Question

A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-6 22 m. What will be
the length of this wire to make its resistance 10 ohm? How much does the resistance
change if the diameter is doubled?​

Answer 1

Solution :-

Initially ,

• Diameter of the copper wire (d)= 0.5 mm = 0.5 x 10 -³ m

• Resistivity of the copper wire (ρ) = 1.6 x 10^-6 m

• Resistance of the wire(R) = 10 Ω

• Radius of the wire = 0.5 x 10 -³ /2 = 0.25 x 10 -³ m

Area of the wire ,

➝ A = π r²

➝ A = 3.14 x 0.0625 x 10^-6

➝ A = 0.196 X 10^-6

we know that ,

➝ R = ρ L /A

On rearranging ,

➝ L = RA /ρ

➝ L = 10 x 0.196 x  10^-6 /  1.6 x 10^-6

➝ L = 1.96 / 1.6

➝ L = 1.225 m

From the formula we can conclude that ,

➝ R ∝  1 / d²

Hence ,

➝ R / R ' = d '²/ d  ²

A per the given condition ,

➝ d' = 2d

➝ R /R ' = (2d)² / d ²

➝ R /R ' = 4d² / d ²

➝ R / R' = 4 / 1

➝ R' = R / 4

➝ R' = 10 / 4

➝ R' = 2.5  Ω

The new resistance will be one -fourth of the original resistance

Answer 2

Given, diameter, d=0.5 mm

resistivity, ρ=1.6×10

−8

Ωm

Resistance, R=10 Ω

Let the length of wire be l.

A=πd

2

/4

R=ρl/A

=

πd

2

/4

ρl

⟹l=

4ρ

Rπd

2

l=

4×1.6×10

−8

10×3.14×(0.5×10

−3

)

2

l=122.7 m

R∝1/d

2

If the diameter is doubled, resistance will be one-fourth.

Hence, new resistance =2.5Ω

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