A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is n×10⁻³Ω. If the resistivity of the material is 2.4×10⁻⁸Ω m. The value of n is ___________.

Solution

Physics sir · 4 steps

Current flows only through the metal ring, so the area is the outer circle minus the hollow inside.

1. Find the radii. Halve the diameters first. The inner radius is r₁ = 2 mm and the outer radius is r₂ = 4 mm.
2. Ring area. The cross-section is a ring, so A = π(r₂² - r₁²) = π(16 - 4) × 10⁻⁶ = 12π × 10⁻⁶ m².

   R = (ρ L)/A = (2.4 × 10⁻⁸ × 3.14)/(12 × 3.14 × 10⁻⁶)

3. Simplify. The length 3.14 m cancels the π in the area. This leaves R = (2.4 × 10⁻⁸)/(12 × 10⁻⁶) = 0.2 × 10⁻² = 2 × 10⁻³ Ω.
4. So. The resistance is 2 × 10⁻³ Ω, so n = 2.

The common slip: 0.5. Putting the diameters straight in as radii gives an area of 48π × 10⁻⁶ m², four times too big, and n = 0.5 instead of 2.

The idea to keep. For a hollow conductor, use the area of the ring, π(r₂² - r₁²), in R = ρ L/A. Always turn diameters into radii before you square them.

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