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Question

A wire of resistance 2 Ω is stretched to double its length such that its volume remains constant. What is the resistance of the stretched wire?
  1. 8

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Solution

The correct option is A 8

Volume of the wire, V= Length (l) × Area of cross-section (A)
i.e. V=lA

Let the new volume be V and the new area be A

So, V=2l×A
Since the volume remains constant, l×A=2l×A
A=A2

We know, R=ρlA
So, new resistance R=ρ×2lA2
R=4ρlA=4R=4×2=8 Ω
Hence, the resistance of the stretched wire is 8 Ω.


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