A wire of length 25 m and cross-sectional area 5 ~mm^2 having resistivity of 2 × 10^ -6 ~m is bent into a complete circle. The resistance between diametrically opposite points will be

Correct answer: D. 25 aligned & L=25 m, A=5 mm^2=5 × 10^ -6 m^2 & =2 × 10^ -6 m & R_ wire = L A = 2 × 10^ -6 × 25 5 × 10^ -6 =10 & R_ eq = R 4 =(10)/(4)=2.5 aligned Answer does not match with NTA option.

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Physics
Chapter: Current Electricity

JEE Main 2025 · 3 April · Morning Shift · Physics · Current Electricity

A wire of length 25 m and cross-sectional area 5 ~mm^2 having...

A wire of length 25 m and cross-sectional area

5 \mathrm{~mm}^2

having resistivity of

2 \times 10^{-6} \Omega \mathrm{~m}

is bent into a complete circle. The resistance between diametrically opposite points will be

Options

A $12.5 \Omega$
B $50 \Omega$
C $100 \Omega$
D $25 \Omega$

Correct answer

D. $25 \Omega$

Step-by-step solution

\begin{aligned}& \mathrm{L}=25 \mathrm{ m}, \mathrm{ A}=5 \mathrm{ mm}^2=5 \times 10^{-6} \mathrm{ m}^2 \\ & \rho=2 \times 10^{-6} \Omega \mathrm{ m} \\ & \mathrm{R}_{\text{wire }}=\frac{\rho \mathrm{L}}{\mathrm{ A}}=\frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}}=10 \\ & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{4}=\frac{10}{4}=2.5 \Omega\end{aligned}

Answer does not match with NTA option.

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