Two wires A and B are made up of the same material and have the same mass. Wire A has radius of 2.0 mm and wire B has radius of 4.0 mm. The resistance of wire B is 2 Ω. The resistance of wire A is ________ Ω.

Solution

Physics sir · 4 steps

Same mass and same material means same volume, so a thinner wire is also a longer one.

1. Fix the volume. Same material means same density, so the same mass gives the same volume V = A L. If the cross-section A shrinks, the length L must grow to keep V fixed.
2. Rewrite resistance. Put L = V/A into R = ρ L / A. Both the extra length and the smaller area raise the resistance, so A appears twice.

   R = (ρ V)/A² = (ρ V)/(π² r⁴) ⇒ R ∝ 1/r⁴

3. Compare the wires. R_A/R_B = (r_B/r_A)⁴ = (4.0/2.0)⁴ = 2⁴ = 16.
4. So. R_A = 16 × 2 Ω = 32 Ω.

The common slip: 8. Keeping the length the same and using R ∝ 1/r² gives a ratio of 4 and R_A = 8 Ω. The thinner wire is also 4 times longer, which adds another factor of 4.

The idea to keep. When a wire is reshaped or the mass is fixed, the volume stays constant, so R ∝ 1/A² ∝ 1/r⁴. Always ask what is held constant before using R = ρ L/A.

Answer from the NTA final answer key. Solution by PrepWiser.