A wire has a resistance of 3 . Its cross-sectional area is 0.60 mm2 and its resistivity is 7.2 x 10⁻⁷ * m. What is its length?

The formula for resistance is $R = \rho \frac{L}{A}$, where $R$ is resistance, $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

Given values:

• Cross-sectional area, $A = 0.60 \text{ mm}^2$.
• Resistivity, $\rho = 7.2 \times 10^{-7} \Omega \cdot \text{m}$. (Note: The question states the unit as $\Omega$, but resistivity is typically measured in $\Omega \cdot \text{m}$. We assume the intended unit is $\Omega \cdot \text{m}$ for dimensional consistency.)
• Resistance, $R = 3 \Omega$.

Convert the cross-sectional area from $\text{mm}^2$ to $\text{m}^2$:
Since $1 \text{ mm} = 10^{-3} \text{ m}$, then $1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$.

$$A = 0.60 \text{ mm}^2 = 0.60 \times 10^{-6} \text{ m}^2$$

$$R = \rho \frac{L}{A}$$

Multiply both sides by $A$:

$$R \cdot A = \rho \cdot L$$

Divide both sides by $\rho$:

$$L = \frac{R \cdot A}{\rho}$$

$$L = \frac{(3 \Omega) \cdot (0.60 \times 10^{-6} \text{ m}^2)}{7.2 \times 10^{-7} \Omega \cdot \text{m}}$$ $$L = \frac{1.8 \times 10^{-6} \Omega \cdot \text{m}^2}{7.2 \times 10^{-7} \Omega \cdot \text{m}}$$ $$L = \frac{1.8}{7.2} \times \frac{10^{-6}}{10^{-7}} \text{ m}$$ $$L = 0.25 \times 10^{(-6 - (-7))} \text{ m}$$ $$L = 0.25 \times 10^{1} \text{ m}$$ $$L = 2.5 \text{ m}$$

The length of the wire is $\text{2.5 m}$.