using the formula:
\[ \rho = \frac{m}{V} \implies m = \rho \cdot V \]

Substituting the values:
- Density of copper \( \rho = 8.9 \times 10^3 \, kg/m^3 \)

Calculating:
\[ m = (8.9 \times 10^3) \cdot (1.684 \times 10^{-6}) \]
\[ m = 0.015 \, kg \quad \text{or} \quad 15 \, g \]

### Final Answer:
The mass of each of these wires is **15 g**.

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