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Question
How will the resistance of a wire be affected if its
- length is doubled, and
- radius is also doubled ?
Give justification for your answer.
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Solution
R `=(rho"l")/"A"`
Where, `rho` = electrical resistivity
l = length of the conductor
A = cross-sectional area of the conductor
Hence if the length is double then
⇒ `"R"_1=rho((2"l"))/"A"`
`therefore "R"_1 = 2("R")`
So, if the length of the resistance gets doubled then resistance also gets doubled.
Now when the radius is double then
⇒ `"R"_2 = (rho"l")/"A"`
⇒`"R"_2 = (rhol)/(pi(2"r")^2`
`therefore "R"_2 = 1/4("R")`
So if the radius gets doubled then resistance will be `(1/4)^"th"` of initial resistance.
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RELATED QUESTIONS
Use the data in the Table given below to answer the following –
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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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