Question

A copper wire has a diameter of 0.5 mm and a resistivity of 1.6x10 -8 Ω m. What will be the length of this wire to make its resistance 10 Ω?

Answer 1

Given :-

→ Diameter of the copper wire = 0.5 mm

→ Resistivity = 1.6 × 10⁻⁸ Ω m

→ Resistance = 10 Ω

To find :-

→ Length of the wire.

Solution :-

Radius of the wire = Diameter/2

= 0.5/2

= 0.25 mm

= 2.5 × 10⁻⁴ m

Area of cross section of the wire :-

= πr²

= 22/7 × (2.5 × 10⁻⁴)²

= 22/7 × 6.25 × 10⁻⁸

= 19.6 × 10⁻⁸

= 1.96 × 10⁻⁷ m²

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Now, we know that :-

R = ρL/A

Substituting values, we get :-

⇒ 10 = [1.6 × 10⁻⁸ × L]/[1.96 × 10⁻⁷]

⇒ 10 = 0.81 × 10⁻¹ × L

⇒ 10 = 0.081L

⇒ L = 10/0.081

⇒ L = 123.45 m

Thus, length of the wire is 123.45 m .

Answer 2

Given :-

Diameter = 0.5 mm

resistivity of 1.6x10 -8 Ω m.

To Find :-

The length of this wire to make its resistance 10 Ω?

Solution :-

We know that

Radius = Diameter/2

Radius = 0.5/2

Radius = 05/20

Radius =  0.25 mm

It may also be written as

2.5 × 10⁻⁴ m

Now

The wire is in the shape of circle. The area of circle is πr²

$\sf 3.14 \times \bigg(2.5 \times 10^{-4}\bigg)^2$

$\sf 3.14 \times 6.25 \times 10^{-8}$

$\sf 19.6 \times 10^{-8}$

$\sf 1.96 \times 10^{-7}$

Now

$\sf R = \dfrac{\rho L}{A}$

$\sf 10 = \dfrac{1.6 \times 10^{-8} \times l}{1.96 \times 10^{-7}}$

$\sf 10 = \dfrac{1.6 \times 10^{-1}\times l}{1.96}$

$\sf 10 = 0.081L$

$\sf L = 123.5 \; m$