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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2025
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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:

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Answer & Solution

Correct Answer

(D) \(25 \Omega\)

Step-by-step Solution

Detailed explanation


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