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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2025 · 28 January · Morning Shift · Physics · Current Electricity

A wire of resistance R is bent into an equilateral triangle and an...

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

Options

  1. A8/9
  2. B27/3227 / 32
  3. C32/2732 / 27
  4. D9/89 / 8

Correct answer

C. 32/2732 / 27

Step-by-step solution

For the wire bent into an equilateral triangle, each side has a resistance R3\frac{R}{3}. Req =(2R3)(R3)2R3+R3=2R9=R1R_{\text{eq }}=\frac{\left(\frac{2 R}{3}\right)\left(\frac{R}{3}\right)}{\frac{2 R}{3}+\frac{R}{3}}=\frac{2 R}{9}=R_1 (lets say) For the wire bent into a square, each side has a resistance R4\frac{R}{4}. Req =(3R4)(R4)3R4+R4=3R16=R3 (lets say) ⇒R1R3=2R93R16=3227\begin{aligned} & R_{\text{eq }}=\frac{\left(\frac{3 R}{4}\right)\left(\frac{R}{4}\right)}{\frac{3 R}{4}+\frac{R}{4}}=\frac{3 R}{16}=R_3 \text{ (lets say) } \\ & \Rightarrow \frac{R_1}{R_3}=\frac{\frac{2 R}{9}}{\frac{3 R}{16}}=\frac{32}{27}\end{aligned}
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