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 .
If the resistivity of the material is m. The value of is ___________.
If the resistivity of the material is m. The value of is ___________.
Show answer and solution
Answer: 2
Solution
The cross-sectional area of the hollow cylinder is A = π(r_outer² - r_inner²) = π((4mm)² - (2mm)²) = π(16 - 4) × 10⁻⁶ = 12π × 10⁻⁶ m².
R = ρL/A = (2.4 × 10⁻⁸ × 3.14) / (12π × 10⁻⁶).
Since 3.14 ≈ π: R = (2.4 × 10⁻⁸ × π) / (12π × 10⁻⁶) = 2.4 × 10⁻⁸ / (12 × 10⁻⁶) = 0.2 × 10⁻² = 2 × 10⁻³ Ω.
Therefore n = 2.
R = ρL/A = (2.4 × 10⁻⁸ × 3.14) / (12π × 10⁻⁶).
Since 3.14 ≈ π: R = (2.4 × 10⁻⁸ × π) / (12π × 10⁻⁶) = 2.4 × 10⁻⁸ / (12 × 10⁻⁶) = 0.2 × 10⁻² = 2 × 10⁻³ Ω.
Therefore n = 2.
[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]