NDA (I) 2017 General Ability Test · Question 86 of 149

Correct — A, 5 Ω. The two ends of a diameter divide the single-turn ring into two semicircular arcs of equal length. Resistance is proportional to length for a wire of uniform thickness and material, so each arc carries half of the coil's 20 Ω, that is 10 Ω. Between the two chosen points those arcs are two separate paths joining the same pair of terminals, which is a parallel combination, and two equal resistances of 10 Ω in parallel give (10 × 10) ÷ (10 + 10) = 5 Ω.

• (b)10 Ω — This is the resistance of just one of the two semicircular halves. Current entering at one end of the diameter has both halves open to it, so the second half must be counted as a parallel path.
• (c)20 Ω — Twenty ohms is the resistance of the whole length of wire measured end to end before it was bent into a ring. Once the ends are joined, no two points on the ring are separated by all 20 Ω.
• (d)40 Ω — Doubling the coil's resistance would need the two halves in series and each still worth 20 Ω. Cutting a wire into two equal parts halves each part's resistance, and joining paths side by side lowers the total further rather than raising it.

For a uniform wire, resistance R = ρL/A, so with the material and cross-section fixed, resistance simply follows length. Bending the wire into a circle changes nothing electrically. Any two points on the ring split it into two arcs whose resistances add up to the resistance of the whole wire, and those two arcs always sit in parallel between those points.

This is the standard 'ring across a diameter' problem, and the diameter is the special case that makes the two arcs equal. Note the general rule too: if the two points are not diametrically opposite, the arcs are unequal — say x and 20 − x — and the answer becomes x(20 − x)/20, which is largest at the diameter. That is a useful sanity check, because 5 Ω is the biggest value the ring can offer between any pair of points, and every other option is larger than that.

• Resistance of a uniform wire is proportional to its length, so cutting it into two equal parts halves each part's resistance.
• Two resistances in parallel combine as the product over the sum, so two equal resistances R give R/2.
• Any two points on a closed loop of wire are connected by two arcs in parallel.
• For a ring of total resistance R, the resistance across a diameter is R/4 — here 20 Ω gives 5 Ω.

• Quoting the coil's stated resistance as the answer without noticing that the question asks for the resistance between two points on it.
• Counting only one arc and forgetting that the other arc is a second path in parallel.
• Assuming a parallel combination raises resistance — it always gives less than the smaller branch.

NDA sets this as a one-step numerical: a wire cut or bent, the parts recombined, and the equivalent resistance asked for.

• practice — not a real PYQ

  A uniform wire of resistance 12 Ω is bent into a circle. The resistance between the two ends of a diameter is

  • (a)12 Ω
  • (b)6 Ω
  • (c)3 Ω
  • (d)24 Ω

  Answer(c) 3 Ω — each half is 6 Ω and the two halves in parallel give 3 Ω, one quarter of the total.

• practice — not a real PYQ

  A wire of resistance 8 Ω is cut into two equal parts, which are then joined in parallel. The resistance of the combination is

  • (a)8 Ω
  • (b)4 Ω
  • (c)2 Ω
  • (d)16 Ω

  Answer(c) 2 Ω — each half is 4 Ω, and two 4 Ω resistances in parallel give 2 Ω.