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A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire (A) nR (B) n2 R (C) 2nR (D) 2n2 R

A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire
(A) nR (B) n2
R (C) 2nR (D) 2n2
R

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2 Answers

Arun
25750 Points
3 years ago
Resistance of wire initially          R=Aρl​            
Using  A=V/l        ⟹R=Vρl2​           
New length wire after stretching    l′=nl
New resistance      R′=Vρ(nl)2​=
Vikas TU
14149 Points
3 years ago
Wire of resistance R is elongated n fold to make a new uniform wire.So its radius is reduced by n times.
Let l  and r be the original length and radius of the original wire.So,
Original Resistance of the wire , R = ρl/πr^2
Now radius is reduced by n times ,so new radius is r' = r/n
New resistance,R' = ρl/πr'^2 = ρl/π(r/n)^2
= n^2ρl/πr^2 = n^2R

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