Question

# We have a copper wire resistance R this wire is pulled so it's length get double find the new resistance in term of original resistance.

Solution

## R1=pL1/a1 volume of wire renains same so A1*L1=A2*2L1 A1=2A2 A1/2=a2 Rnew=p2L1/A1/2    =p4L1/A2     =4pL1/A2     =4*R1 so new resistance is 4 times that of old resistance Another method  To do this, we the following formula R=ρlA .....(1) where R is resistance wire of length l, A its cross-sectional area and ρspecific resistance of material of wire, copper in the given question. When length is doubled, volume remains same. If Anew is new area of cross section then l×A=(2l)×Anew ⇒Anew=lA2l=A2 Remembering that ρ remains same, inserting values in (1) we get Rnew=ρ2lA2 ⇒Rnew=4ρlA Using (1) we get Rnew=4R

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