JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 3)
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
$100 \Omega$
$2.5 \Omega$
$12.5 \Omega$
$50 \Omega$
Explanation
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$$\begin{aligned} &\begin{aligned} & \mathrm{L}=25 \mathrm{~m}, \mathrm{~A}=5 \mathrm{~mm}^2=5 \times 10^{-6} \mathrm{~m}^2 \\ & \rho=2 \times 10^{-6} \Omega \mathrm{~m} \\ & \mathrm{R}_{\text {wire }}=\frac{\rho \mathrm{L}}{\mathrm{~A}}=\frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}}=10 \\ & \mathrm{R}_{\text {eq }}=\frac{\mathrm{R}}{4}=\frac{10}{4}=2.5 \Omega \end{aligned}\\ &\text { Answer does not match with NTA option. } \end{aligned}$$
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