A wire of uniform resistance $\lambda \Omega / \mathrm{m}$ is bent into a circle of radius $r$ and another piece of wire with length $2 r$ is connected between points $A$ and $B(A O B)$ as shown in figure. The equivalent resistance between points $A$ and $B$ is $\_\_\_\_$ $\Omega$. ![JEE Main 2026 Physics, Current Electricity — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Current_Electricity/6983c1938bfdbb243563967a/question_1__q_6983c1938bfdbb243563967a__cdn-question-pool.getmarks.app__23S1_q_31_1__2a57d5ea3c_final_ppt_sync.png)

JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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

JEE Main 2026 Physics, Current Electricity — question figure

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\frac{3 \pi \lambda r}{8}$
B
$\frac{6 \pi \lambda r}{3 \pi+16}$
C
$2 \pi \lambda r$
D
$(\pi+1) 2 r \lambda$

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Solution

The wire bent into a circle of radius $r$ has resistance $\lambda \cdot \pi r$ for the semicircular arcs, and the straight wire of length $2r$ has resistance $\lambda \cdot 2r$ . The three paths between A and B (two semicircular arcs of the circle and one straight wire through center) are in parallel.

$\frac{1}{R_{AB}} = \frac{2}{\lambda \pi r} + \frac{1}{\lambda \cdot 2r} + \frac{2}{\lambda \cdot 3\pi r}$

$= \frac{1}{\lambda r} \left[\frac{2}{\pi} + \frac{1}{2} + \frac{2}{3\pi}\right]$

$= \frac{1}{\lambda r} \left(\frac{12 + 3\pi + 4}{6\pi}\right) = \frac{1}{\lambda r} \cdot \frac{16 + 3\pi}{6\pi}$

$R_{AB} = \lambda r \left(\frac{6\pi}{16 + 3\pi}\right) = \frac{6\pi \lambda r}{3\pi + 16}$

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Solution

$\frac{1}{R_{AB}} = \frac{2}{\lambda \pi r} + \frac{1}{\lambda \cdot 2r} + \frac{2}{\lambda \cdot 3\pi r}$

$= \frac{1}{\lambda r} \left[\frac{2}{\pi} + \frac{1}{2} + \frac{2}{3\pi}\right]$

$= \frac{1}{\lambda r} \left(\frac{12 + 3\pi + 4}{6\pi}\right) = \frac{1}{\lambda r} \cdot \frac{16 + 3\pi}{6\pi}$

$R_{AB} = \lambda r \left(\frac{6\pi}{16 + 3\pi}\right) = \frac{6\pi \lambda r}{3\pi + 16}$

JEE Main 2026 — Physics Electromagnetism

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JEE Main 2026 — Physics Electromagnetism

Options

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Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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