JEE Main 2026 · 23 January · morning · Single correct · +4 / −1
A wire of uniform resistance λΩ/m is bent into a circle of radius r and another piece of wire with length 2r is connected between points A and B(AOB) as shown in figure. The equivalent resistance between points A and B is ____ Ω.
- (a) (π+1)2rλ
- (b) (6πλ r)/(3π+16)
- (c) (3πλ r)/8
- (d) 2πλ r
Answer and solution
Answer (b) (6πλ r)/(3π+16)
Solution
Physics sir · 2 steps
Three separate wires join A and B, so they are three resistors in parallel. Find each length first.
- Spot the paths. ∠ AOB = 90°. Current can go from A to B along the short quarter arc, along the long three-quarter arc, or straight through the bent wire A → O → B. All three start at A and end at B, so they are in parallel.
Path Length Resistance Quarter arc (π r)/2 (π λ r)/2 Three-quarter arc (3π r)/2 (3π λ r)/2 Wire AOB r + r = 2r 2λ r 1/R = 2/(πλ r) + 2/(3πλ r) + 1/(2λ r) = (12 + 4 + 3π)/(6πλ r) = (3π + 16)/(6πλ r)
- So. Turning it over gives R = (6πλ r)/(3π + 16), which is option (b).
The idea to keep. When a wire network has only two junctions, every wire between them is a parallel branch. Turn each length into resistance with R = λ × length, then add the reciprocals.
Answer from the NTA final answer key. Solution by PrepWiser.