An aluminium cable of diameter 4 x 10-3m and resistivity 3.0 x 10-8$\omega$m has a resistance of 21$\Omega$. Calculate the length of the cable. ($\pi = 3.14$)

Analyze the question: We're given the diameter, the resistivity and the resistance of a cable, and we need its length.

Concept: Resistance of a conductor is

R = ρL/A, so L = RA/ρ

where A = πd²/4 is the cross-sectional area (the cable is circular).

Solution:

1. Radius: r = d/2 = 2 x 10⁻³ m

2. Area: A = πr² = 3.14 x (2 x 10⁻³)² = 3.14 x 4 x 10⁻⁶ = 1.256 x 10⁻⁵ m²

3. Length: L = RA/ρ = (21 x 1.256 x 10⁻⁵) / (3.0 x 10⁻⁸)

Numerator: 21 x 1.256 x 10⁻⁵ = 26.376 x 10⁻⁵ = 2.6376 x 10⁻⁴

L = 2.6376 x 10⁻⁴ / 3.0 x 10⁻⁸ = 0.8792 x 10⁴ ≈ 8.8 x 10³ m

Answer: This matches option B (8.

Tip: Always convert diameter to radius before using A = πr². A common mistake is using the diameter directly, which gives an answer 4 times too large. Also, handle the powers of 10 separately from the numbers to avoid slips: 10⁻⁴ ÷ 10⁻⁸ = 10⁴.