A wire of uniform resistance λ Ω/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 Ω. (a) (π + 1)2rλ (b) (6πλ r)/(3π+16) (c) (3πλ r)/(8) (d) 2πλ r

Answer: (b). A and B are 90° apart, so three paths join A and B in parallel. Minor arc: (π rλ)/(2); major arc: (3π rλ)/(2); straight wire AOB: 2rλ. 1/R = (1)/(rλ)(2/π + 2/3π + 1/2) = (1)/(rλ) · (16 + 3π)/(6π) R = (6πλ r)/(3π + 16) So the answer is (b).

A wire of uniform resistance λ Ω/m is bent into a circle of radius r… | JEE Main 2026 (23 Jan, Shift 1)

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Combination of Resistors

A wire of uniform resistance $\lambda\ \Omega/\text{m}$ is bent into a circle of radius $r$ and another piece of wire with length $2r$ is connected between points $A$ and $B(\text{AOB})$ as shown in figure. The equivalent resistance between points $A$ and $B$ is $\underline{\qquad}\ \Omega$ $\Omega$ .

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Answer: (b)

A and B are $90^\circ$ apart, so three paths join A and B in parallel.
Minor arc: $\dfrac{\pi r\lambda}{2}$ ; major arc: $\dfrac{3\pi r\lambda}{2}$ ; straight wire AOB: $2r\lambda$ .
$\dfrac{1}{R} = \dfrac{1}{r\lambda}\left(\dfrac{2}{\pi} + \dfrac{2}{3\pi} + \dfrac{1}{2}\right) = \dfrac{1}{r\lambda} \cdot \dfrac{16 + 3\pi}{6\pi}$
$R = \dfrac{6\pi\lambda r}{3\pi + 16}$
So the answer is (b).

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