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I. Multiple Choice Questions · Q2

Q.A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B, is

(a) π Ω\pi\ \Omega
(b) π2 Ω\dfrac{\pi}{2}\ \Omega
(c) 2π Ω2\pi\ \Omega
(d) π4 Ω\dfrac{\pi}{4}\ \Omega
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Step 1. The wire's total resistance is (resistance per metre) ×\times (circumference) =2 Ω/m×2π(1 m)=4π Ω=2\ \Omega/\text{m}\times2\pi(1\ \text{m})=4\pi\ \Omega.

Step 2. Because A and B are diametrically opposite points on the circle, the wire splits into two equal semicircular arcs between them, each carrying half the total resistance: 4π/2=2π Ω4\pi/2=2\pi\ \Omega per arc.

Step 3. These two equal 2π Ω2\pi\ \Omega arcs form two parallel paths between A and B. For two equal resistors R in parallel, the equivalent resistance is R/2R/2, so Req=2π/2=π ΩR_{eq}=2\pi/2=\pi\ \Omega.

Step 4. Eliminating the others: (b) π/2\pi/2 would come from an extra, unjustified halving; (c) 2π2\pi is just a single arc's resistance, without combining the two arcs in parallel at all; (d) π/4\pi/4 comes from halving twice by mistake.

✓Final answer

(a) π Ω\pi\ \Omega.

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