wire is folded and doubled upon itself from centre.what is percentage change in its resistance
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Let initial length and area of wire be \’l\’ and \’A\’.Then,initial resistance,R=p (l÷A).——–(1)Now,when the wire is folded and doubled,its length becomes half and area gets doubled.Thus,l\’=l÷2 and A\’=2A.Then,New resistance=p l\’÷A\’ =p l/2÷2A =p l÷4A =p l÷A ×1/4 =R/4 (from 1) Thus,change in resistance=R-R/4 =3R/4Percentage change=change in resistance/initial resistance×100 =3R/4÷R×100 =3/4×100 =75%(decrease)
Remains same