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Question
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8Ω 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?
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Solution
Given diameter of the wire,
D = 0.5 × 10−3 m
resistivity of copper, ρ = 1.6 x 10−8 ohm m required resistance, R = 10 ohm
As R = `(ρl)/A, l = (RA)/ρ = (R((piD^2)/4))/ρ = (πRD^2)/(4ρ)` ...`[∵ A = pir^2 = pi(D/2)^2 = (piD^2)/4]`
∴ `l = (3.14 xx 10 xx (0.5 xx 10^-3)^2)/(4 xx 1.6 xx 10^-8)`
m = 122.7 m
Since `R = (ρl)/((πD^2)/4)`
= `(4ρl)/(πD^2), R∝1/D^2`.
When D is doubled, R becomes `1/4` times.
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Table give below Electrical resistivity of some substances at 20°C
| Electrical resistivity of some substances at 20°C | ||
| − | Material | Resistivity (Ω m) |
| Conductors |
Silver | 1.60 × 10−8 |
| Copper | 1.62 × 10−8 | |
| Aluminium | 2.63 × 10−8 | |
| Tungsten | 5.20 × 10−8 | |
| Nickel | 6.84 × 10−8 | |
| Iron | 10.0 × 10−8 | |
| Chromium | 12.9 × 10−8 | |
| Mercury | 94.0 × 10−8 | |
| Manganese | 1.84 × 10−6 | |
| Alloys |
Constantan (alloy of Cu and Ni) |
49 × 10−6 |
| Manganin (alloy of Cu, Mn and Ni) |
44 × 10−6 | |
| Nichrome (alloy of Ni, Cr, Mn and Fe) |
100 × 10−6 | |
| Insulators | Glass | 1010 − 1014 |
| Hard rubber | 1013 − 1016 | |
| Ebonite | 1015 − 1017 | |
| Diamond | 1012 − 1013 | |
| Paper (dry) | 1012 | |
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