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

To solve this problem, we need to calculate the resistance between diametrically opposite points on a circular wire. Given are the following:

• Length of the wire, \(L = 25 \, \text{m}\)
• Cross-sectional area, \(A = 5 \, \text{mm}^2 = 5 \times 10^{-6} \, \text{m}^2\)
• Resistivity, \(\rho = 2 \times 10^{-6} \, \Omega \cdot \text{m}\)

Let's calculate the resistance using the formula for resistance of a wire:

\(R = \frac{\rho \cdot L}{A}\)

Substituting the values,

\(R = \frac{2 \times 10^{-6} \cdot 25}{5 \times 10^{-6}}\)

Simplifying this gives:

\(R = \frac{50 \times 10^{-6}}{5 \times 10^{-6}} = 10 \, \Omega\)

This resistance is for the full length of the wire. Since the wire is bent into a circle, we treat it as two equal halves. Therefore, the resistance between the two diametrically opposite points is equivalent to two halves of \(10 \, \Omega\) in parallel:

The resistance of each half-circle is \(R_{\text{half}} = \frac{R}{2} = \frac{10}{2} = 5 \, \Omega\).

The total resistance \(R_{\text{total}}\) across the diameter is then given by the parallel resistance formula:

\(\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{half}}} + \frac{1}{R_{\text{half}}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\)

This simplifies to:

\(R_{\text{total}} = \frac{5}{2} = 2.5 \, \Omega\)

Thus, the resistance between the diametrically opposite points is 2.5 \( \Omega \), which matches the given correct answer.

Answer 2

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: \( R_{eq} = \frac{R/2}{2} = \frac{10}{4} = 2.5 \Omega \)