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Question
A given wire of resistance 1 Ω is stretched to double its length. What will be its new resistance?
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Solution
Let ‘I’ be the length and ‘a’ be the area of cross – section
of the resistor with resistance, R = 1Ω
`"R" = ∝ l/r^2`
when the wire is stretched to double its length,
the new length I’ = 2I and the new area of cross section,
`a^1 = a/2 "and" l^1 = 2l`
`"R"^1 ∝ l^1/a^1`
∴ `"R"^1 ∝ (2l)/(a/2)`
`"R"^1 ∝ (4l)/a`
R1 ∝ 4R
Resistance increases to 4 times original resistance = R1 = 4Ω
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