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Physics · Current Electricity

JEE Main 2025 — 3 April, Morning Shift — Question 62

A wire of length 25 m and cross-sectional area 5 mm25 \mathrm{~mm}^{2} having resistivity 2×10−6Ω m2 \times 10^{-6} \Omega \mathrm{~m} is bent into a complete circle. The resistance between diametrically opposite points will be

  1. Option A:

    100Ω100 \Omega

  2. Option B:

    50Ω50 \Omega

  3. Option C:

    12.5Ω12.5 \Omega

  4. Option D:

    2.5Ω2.5 \Omega

    Correct

Answer: D

Step-by-step solution

Let RR be total resistance across ends of wire, then

Req =R4=ρℓ4A=2×10−6×254×5×10−6=2.5Ω\begin{aligned} R_{\text {eq }} & =\frac{R}{4}=\frac{\rho \ell}{4 A}=\frac{2 \times 10^{-6} \times 25}{4 \times 5 \times 10^{-6}} & =2.5 \Omega \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Physics
Chapter
Current Electricity
Topic
Ohm's Law and Calculation of Resistance

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