Concept:Resistance is determined by the length of the path the current travels through and the cross-sectional area perpendicular to that flow.
Formula:$$ R = \rho \frac{L}{A_{cross}} $$
Solution:- The film is a square of area \( 1 \text{ mm}^2 \), so each side length \( L = 1 \text{ mm} = 10^{-3} \text{ m} \).
- Current flows between opposite faces (edges) of this square, meaning it travels a distance \( L = 10^{-3} \text{ m} \).
- The cross-sectional area \( A_{cross} \) the current sees is the width \( \times \) thickness.
- \( A_{cross} = (1 \text{ mm}) \times (1 \, \mu\text{m}) = (10^{-3} \text{ m}) \times (10^{-6} \text{ m}) = 10^{-9} \text{ m}^2 \).
- Given resistivity \( \rho = 10^{-6} \, \Omega\text{m} \).
- \( R = \frac{10^{-6} \times 10^{-3}}{10^{-9}} = \frac{10^{-9}}{10^{-9}} = 1 \, \Omega \).
Why other options are incorrect:- Opt A, C, D: Result from substituting incorrect length or cross-sectional area (e.g., using the face area \(10^{-6}\) as the cross-section).
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