Question

A wire of resistance 1000 O is uniformly
stretched so that its length increases by 5%
Assuming that the volume of the wire remains
the same on stretching, find the resistance of
the wire after stretching.

Answer is 1102.5

hey pls help I know answer but I want steps to this answer
plse help me​​

Answer 1

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Let l be the original length of A.

Original area of cross-section, then

Original resistance R= Aρl

Now, length of the wire =l ′ =nlIf A ′

be the new cross-sectional area, then l ′ A ′ =lA

(∵ volume of metal is a constant).

A = l ′lA = nllA = nA

New resistance of the wire is

R ′ = Alρl ′ = nAρ(nl) = lρnl × An =n 2 ( Aρl )=n 2 R