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JEE Main 2019 (9 Apr, Shift 2) · Combination of Resistors

Combination of Resistors question from JEE Main 2019 (9 Apr, Shift 2)

Q62JEE Main 2019 · 9 APR · SHIFT 2Medium

Combination of Resistors

A metal wire of resistance 3 Ω\Omega is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :-

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Answer: (d)

  1. Doubling length (volume constant): R′=4×3=12 ΩR' = 4 \times 3 = 12\ \Omega.
  2. A 60∘60^\circ arc is 16\dfrac{1}{6} of the circle: 2 Ω2\ \Omega; the rest is 10 Ω10\ \Omega.
  3. Req=2×1012=53 ΩR_{eq} = \dfrac{2 \times 10}{12} = \dfrac{5}{3}\ \Omega
  4. So the answer is (d) 5/3 Ω5/3\ \Omega.
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