Question

When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be

• A. R/4
• B. 4R
• C. R/8
• D. R/2

Answer 1

The Correct Option is A

To solve this problem, we need to determine the resistance between two diametrically opposite points on a wire that forms a circle. Let's break down the solution step-by-step: 

Step 1: Understanding the Problem

A wire of uniform cross-section \(a\) and length \(l\) is bent into a complete circle. The resistance of the entire wire is given as \(R\). We are to find the resistance between two diametrically opposite points on this circle.

Step 2: Conceptualizing the Wire as a Circle

When the wire is bent into a circle, it forms two semicircular halves of equal resistance between any two diametrically opposite points.

Step 3: Resistance of Each Semicircle

The entire circle has a resistance \(R\), and it is composed of two semicircles. Each semicircle will have half the resistance of the entire circle:

\(R_{\text{semi}} = \frac{R}{2}\)

Step 4: Calculating the Effective Resistance

The two semicircles are in parallel because these form two parallel paths between the diametrically opposite points. The formula for the total resistance \(R_{\text{total}}\) of two resistors \(R_1\) and \(R_2\) in parallel is given by:

\(\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}\)

Here, \(R_1\) and \(R_2\) are both \(\frac{R}{2}\), so:

\(\frac{1}{R_{\text{total}}} = \frac{1}{\frac{R}{2}} + \frac{1}{\frac{R}{2}} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}\)

Thus, the effective resistance between two diametrically opposite points is:

\(R_{\text{total}} = \frac{R}{4}\)

Conclusion

Therefore, the resistance between any two diametrically opposite points when the wire is bent into a complete circle is \(\frac{R}{4}\).

The correct option is

R/4

.