Question

A wire with 15 Q resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will be

Answer 1

Answer:

If the wire is stretched by (1/10)th(1/10)th of its original length then the new length of wire become

l2=l+l10=11l10....(i)l2=l+l10=11l10....(i)

As the volume of wire remains constant then

πr21l=πr22l1=πr22l(11l10)(using(i))πr12l=πr22l1=πr22l(11l10)(using(i))

⇒r22=1011r21....(ii)⇒r22=1011r12....(ii)

Now the resistance of stretched wire.

R2=p(1110l)πr22=(1110ρl)π×1011r21×ρlπr21R2=p(1110l)πr22=(1110ρl)π×1011r12×ρlπr12 (