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A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
d= diameter = 0.5 mm, ρ = 1.6 × 10–8 Ωm, R = 10Ω
Assume wire length as l
r = d/2
r = (0.5 x 10-3)/2 // 1 m = 1000 mm
A = πd2/4
R = ρ (l/A)
= ( ρl)/(πd2/4)
⟹ l = (Rπd2)/4ρ
l = 10 x 3.14 x ((0.5 x 10-3)/2 )2 / (4 x 1.6 × 10–8 ) // π = 3.14
l = 10 x 3.14 x ((0.5 x 10-3)/2 )2 / (4 x 1.6 × 10–8 )
l = (10 x 314 x 10-2 x 25 x 10-6)/ (4 x 16 x 10-1 x 10-8)
l = 3140 x 25 x 10-8/ 64 x 10-9
l = 122.7 m
∴ Length of wire = 122.7 m
If diameter is doubled then d = 0.25mm
d = 25 x 10-2 x 10-3 m
d = 25 x 10-5m
and R = ( ρl)/(πd2/4)
R = (1.6 × 10–8 X 122.7) / ((3.14 X 25 x 10-5) / 4)
R = 490.5m
Thus, we can say,
Since R ∝ 1/d2
So, if d is doubled, then R becomes 1/4th time.
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