Question

The length and area of cross-section of a copper wire are respectively 30 m and \( 6 \times 10^{-7} \, \text{m}^2 \). If the resistivity of copper is \( 1.7 \times 10^{-8} \, \Omega \, \text{m} \), then the resistance of the wire is

• A. \( 0.51 \, \Omega \)
• B. \( 0.68 \, \Omega \)
• C. \( 0.85 \, \Omega \)
• D. \( 0.75 \, \Omega \)

Hint:

Resistance of a conductor is given by \( R = \frac{\rho L}{A} \), where: - \( \rho \) is the resistivity of the material (in \( \Omega \cdot \text{m} \)). - \( L \) is the length of the conductor (in m). - \( A \) is the cross-sectional area of the conductor (in \( \text{m}^2 \)). Ensure all units are consistent before calculation.

Answer 1

The Correct Option is C

Length of the wire \( L = 30 \) m.
Area of cross-section \( A = 6 \times 10^{-7} \, \text{m}^2 \).
Resistivity of copper \( \rho = 1.
7 \times 10^{-8} \, \Omega \, \text{m} \).
The formula for resistance \(R\) of a wire is \( R = \frac{\rho L}{A} \).
Substitute the given values: \[ R = \frac{(1.
7 \times 10^{-8} \, \Omega \, \text{m}) \times (30 \, \text{m})}{6 \times 10^{-7} \, \text{m}^2} \] \[ R = \frac{1.
7 \times 30}{6} \times \frac{10^{-8}}{10^{-7}} \, \Omega \] \[ R = \frac{1.
7 \times 5}{1} \times 10^{-8 - (-7)} \, \Omega \] \[ R = 1.
7 \times 5 \times 10^{-1} \, \Omega \] \[ R = 8.
5 \times 10^{-1} \, \Omega = 0.
85 \, \Omega \] This matches option (3).