This problem combines two ideas about resistance: how resistance depends on the length of a wire, and how resistances combine in parallel.

Step 1 — Cutting the wire. The resistance of a uniform wire is directly proportional to its length (R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area). When a 10 Ω wire is cut into two equal pieces, each piece has half the original length while ρ and A stay the same, so each piece has half the resistance:

Each half = 10 ÷ 2 = 5 Ω.

Step 2 — Connecting the halves in parallel. For resistors in parallel the reciprocals add:

1/R = 1/5 + 1/5 = 2/5, therefore R = 5/2 = 2.5 Ω.

(Shortcut: for two equal resistors in parallel the result is simply half of one, i.e. 5 ÷ 2 = 2.5 Ω.)

So the equivalent resistance is 2.5 ohms.

The distractors correspond to common errors: 10 ohms ignores the cut altogether; 5 ohms stops after halving the wire but forgets the parallel combination; and 20 ohms is what you would get by wrongly adding the two 5 Ω halves in series (which would in fact just rebuild the original 10 Ω wire, not 20 Ω). Parallel connection always gives a value smaller than the smallest branch, confirming 2.5 Ω.