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Physics · Current Electricity

JEE Main 2025 — 28 January, Morning Shift — Question 61

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

  1. Option A:

    9/8

  2. Option B:

    8/9

  3. Option C:

    27/32

  4. Option D:

    32/27

    Correct

Answer: D

Step-by-step solution

R=ρℓAR=\frac{\rho \ell}{A}

So, R∝ℓ\mathrm{R} \propto \ell

Side length of triangle is 1/31 / 3 of total length.

(Req)1=2r/3×r/32r/3+r/3(Req)2=3r/4×r/43r/4+r/4\left(R_{e q}\right)_{1}=\frac{2 r / 3 \times r / 3}{2 r / 3+r / 3} \quad\left(R_{e q}\right)_{2}=\frac{3 r / 4 \times r / 4}{3 r / 4+r / 4}

(Req )1=2r/9\left(R_{\text {eq }}\right)_{1}=2 r / 9

(Req)2=3r/16\left(\mathrm{R}_{\mathrm{eq}}\right)_{2}=3 \mathrm{r} / 16

(Req)1(Req)2=2r/93r/16=3227\frac{\left(\mathrm{R}_{\mathrm{eq}}\right)_{1}}{\left(\mathrm{R}_{\mathrm{eq}}\right)_{2}}=\frac{2 \mathrm{r} / 9}{3 \mathrm{r} / 16}=\frac{32}{27}

IMAGES

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Morning Shift
Subject
Physics
Chapter
Current Electricity
Topic
Combination of Resistors and cells, Wheatstone Bridge