Question

A wire of length $ 25 \, \text{m} $ and cross-sectional area $ 5 \, \text{mm}^2 $ having resistivity $ 2 \times 10^{-6} \, \Omega \cdot \text{m} $ is bent into a complete circle. The resistance between diametrically opposite points will be:

• A. 12.5 \( \Omega \)
• B. 50 \( \Omega \)
• C. 100 \( \Omega \)
• D. 2.5 \( \Omega \)

Hint:

When a wire is bent into a circle, the resistance between diametrically opposite points is the parallel combination of the resistances of the two semicircles.

Answer 1

The Correct Option is D

The wire is bent into a circle. The resistance between diametrically opposite points will be the resistance of two semicircles in parallel. 

\( L = 25 \) m, \( A = 5 mm^2 = 5 \times 10^{-6} m^2 \) 

\( \rho = 2 \times 10^{-6} \Omega m \) \( R_{wire} = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10 \Omega \) 

The resistance of each semicircle is \( \frac{R_{wire}}{2} = \frac{10}{2} = 5 \Omega \) 

The equivalent resistance of two semicircles in parallel is: \( R_{eq} = \frac{R/2}{2} = \frac{10}{4} = 2.5 \Omega \)