Question

Aluminium wire has radius 0.25mm and length of 75m of the resistance of the wire is 10 ohm.calculate the resistivity of aluminium

Answer 1

Resistivity = (R x A)/length
Radius = 0.25mm = 0.00025 m
length = 75m
Resistance = 10 ohms
Area of cross section = πr^2
= 22/7 x 0.00025 x 0.00025
Resistivity =
(10 x 22/7 x 0.00025 x 0.00025 ) ÷ 75
= (220/7 x 0.00025 x 0.00025)/75
= 2.63 x 10 to the power (-8) ohmmeter

Answer 2

Given :

• Radius of wire, r = 0.25 mm

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀⠀ ⠀⠀ ⠀ = $\sf \dfrac{0.25}{1000}m$

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀⠀ ⠀⠀ ⠀ = $\sf 0.25 × 10^{-3}m$

• Length, l = 75 m
• Resistance, R = 10 Ω

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀

To Find :

• Resistivity of aluminium, ρ

⠀⠀ ⠀ ⠀⠀

Solution :

$\sf Now, \: we \: know \: that, \: R = \dfrac{ρl}{A}$

$\sf \implies ρ = \dfrac{RA}{l}$

$\sf \implies ρ = \dfrac{Rπr²}{l} \: ( \because A = πr²)$

Now, put values :

$\sf ρ = \dfrac{10 × 3.14 × (0.25×10^{-3})²}{75}$

$\sf ρ = \dfrac{10 × 314 × 0.25 × 0.25×10^{-3}×10^{-3}}{100×75}$

$\sf ρ = \dfrac{10 × 314 × 25 × 25×1×1}{100×75×100×100×1000×1000}$

$\sf ρ = \dfrac{10 × 314 × \cancel{25}^{1} × 25}{100× \cancel{75}_{3}×10000000000}$

$\sf ρ = \dfrac{10× 314 × 25}{100× 3×10000000000}$

$\sf ρ = \dfrac{785\cancel{0}\cancel{0}}{100× 3×100000000\cancel{0}\cancel{0}}$

$\sf ρ = \dfrac{785}{100× 3×100000000}$

$\sf ρ = \dfrac{785}{100× 3×10^{8}}$

$\sf ρ = \dfrac{785}{300}×10^{-8}$

$\sf ρ = 2.6166666... ×10^{-8}$

$\sf ρ = 2.62 ×10^{-8} \: (approx.)$

$\sf \implies ρ = 2.62 × 10^{-8}Ωm \: (approx.)$

⠀ ⠀⠀ ⠀

$\sf \color{fuchsia} Therefore, \: Resistivity \: of \: aluminium \: is$

$\sf \color{fuchsia} approximately \: 2.62 × 10^{-8}Ωm$