Question

A piece of wire of resistance 20 ohm is drawn out so that its length is increased to twice its original length ? Calculate the resistance of the wire in the new situation?​

Answer 1

Answer:

80 ohm

Step-by-step explanation:

Given,

R=20 Ω  

Let,Length of wire is L and when it drawn out, its new length is 2L

The volume of wire in both the situations remains constant.So, it the length of wire doubled, the area of cross section of wire becomes half.

So, the new resistance of wire-

Rnew = 2l/(A/2)

=4L/A

=4×Old resistance of wire           (Because, R=L/A)

=4×20 = 80 Ω

Answer 2

Answer:

Suppose the length of R1 = 20Ω resistance wire is I, its area of cross-section is,4 and its resistivity is ρ Then, - in ist attachment

When length becomes twice (2l), then its area of cross-section will become half (A/2). The new resistance -  in second attachment

Answer is 80 ohms.

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