An aluminium cable of diameter 4 x 10-3m and... | WAEC SSCE - Physics - 2016 (Objective)

Question 1 Report

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. (\(\pi = 3.14\))

8.0 x 10²m

8.8 x 10³m

8.0 x 10⁴m

8.8 x 10⁴m

Answer Details

The resistance of a conductor is given by the formula:

R = ρL/A

where R is resistance, ρ (rho) is resistivity, L is the length of the conductor, and A is the cross-sectional area of the conductor.

We are given the resistance of the cable (R = 21 Ω), its diameter (d = 4 x 10^-3 m), and its resistivity (ρ = 3.0 x 10^-8 Ωm). We can use the diameter to find the cross-sectional area of the cable, using the formula:

A = πd²/4

Substituting the given values:

A = π x (4 x 10^-3)²/4 = 1.26 x 10^-5 m²

Now we can rearrange the formula for resistance to solve for the length of the cable (L):

L = RA/ρ

Substituting the given values:

L = 21 x 1.26 x 10^-5 / 3.0 x 10^-8

Simplifying the expression, we get:

L = 8.82 x 10³ m

Therefore, the length of the cable is 8.82 x 10³ m.

So the correct answer is) 8.8 x 10³ m.

Read lesson note on Capacitors (JAMB)