A metallic wire having resistance of 20 Ω is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to
- (a)20 Ω
- (b)10 Ω
- (c)5 Ω
- (d)15 Ω
Correct — C, 5 Ω. Resistance is proportional to length (R = ρL/A), so cutting the 20 Ω wire into two equal lengths gives two pieces of 10 Ω each. Connecting two equal 10 Ω resistors in parallel gives R = (10 × 10)/(10 + 10) = 100/20 = 5 Ω. As a shortcut, two equal resistances R in parallel always give R/2, so 10 Ω → 5 Ω.
- (a)20 Ω — This is the original wire's resistance — it ignores both the halving from the cut and the reduction from the parallel connection.
- (b)10 Ω — This is the resistance of just one half of the wire; it forgets that the two halves are then joined in parallel, which halves it again to 5 Ω.
- (d)15 Ω — There is no combination of the cut-and-parallel steps that yields 15 Ω; it does not follow from R ∝ length or the parallel formula.
The resistance of a uniform wire is R = ρL/A, so it is directly proportional to length and inversely proportional to cross-sectional area. Cutting a wire into equal parts divides its resistance among the parts. For resistors in parallel, 1/R = 1/R₁ + 1/R₂; two equal resistors in parallel give half the value.
This is a two-step problem, and each step must be applied in order: first the cut (halving each piece to 10 Ω), then the parallel combination (halving again to 5 Ω). Stopping after either single step gives one of the tempting wrong answers (20 Ω or 10 Ω).
- R = ρL/A: resistance is proportional to length.
- Cutting a wire into two equal parts gives two pieces of half the resistance each (10 Ω).
- Two equal resistances R in parallel give R/2.
- Final result: 10 Ω ∥ 10 Ω = 5 Ω.
Length halved gives 10 Ω per piece; parallel of two equal 10 Ω gives 5 Ω.
- Forgetting that cutting the wire halves each piece's resistance before you combine them.
- Adding the two 10 Ω resistances (giving 20 Ω) instead of combining them in parallel.
Asked as a cut-then-combine problem — track the length change first (R ∝ L), then apply the parallel formula.
Two wires have their lengths, diameters and resistivities, all in the ratio of 1 : 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is
- (a) 10 ohms
- (b) 5 ohms
- (c) 20 ohms
- (d) 40 ohms
Answer(a) 10 ohms
Same core idea — R = ρL/A. UPSC 2001 makes you track how length, area (diameter²) and resistivity together change resistance; the NDA item makes you track how cutting a wire changes its length (and hence resistance) before combining the parts.
If three resistors of 1 Ohm each connect in parallel to each other the resultant resistance is
- (a) 1 Ohm
- (b) 1⁄3 Ohm
- (c) 3 Ohm
- (d) 9 Ohm
Answer(b) 1⁄3 Ohm
Directly the same parallel-combination step. Three equal 1 Ω resistors in parallel give 1/3 Ω; here two equal 10 Ω pieces in parallel give 5 Ω. Both use 1/R = Σ(1/Rᵢ) for equal resistors.
- practice — not a real PYQ
A 12 Ω wire is cut into three equal parts, which are then joined in parallel. The equivalent resistance is
- (a)12 Ω
- (b)4 Ω
- (c)1.33 Ω
- (d)36 Ω
Answer(c) 1.33 Ω — each part is 4 Ω; three 4 Ω in parallel give 4/3 ≈ 1.33 Ω.
- practice — not a real PYQ
Two equal resistors of R each are connected in parallel. Their equivalent resistance is
- (a)2R
- (b)R
- (c)R/2
- (d)R/4
Answer(c) R/2 — two equal resistances in parallel always give half the value.