A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is $$8.92 \times 10^{-3}$$ kg, density is $$8.92 \times 10^3$$ kg m$$^{-3}$$ and resistivity is $$1.7 \times 10^{-8}$$ $$\Omega$$ - m. The length of wire is:

Find the length of a uniform metallic wire carrying 2 A current when connected to a 3.4 V battery, given mass $$= 8.92 \times 10^{-3}$$ kg, density $$= 8.92 \times 10^3$$ kg m$$^{-3}$$, and resistivity $$= 1.7 \times 10^{-8}$$ $$\Omega$$-m.

Find the resistance.

$$R = \frac{V}{I} = \frac{3.4}{2} = 1.7 \; \Omega$$

Find the volume of the wire.

$$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{8.92 \times 10^{-3}}{8.92 \times 10^3} = 10^{-6} \text{ m}^3$$

Relate resistance to length.

For a wire: $$R = \frac{\rho L}{A}$$ and $$\text{Volume} = A \times L$$, so $$A = \frac{V_{\text{vol}}}{L}$$.

$$R = \frac{\rho L}{V_{\text{vol}}/L} = \frac{\rho L^2}{V_{\text{vol}}}$$

Solve for L.

$$L^2 = \frac{R \times V_{\text{vol}}}{\rho} = \frac{1.7 \times 10^{-6}}{1.7 \times 10^{-8}} = 100$$

$$L = 10 \text{ m}$$

The correct answer is $$l = 10$$ m.