JEE Mains · Physics · STD 12 - 3. current electricity
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:
- A \(12.5 \Omega\)
- B \(50 \Omega\)
- C \(100 \Omega\)
- D \(25 \Omega\)
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 \\ &…
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